Definition of viscosity

In the article Viscosity, the cause of viscosity was mainly attributed to attractive forces between the layers of a fluid. These forces act similar to frictional forces, so that the individual fluid layers try to slow each other down. For the definition of viscosity one can imagine a fluid between two plates. The lower plate is at rest and the upper plate is moved at constant speed.

Due to the no-slip condition, both the upper and the lower fluid layer adhere to these plates. The lower fluid layer thus remains at rest and the upper fluid layer moves at the same speed as the upper plate. A linear velocity profile is formed between the plates. A considered fluid layer thus always moves slower than the fluid layer above it. Due to the molecular forces between the molecules, a lower fluid layer always tries to slow down the fluid layer above.

These frictional forces between the layers must be compensated if the uppermost layer is to be moved at a constant speed. The more viscous a fluid is, the stronger the internal friction forces and the greater the force required to move the top plate. The viscosity η shows the relationship between the area-related force F/A (called shear stress τ), which is required to move the fluid layers, and the slope of the velocity profile dv/dy (called velocity gradient):

\begin{align}
\label{t}
&\frac{F}{A}=\boxed{\tau= \eta \cdot \frac{\text{d} v}{\text{d} y}} ~~~~~\text{Newton’s law of fluid friction}\\[5px]
\end{align}

Viscosity of gases (momentum transfer)

The cause of viscosity due to frictional forces acting between the fluid layers can be clearly understood for liquids. In gases, however, the molecules exert almost no molecular forces on each other. Frictional forces between the fluid layers due to intermolecular cohesion are therefore almost non-existent. However, practice shows that even gases have a considerable viscosity and that fluid layers are slowed down in flows. How can this behavior be explained without the presence of inter-molecular forces?

The slowing down effect of fluid layers in gases is mainly due to the momentum transfer of the gas molecules when they diffuse from a slower layer into a faster layer. This process also takes place in liquids, but it is negligible compared to the decelerating effect due to the intermolecular forces.

Let us again consider the already mentioned laminar flow, which this time consists of an ideal gas as fluid. If one gas molecule collides with another, an momentum exchange takes place, i.e. a slower molecule takes up part of the momentum of the faster molecule. However, such a momentum transfer does not only take place within a fluid layer. Due to the random molecular motion (Brownian motion), molecules also diffuse into adjacent fluid layers. What happens if a slower molecule diffuses into a layer of faster particles? The fast molecules are slowed down by this diffused gas molecule and the layer slows down.

One can illustrate the situation with a cart and a ball. The cart is to be moved at a constant speed when suddenly the heavy ball is put in while passing by. The ball represents the slower gas molecule (in this case even stands still), which diffuses into the faster fluid layer (illustrated by the cart). Since the ball has a lower speed than the cart when put in, the ball must be accelerated to the speed of the cart if the speed is to remain constant. This requires a force corresponding to the mass of the ball (force = mass x acceleration). This means that if this force would not be applied, the cart would be slowed down, similar to a frictional force.

The diffusion of gas molecules between the layers of a laminar flow leads to a momentum transfer on which the viscosity of gases is mainly based!

Faster layers thus transfer part of their momentum by diffusion into slower layers. All in all there is a transport of momentum from the moving plate at the top of the fluid to the resting plate at the bottom of the fluid. This momentum transport (which ultimately corresponds to a force between the layers) is directed in the direction of decreasing velocity, so to speak, i.e. against the velocity gradient. The transport of momentum becomes particularly clear when the upper plate is set in motion from rest. First, only the layer directly adhering to the upper plate is set in motion. By the transport of momentum onto the layer underneath, this layer starts to move, etc. The momentum thus gradually spreads through the layers, so to speak.

The resistance that one experiences when moving the upper plate is caused by the fact that the movement of the gas layers is hindered because the lower plate is fixed. Holding the bottom plate in place ultimately requires the same force as maintaining the movement of the upper plate. The momentum is, so to speak, introduced at the upper plate and is transferred from layer to layer and exits at the lower plate. Finally, in a state of equilibrium, there is no net momentum flow, which is transferred to the fluid layers, so that these finally move at constant but different speeds according to the linear velocity profile.

Derivation of the viscosity of ideal gases

Shear stress as momentum flux

A force F can generally be determined from the change in momentum per unit time:

\begin{align}
&F= \frac{\text{d} p}{\text{d} t} = \dot p ~~~~~\Rightarrow~~~~~\boxed{\text{force = momentum flow rate}}\\[5px]
\end{align}

The force acting on the individual fluid layers is thus caused by the change in momentum and can therefore also be understood as momentum flow rate p*. If one relates the force and thus the momentum flow rate to the surface area, this corresponds to a shear stress, which in turn can be interpreted as momentum flux p*_A (change in momentum per unit time and unit area):

\begin{align}
&\tau = \frac{F}{A}= \frac{\dot p}{A} = \dot p_\text{A} ~~~~~\Rightarrow~~~~~\boxed{\text{shear stress = momentum flux}}\\[5px]
\end{align}

Newton’s law of fluid friction (\ref{t}) can thus also be represented as follows:

\begin{align}
\label{tt}
&\boxed{\dot p_\text{A} = – \eta \cdot \frac{\text{d} v}{\text{d} y}} \\[5px]
\end{align}

The negative sign was introduced to account for the fact that the momentum flux is transferred away from faster layers to slower layers, i.e. in the direction of decreasing velocity gradient. At this point an interesting analogy can be drawn to other transport mechanisms like heat transport and mass transport, which are ultimately described in a similar way:

Momentum transfer between the layers

With the help of the kinetic theory of gases, the viscosity of ideal gases can be calculated. To derive a formula, we consider a laminar flow, where an ideal gas is placed between two plates. The lower plate is fixed and the upper plate moves with a constant speed.

We observe the gas flow on a microscopic level and move along with a fluid layer. The mean distance a gas molecule travels between two collisions is called mean free path λ. We therefore consider gas layers that have a distance λ from each other, so that diffusion processes result in a collision inside those layers and thus in a momentum transfer. We now look at a layer at any height y. The mean velocity of the gas molecules in x-direction with respect to a coordinate system fixed at the resting plate is denoted by v_x(y).

The mean velocityv_x(y+λ) of the gas molecules in the layer above at the distance λ can be determined using the velocity gradient dv/dy:

\begin{align}
&v_{x}(y+\lambda)= v_{x}(y) + \lambda \cdot \frac{\text{d}v}{\text{d}y} \\[5px]
\end{align}

In the same way, the mean velocity v_x(y-λ) of the gas particles in the layer below at the distance λ can be determined:

\begin{align}
&v_{x}(y-\lambda)= v_{x}(y) – \lambda \cdot \frac{\text{d}v}{\text{d}y} \\[5px]
\end{align}

If n*_A denotes the particle flux, i.e. the number of particles per unit time and unit area that diffuse from the upper layer or lower layer into the middle layer, then the respective momentum fluxes can be determined with the following formulas. Note that the particle flux is identical for both layers if we assume an incompressible gas flow where the particle density is the same at every point in the flow.

\begin{align}
&\dot p_{A}(y+\lambda) = \dot n_\text{A} \cdot \overbrace{m \cdot v_{x}(y+\lambda)}^{\text{momentum of one molecule}} = \dot n_\text{A} \cdot m \cdot \left(v_{x}(y) + \lambda \cdot \frac{\text{d}v}{\text{d}y} \right) \\[5px]
&\dot p_{A}(y-\lambda) = \dot n_\text{A} \cdot m \cdot v_{x}(y-\lambda) = \dot n_\text{A} \cdot m \cdot \left(v_{x}(y) – \lambda \cdot \frac{\text{d}v}{\text{d}y} \right) \\[5px]
\end{align}

The net momentum flux p*_A(y) in the layer at the height y is finally the sum of both momentum fluxes. According to the chosen coordinate system the bottom-up momentum flux (pointing in positive y direction) corresponds to a positive value and the downward facing momentum flux to a negative value.

\begin{align}
\dot p_\text{A}(y) &= \dot p_{A}(y-\lambda) ~-~ \dot p_{A}(y+\lambda) \\[5px]
&= \dot n_\text{A} \cdot m \cdot \left(v_{x}(y) – \lambda \cdot \frac{\text{d}v}{\text{d}y} \right)- \dot n_\text{A} \cdot m \cdot \left(v_{x}(y) + \lambda \cdot \frac{\text{d}v}{\text{d}y} \right) \\[5px]
&= \dot n_\text{A} \cdot m \cdot \left(v_{x}(y) ~- \lambda \cdot \frac{\text{d}v}{\text{d}y} ~-~ v_{x}(y) ~- \lambda \cdot \frac{\text{d}v}{\text{d}y}\right) \\[5px]
\end{align}

\begin{align}
&\boxed{\dot p_\text{A}= – 2~ \dot n_\text{A} \cdot m \cdot \lambda \cdot \frac{\text{d}v}{\text{d}y}}~~~\text{net momentum flux} \\[5px]
\end{align}

Viscosity of ideal gases as a function of particle flux

According to the above equation, the net momentum flux is obviously no longer a function of the variable y and thus identical in every point of the flow! If one compares this formula with Newton’s law of fluid friction (\ref{tt}), it is apparent that the expression 2⋅n*_A⋅m⋅λ obviously corresponds to the viscosity η:

\begin{align}
&\dot p_\text{A} = – \eta \cdot \frac{\text{d} v}{\text{d} y} \\[5px]
&\dot p_\text{A}=- \underbrace{2~ \dot n_\text{A} \cdot m \cdot \lambda}_{\eta} \cdot \frac{\text{d}v}{\text{d}y} \\[5px]
\label{eta}
&\boxed{\eta= 2~ \dot n_\text{A} \cdot m \cdot \lambda} ~~~\text{viscosity of ideal gases}\\[5px]
\end{align}

The viscosity of an (ideal) gas is therefore only dependent on the mass of a gas particle, the mean free path and the particle flux. The area-related particle flow n*_A, which diffuses in from a layer above or below, depends in turn on how strongly the gas molecules move due to the random diffusion motion (Brownian motion). This in turn is determined by the temperature.

If the temperature is high, diffusion take place more strongly and more particles diffuse between the layers. Thus, the particle flux is high and so is the momentum transfer. This results in an increasing force, which is necessary to maintain the macroscopic flow (movement of the plate)! The viscosity of gases therefore generally increases with temperature and not decreases as with liquids!

At this point it is also evident that with ideal gases, pressure has no influence on viscosity. Although the particle density and thus the diffusing particle flow increases proportionally with increasing pressure, the mean free path decreases to the same extent. Both effects cancel each other out.

With ideal gases the viscosity is independent of pressure and increases with increasing temperature!

Viscosity of ideal gases as a function of temperature

At this point we would like to explicitly derive the dependence of the viscosity of ideal gases on temperature. For this purpose, a correlation between the particle flux diffusing perpendicular to the flow and the temperature must be found.

For this we move in thoughts with a layer, so that this layer rests relative to us. The gas molecules themselves, however, are by no means at rest on the microscopic level. Due to Brownian motion, they move in all directions in a completely random manner. According to the Maxwell-Boltzmann distribution, the mean speed of a gas particle v_T is linked to the temperature T of the gas as follows

\begin{align}
\label{a}
&\boxed{ \overline{v_\text{T}} = \sqrt{\frac{8 k_B T}{\pi m}}} ~~~\text{arithmetic mean speed} \\[5px]
\end{align}

In this equation m denotes the mass of a gas molecule and k_B is the Boltzmann constant. Note that the speed v_T represents the mean speed relative to the moving fluid layers and does not include the superposition of the macroscopic flow motion. The latter has no influence on the temperature anyway; after all, the temperature of a gas does not depend on whether the gas is at rest or moving.

Let us now consider a directed flow in which all particles move in the same direction with the (mean) velocity v. The number of particles flowing per unit time and area (particle flux) can be determined as follows. We consider an area element dA through which the particles flow with the (mean) velocity v within a time dt. The particles cover the distance dl=v⋅dt. Thus the particles obviously flow through the volume element dV:

\begin{align}
&\text{d}V =\text{d}A \cdot \text{d}l = \text{d}A \cdot \overline{v} \cdot \text{d}t \\[5px]
\end{align}

For a given particle density n (number of particles per unit volume), the following number of particles dN will be found this volume element:

\begin{align}
&\text{d}N = n \cdot \text{d}V =n \cdot \text{d}A \cdot \overline{v} \cdot \text{d}t \\[5px]
\end{align}

The number of particles per unit time and area (particle flux n*_A) passing through an area perpendicular to the flow can thus be calculated with the following formula:

\begin{align}
&\dot n_\text{A} = \frac{\text{d}N}{\text{d}A \cdot \text{d}t} =n \cdot \overline{v} \\[5px]
&\boxed{\dot n_\text{A} =n \cdot \overline{v} } ~~~\text{particle flux of a directed motion}\\[5px]
\end{align}

Let us now look again at our laminar flow and we move along in our thoughts with a fluid layer. From this point of view, the flow is no longer directed, but completely random with a mean molecular velocity denoted by v_T. The molecules do not move in any preferred direction. This means that only a sixth of the particles move downwards and diffuse into a gas layer below. The particle flux directed perpendicular to the main flow (bulk motion) is thus only one-sixth as large:

\begin{align}
\label{na}
&\boxed{\dot n_\text{A} = \frac{1}{6} n \cdot \overline{v_\text{T}} } ~~~\text{particle flux of a random motion}\\[5px]
\end{align}

Using (\ref{a}) in equation (\ref{na}) results in the following diffusing particle flux as a function of temperature:

\begin{align}
&\dot n_\text{A} = \frac{1}{6} n \cdot \underbrace{\sqrt{\frac{8 k_B T}{\pi m}}}_{\overline{v_\text{T}}} \\[5px]
\end{align}

This equation used in formula (\ref{eta}) finally shows the following relationship between viscosity and temperature:

\begin{align}
&\eta= 2~ \dot n_\text{A} \cdot m \cdot \lambda \\[5px]
&\eta= 2~ \frac{1}{6} n \cdot \sqrt{\frac{8 k_B T}{\pi m}} \cdot m \cdot \lambda \\[5px]
\label{ac}
&\boxed{\eta= \frac{1}{3} n \cdot \sqrt{\frac{8 k_B m T}{\pi}} \cdot \lambda} \\[5px]
\end{align}

Last but not least, the mean free path λ can be expressed by the particle density n and the diameter of the gas molecules d (for the derivation of this formula see article Mean free path & collision frequency):

\begin{align}
& \boxed{\lambda = \frac{1}{\sqrt{2}~n ~\pi d^2}} \\[5px]
\end{align}

Using this formula in equation (\ref{ac}), the following formula for calculating the viscosity of ideal gases is finally obtained:

\begin{align}
&\eta= \frac{1}{3} n \cdot \sqrt{\frac{8 k_B m T}{\pi}} \cdot \lambda \\[5px]
&\eta= \frac{1}{3} n \cdot \sqrt{\frac{8 k_B m T}{\pi}} \cdot \frac{1}{\sqrt{2}~n ~\pi d^2} \\[5px]
&\boxed{\eta= \sqrt{\frac{4 k_B m~T}{9\pi^3~d^4}}} \\[5px]
&\boxed{\eta \sim \sqrt{T}} \\[5px]
\end{align}

This formula shows that the particle density and thus the pressure has no influence on the viscosity of ideal gases. Only the temperature as a variable quantity influences the viscosity. The viscosity increases proportionally with the square root of the temperature!

Note that this formula only applies to laminar flows where the gas layers do not mix macroscopically and diffusion between the layers only occur at the microscopic level. In turbulent flows, the momentum exchange through the turbulence is greater and the viscosity is higher.

Comparing viscosity and thermal conductivity

Equation (\ref{na}) can also be put directly into the formula (\ref{eta}) for the viscosity, resulting in the following relationship:

\begin{align}
&\eta= 2~ \dot n_\text{A} \cdot m \cdot \lambda \\[5px]
&\eta= 2~ \frac{1}{6} n \cdot \overline{v_\text{T}} \cdot m \cdot \lambda \\[5px]
&\eta= \frac{1}{3} \underbrace{n \cdot m}_{\rho} \cdot \lambda \cdot \overline{v_\text{T}} \\[5px]
&\boxed{\eta= \frac{1}{3} \cdot \rho \cdot \lambda \cdot \overline{v_\text{T}}} \\[5px]
\end{align}

This derivation exploited the fact that the product of particle density and mass of a single particle equals the density ϱ of the gas. At this point an interesting analogy to thermal conductivity k of ideal gases can be seen (to avoid confusion with the mean free path, the thermal conductivity was not denoted by λ but k):

\begin{align}
& \boxed{k= \frac{1}{3} \cdot \rho \cdot \lambda \cdot c_v \cdot \overline{v_\text{T}} } \\[5px]
\end{align}

The thermal conductivity thus obeys in principle the same laws as the viscosity, i.e. it increases with increasing mean particle speed (increasing temperature). This is not surprising, since the diffusion of particles is not only associated with an momentum transfer, but also with a energy transfer in terms of heat.

Introduction

In the article Thermal conduction in solids and ideal gases, the following formula for estimating the thermal conductivity λ of ideal gases was derived:

\begin{align}
\label{l}
& \boxed{\lambda = \frac{1}{3} \cdot c_v \cdot \rho \cdot v \cdot l}  \\[5px]
\end{align}

In this formula c_v denotes the specific heat capacity at constant volume, ϱ the density of the gas, v the mean speed of the gas molecules and l the mean free path. This formula will be explained in more detail in this article and the resulting conclusions for gases will be discussed.

Using the formula, one could assume that the thermal conductivity depends on the pressure, because the higher the pressure, the higher the density of the gas. This argument can also be clearly understood with the help of the particle model of matter, because the more particles there are, the more energy the particles can transport in total. Note that according to the kinetic theory of gases each gas molecule carries the energy ½⋅k_B⋅T per degree of freedom (with k_B as the Boltzmann constant).

However, to the same extent as the density increases with increasing pressure, the mean free path decreases! In fact, the thermal conductivity of ideal gases is therefore independent of pressure or particle density (for the limitation of this statement, later more)!

The thermal conductivity of gases is not dependent on pressure for not too low pressures!

Independence of thermal conductivity from pressure for high pressures (dense gases)

The independence of thermal conductivity from pressure can also be shown mathematically. For this purpose the density ϱ in equation (\ref{l}) is first expressed by the quotient of gas mass m_gas and gas volume V_gas. Then the gas mass can be expressed by the amount of substance n_gas (chemical amount) and the molar mass M_gas of the gas.

\begin{align}
\lambda &= \frac{1}{3} \cdot c_v \cdot \frac{m_{gas}}{V_{gas}} \cdot v \cdot l  \\[5px]
&= \frac{1}{3} \cdot c_v \cdot \frac{n_{gas} \cdot M_{gas}}{V} \cdot v \cdot l  \\[5px]
\end{align}

The product of mass-specific heat capacity and molar mass equals the so-called molar heat capacity C_m,v, whereby the molar heat capacity is only dependent on the degrees of freedom f and the molar gas constant R_m (C_m,v=f/2⋅R_m). In addition, the amount of substance n_gas can be expressed by the ratio of the number of particles N and Avogadro constant N_A (n_gas=N/N_A):

\begin{align}
\lambda &= \frac{1}{3} \cdot \underbrace{c_v \cdot M_{gas}}_{C_{m,v}} \cdot \frac{n_{gas}}{V} \cdot v \cdot l  \\[5px]
& = \frac{1}{3} \cdot \underbrace {C_{m,v}}_{=\frac{f}{2}R_m} \cdot \frac{N}{N_A \cdot V} \cdot v \cdot l  \\[5px]
& = \frac{1}{3} \cdot \frac{f}{2} R_{m} \cdot \frac{1}{N_A} \cdot \frac{N}{V} \cdot v \cdot l  \\[5px]
& = \frac{f}{6} \frac{R_m}{N_A} \cdot \frac{N}{V} \cdot v \cdot l  \\[5px]
\end{align}

The ratio of particle number and gas volume corresponds to the particle density n and the quotient of molar gas constant and Avogadro constant corresponds to the Boltzmann constant k_B (for this relationship see article Internal Energy & Heat Capacity):

\begin{align}
\lambda & = \frac{f}{6} \underbrace{\frac{R_m}{N_A}}_{k_B} \cdot \underbrace{\frac{N}{V}}_{n} \cdot v \cdot l  \\[5px]
\label{ll}
& = \frac{f}{6} k_B \cdot n \cdot v \cdot l  \\[5px]
\end{align}

Now we only need the dependencies of the mean speed and the mean free path. According to the Maxwell-Boltzmann distribution, the mean speed v of the gas molecules depends on the temperature T of the gas and the mass m of a molecule (m denotes the mass of a single gas particle and not the entire gas mass!)

\begin{align}
& \boxed{v = \sqrt{\frac{8 k_B T}{\pi m}}}  \\[5px]
\end{align}

The mean free path l of the molecules depends not only on the particle density n but also on the particle diameter d and is determined by the following formula:

\begin{align}
& \boxed{l= \frac{1}{\sqrt{2} n \pi d^2}}  \\[5px]
\end{align}

If both formulas are used in equation (\ref{ll}), the thermal conductivity can be determined with the following formula:

\begin{align}
& \lambda = \frac{f}{6} k_B \cdot n \cdot v \cdot l   \\[5px]
& \lambda = \frac{f}{6} k_B \cdot n \cdot \sqrt{\frac{8 k_B T}{\pi m}} \cdot \frac{1}{\sqrt{2} n \pi d^2}   \\[5px]
\require{cancel}
& \lambda = \frac{f}{6} k_B \cdot \cancel{n} \cdot \sqrt{\frac{8 k_B T}{\pi m}} \cdot \frac{1}{\sqrt{2} \cancel{n} \pi d^2}   \\[5px]
& \lambda = \frac{f}{6} k_B \cdot \sqrt{\frac{8 k_B T}{\pi m}} \cdot \frac{1}{\sqrt{2} \pi d^2}   \\[5px]
& \boxed{\lambda = \frac{f}{3d^2} \sqrt{\frac{k_B^3 T}{\pi^3 m }}}   \\[5px]
\end{align}

This formula now clearly shows that the thermal conductivity of ideal gases is not dependent on particle density and thus not on pressure. It also shows that gases with relatively large molecules have a lower thermal conductivity than gases with smaller molecules (this is due to the decrease in the mean free path resulting from the larger collision diameter d). Furthermore, the thermal conductivity of gases with light particles is higher than that of gases with heavier particles. Furthermore, the thermal conductivity depends on the temperature. The thermal conductivity increases with increasing temperature!

The thermal conductivity of gases is the greater, the smaller and lighter the molecules are and the higher the temperature is!

Dependence of thermal conductivity from pressure for low pressures (diluted gases)

If pressure has no influence on the thermal conductivity of gases, why use a vacuum for thermal insulation?

The fact that thermal conductivity does not depend on pressure is only true as long as the mean free path is much smaller than the dimensions of the volume in which the gas is contained. If the pressure (particle density) in a container is reduced more and more, the particles no longer collide with each other, but rather with the container walls. At very low pressures, the mean free path is thus determined by the dimension of the container and no longer by the free path between two particle collisions.

This also applies if the pressure is not reduced, but the container dimension is. This is relevant, for example, for insulating materials where gases are enclosed in small pores. Such situations can also occur with thin layers of foil or small gaps if there is a gas in between.

The mean free path in these cases corresponds approximately to the dimension δ of the volume (e.g. pore diameter or gap distance) and is therefore a constant value. In this case the mean free path is no longer dependent on the particle density: l≈δ=constant. With the mean free path as a constant, equation (\ref{ll}) then indicates a reduction of the thermal conductivity with decreasing particle density (or pressure)!

\begin{align}
\label{a}
&\lambda = \frac{f}{6} k_B \cdot n \cdot v \cdot \delta  \\[5px]
\end{align}

In diluted gases or with small gas volumes, the thermal conductivity depends on the pressure!

In so-called Pirani gauges, this relationship is used to draw conclusions about the pressures in a high vacuum environment on the basis of thermal conductivity.

Knudsen number

As already indicated, the characteristic length δ of the pores or the foil spacing in insulation materials is often much smaller than the mean free path l of the gases contained therein. In this case the gas can no longer be described as a continuum, so that equation (\ref{a}) can no longer be applied in this form (the qualitative statement of this equation does not lose its validity, however).

The characteristic length refers to the dimension/size of a system!

In this context, the so-called Knudsen number indicates whether the gas can still be regarded as a continuum or whether the kinetics of the gas theory must be applied. The unitless Knudsen number Kn describes the ratio of the mean free path l to the characteristic length δ of the gas volume:

\begin{align}
&\boxed{Kn := \frac{l}{\delta}}  \\[5px]
\end{align}

For values much smaller than 1, continuum mechanics still applies and for values much larger than 1 the description by means of the laws of kinetic gas theory are used.

Since for ideal gases the mean free path can be expressed not only by the particle density but also by the pressure p using the ideal gas law (see article here), the Knudsen number can be determined as follows from the pressure and temperature (d describes the diameter of the gas molecules):

\begin{align}
&\boxed{Kn = \frac{k_B \cdot T}{\sqrt{2}\pi \cdot d^2 \cdot p \cdot \delta}}  \\[5px]
\end{align}

In the case of insulation materials, where the Knudsen number is often much smaller than 1, the thermal conductivity of the enclosed gas can be determined with the following formula [see M.G. Kaganer: “Thermal insulation in cryogenic engineering“, 1969]:

\begin{align}
\label{lam}
&\boxed{\lambda = \frac{\lambda_0}{1+2\beta \cdot Kn}}  \\[5px]
\end{align}

In this formula, λ₀ denotes the thermal conductivity under standard conditions (1 atm, 0°C) and β is a weighting factor which will not be discussed further here. Even if the use of equation (\ref{lam}) requires that the Knudsen number is much smaller than 1, it should still be as high as possible, especially for insulating materials! This then results in low thermal conductivity.

For low thermal conductivity, the Knudsen number should be kept as high as possible!

Internal energy

In the article equipartition theorem it has already been explained in detail that the energy of a gas is equally divided among the different microscopic forms of energy. In the case of monatomic ideal gases, this only includes the kinetic energy of the gas particles in terms of the translational motion (motion of the centre of gravity). In contrast to monatomic ideal gases, the molecules of polyatomic gases can also store energy in terms of a rotational motion. In addition, the atoms of gas molecules are connected to each other by “elastic” binding forces. Within certain limits, these binding forces can be considered analogous to an elastic spring. At the atomic level, vibrational energies are also found.

All the forms of energy mentioned above are present at the atomic level, i.e. “inside” the gas. The sum of these energies of each individual molecule thus contributes to the energy content of the entire gas. The energy content of the gas is also called internal energy. The particles represent so to speak “energy storages” for the internal energy.

Internal energy is the “atomic” energy content of a substance, which is contained in the kinetic energy of the particles, among other things!

In fact, other forms of energy are counted as internal energy as well, such as ionization energies or chemically bound energies, which can be released by chemical reactions. In comparison to these forms of energy it is characteristic for the translational, rotational and vibrational energy that these are related to random motions of the molecules. Therefore, these energies, which are related to the random motions of the particles, are referred to as thermal energy in the narrower sense. In this respect, the thermal energy of the random molecular motion is only a part of the internal energy.

Thermal energy refers to the energy stored in the random motion of the molecules (translation, rotation, vibration)!

In the context of the kinetic theory of gases, however, it is sufficient to reduce the internal energy to the thermal energy, since neither chemical reactions nor ionizations or the like are considered. When we talk about the internal energy in the following, we mean exclusively the thermal energy.

Degrees of freedom

The different types of thermal energy can even be further divided. For example, a molecule can move in three independent spatial directions. The translational energy can thus be divided among each direction, i.e. a kinetic energy related to the x-direction, the y-direction and the z-direction. For the rotation, a division of the rotational energy can take place analogously, which is then linked with a rotation around the x-axis, y-axis and z-axis. The same applies to the vibrational energy, which can occur in different ways with molecules. All these possibilities of energy storage are also called degrees of freedom.

In the thermodynamic sense, the term “degrees of freedom” refers to the possibilities of storing energy as translational, rotational and vibrational motion of a molecule!

Translational motion has a total of three degrees of freedom, namely for each spatial direction one (because translational kinetic energy can be stored in terms of the motion in each direction). The same applies to rotation, in which rotational energy can be stored around each of the three spatial axes. Thus, rotational motion also has three degrees of freedom.

In the case of diatomic molecules (or linear molecules in general, i.e. molecules whose atoms are all on a common axis) there is a peculiarity in the degrees of freedom of rotation. One would probably assume three degrees of freedom, since the molecule should be able to rotate both around the x-axis and the y-axis as well as around the z-axis (molecular axis).

In fact, however, linear molecules will not rotate around their molecular axis and thus will not be able to use this degree of freedom for energy storage. This has to do with quantum effects. One can also clarify this fact by idealizing the gas as an ideal gas. If the atoms of a linear molecule are regarded as mass points, then, effectively speaking, there will be no mass in rotation when rotating around the molecular axis and thus no rotational energy (no moment of inertia, since the entire mass is concentrated in the center of rotation).

The missing degree of freedom of rotation around the molecular axis can also be explained by the fact that if the atoms are assumed to be rigid, friction-free spheres, there is no possibility to rotate them around the molecular axis. Because of the sphere-shape, each collision with other particles will be directed directly towards the molecular axis. Thus, no torque (no right-angled lever arm) is possible that could cause a linear molecule to rotate around the molecular axis. Therefore, this degree of freedom is omitted as energy storage for linear atoms.

Linear molecules have only two degrees of freedom for rotation!

In addition to the translational motion and rotational motion, molecules with two or more atoms can also vibrate. The possibilities of vibration (also called normal modes of vibration) depend on the number of atoms and the shape of the molecule. The atoms of diatomic molecules can only oscillate along their molecular axis and thus have only one vibrational degree of freedom (one normal mode of vibration).

Triatomic linear molecules, on the other hand, have four vibrational degrees of freedom. The atoms can either vibrate symmetrically along their bond axes (symmetric stretch mode) or phase-shifted (asymmetric stretch mode). In addition, two bending modes are possible, each of which leads to a change in the bond angle. A distinction is made between vertical and horizontal vibration mode.

Triatomic non-linear molecules, however, have a total of three degrees of freedom of vibration. On the one hand again a symmetrical stretching vibration and on the other hand again a asymmetric stretch mode along their bond axes. In addition, a bending mode is present wihtin the molecular plane.

From an energetic point of view, however, it must be taken into account with all vibrational degrees of freedom that a vibration contains both kinetic energy and potential energy. The vibrational energy is “redistributed” during a period, i.e. potential energy is gradually converted into kinetic energy and vice versa; the energy sum as the actual vibrational energy, however, always remains constant. Thus a normal mode of vibration always has two possibilities to store energy. As kinetic energy and as potential energy. Therefore, from an energetic point of view, each degree of freedom of a vibrational motion must be counted twice!

From an energetic point of view, degrees of freedom of vibrational motions are counted twice, because a vibration contains both potential and kinetic energy!

The table below shows the number of degrees of freedom f for monatomic particles, diatomic molecules (linear molecules) and non-linear molecules, divided into the various forms of motion.

The total number of degrees of freedom f results from the sum of the individual degrees of freedom for translation (f_trans), rotation (f_rot) and vibration (f_vib):

\begin{align}
&\boxed{f = f_{trans} + f_{rot} + 2 \cdot f_{vib}} \\[5px]
\end{align}

Equipartition theorem

There is a particular reason for such a division of energy into different degrees of freedom. Because it has been shown that the (internal) energy of a gas is equally divided among all degrees of freedom (see also main article equipartition theorem). The same amount of energy that, for example, is contained in the translational motion of the particles in the x-direction is also present as rotational energy of the particles around the z-axis. On average, this amount of energy will, for example, also be stored in the potential energy or kinetic energy of a vibrational motion.

Such a uniform distribution of the energy among the different degrees of freedom is also referred to as the equipartition theorem.

The equipartition theorem describes the equal division of the energy among the various degrees of freedom!

The article Pressure and Temperature shows that the temperature T of a gas is linked to the mean translational kinetic energy W of a particle by the following equation (with k_B as the Boltzmann constant):

\begin{align}
\label{bew}
&W_{kin} = \frac{3}{2} k_BT ~~~~~\text{mean translational kinetic energy of a particle}\\[5px]
\end{align}

This relationship is at first only valid between the temperature and the energy that is in the (three-dimensional) translational motion of a particle. Each degree of freedom of the translational motion thus has an average energy of ½⋅k_B⋅T according to the equipartition theorem.

However, the equipartition theorem is not limited to the translational kinetic energy, but applies to all forms of energy. If a degree of freedom of the translational motion has an average energy of ½⋅k_B⋅T, then this also applies to the degrees of freedom of rotation and vibration (because the energy is just equally distributed among all degrees of freedom!).

On average, each degree of freedom (no matter if translational, rotational or vibrational!) has an energy of ½⋅k_B⋅T. If one knows the total number of degrees of freedom f of a molecule, then this molecule carries on average an energy of W=f⋅½⋅k_B⋅T at a given temperature T:

\begin{align}
\label{frei}
&\boxed{W = \frac{f}{2} k_BT~} ~~\text{energy of a molecule}\\[5px]
\end{align}

If, in addition, the number of molecules N in a gas is known, then the total energy of the gas and thus its internal energy can be determined. Because if a particle has on average an energy W, then the total energy U=N⋅W will be present in the gas:

\begin{align}
&\boxed{U = \frac{f}{2}Nk_BT~} ~~~~~\text{internal energy of an (ideal) gas}\\[5px]
\end{align}

The internal energy of the gas can be expressed not only by the number of molecules but also by the amount of substance n=N/N_A, i.e. by the “number of moles” (with N_A as Avogadro constant):

\begin{align}
&U = \frac{f}{2} \cdot n \cdot \underbrace{N_A \cdot k_B}_{R_m} \cdot T \\[5px]
\end{align}

In the equation above, the Avogadro constant N_A and the Boltzmann constant k_B can be combined to a new constant, the so-called molar gas constant R_m (also called universal gas constant or ideal gas constant):

\begin{align}
\label{inn}
&\boxed{U = \frac{f}{2} R_m n T~}~~~~~R_m = 8,314 \frac{\text{J}}{\text{mol}\cdot\text{K}} \\[5px]
\end{align}

Change of internal energy

The understanding of the degrees of freedom and the connection to the internal energy can now be used to calculate the temperature change of an ideal gas when energy in form of heat or mechanical work is supplied.

For this an ideal gas is considered whose degrees of freedom f are assumed to be known. If the internal energy of the gas changes by an amount ΔU, then this results in a temperature change ΔT, since all other values in equation (\ref{inn}) are constant:

\begin{align}
\label{du}
&\boxed{\Delta U = \frac{f}{2} R_m n ~ \Delta T~} ~~~~~\text{valid for any thermodynamic process of an ideal gas} \\[5px]
\end{align}

This fact also becomes clear. If, for example, the internal energy of a gas is increased by adding heat, the increase in energy will be equally divided among all forms of energy. Also the translational kinetic energy will increase, which is directly connected with a temperature increase according to the equation (\ref{bew}). Conversely, a reduction of the internal energy through heat dissipation will lead to a reduction of the kinetic energies and thus lower the temperature.

The temperature increase during a compression of a gas can also be explained with this equation. If a gas is compressed (by mechanical work), the required energy is used to increase the internal energy of the gas. According to the equation (\ref{du}), this is associated with an increase in temperature. This principle of increasing the temperature by compression is used, for example, in diesel engines to heat the fuel-air mixture so strongly that it ignites itself.

The animation below shows an experiment to demonstrate the temperature increase during compression. A piece of a cotton pad is placed in a plexiglass cylinder. The cylinder is closed with a piston. If the piston is then suddenly moved downwards, the air in it is compressed and heats up to such an extent that the cotton pad ignites.

The compression must therefore take place very quickly, otherwise the gas on the relatively cool cylinder walls would immediately cool down again. The rapid compression prevents heat dissipation from the gas (so-called adiabatic process).

The temperature increase during compression can also be explained in another way. If the piston is moved into the cylinder, then the piston collides with gas particles like a tennis racket with tennis balls. The speed of the molecules increases, which means an increase in temperature.

In practice, such a complex experimental setup is not required to demonstrate the temperature increase during compression. The daily experience of inflating a bicycle tyre already shows that air heats up during compression. The heating of the air pump is not due to friction as one might think but mainly to the energy input in terms of mechanical work during compression so that the gas heats up!

Molar heat capacity

The qualitative statements about the connection between an energy input – in particular a heat supply – and the resulting temperature increase will be discussed in more detail in the following.

Molar heat capacity of isochoric processes

If a gas is supplied with a certain amount of heat – and the gas itself does not release any energy by mechanical work or heat dissipation (!) – then the supplied heat completely benefits the internal energy of the gas (law of conservation of energy). The internal energy will therefore increase by the amount of the supplied heat Q, i.e. the heat supply Q corresponds directly to the change of the internal energy ΔU:

\begin{align}
& Q \overset{!}{=} \Delta U \\[5px]
&\boxed{Q = \frac{f}{2}R_m n \cdot \Delta T~} ~~~~~\text{only valid for isochoric processes of ideal gases} \\[5px]
\end{align}

This equation applies not only to a heat supply, but also to a heat dissipation (the heat then has a negative sign in the thermodynamic sense). As already explained, heat dissipation leads to a reduction of the internal energy and thus to a lowering of the temperature (negative temperature change).

What is remarkable about the above equation is that it obviously does not matter which type of gas exactly is involved. There is no gas-specific quantity in this equation. Of importance is only the number of degrees of freedom of the molecules, but not the type of atoms that make up the molecule. For an ideal gas, therefore, only the “shape” of the molecule is important, i.e. whether it is monatomic (f=3), diatomic or linear (f=7) or non-linear (f=12), as this determines the number of degrees of freedom (see table above).

The constant term f/2⋅R_m for a certain number of degrees of freedom is also known as the molar heat capacity C_m,v, because it determines the quantitative relationship between a heat input/output and the temperature change:

\begin{align}
\label{qv}
&\boxed{Q = C_{m,v} \cdot n \cdot \Delta T~} \\[5px]
\label{c}
&\boxed{C_{m,v} := \frac{f}{2}R_m} = \frac{Q}{n ~ \Delta T} ~~~~\left[C_{m,v}\right]=\frac{\text{J}}{\text{mol} \cdot \text{K}}\\[5px]
\end{align}

As can be seen from equation (\ref{c}), the molar heat capacity indicates how much heat must be added to increase the temperature of a gas with an amount of substance of one mole by one Kelvin.

The molar heat capacity indicates how much heat is required to heat one mole of a substance by one Kelvin!

Equation (\ref{qv}) was derived under the condition that the gas would not release any energy during the heat supply. The question then arises how such a thermodynamic process would look in practice.

A simple example is a sealed container filled with gas and heated with a Bunsen burner. Although the vessel absorbs a part of the supplied heat and the gas can release a part of the supplied heat to the environment through the vessel, the equation (\ref{qv}) remains valid if the heat Q is based on the net amount, i.e. the heat effectively supplied to the gas.

However, a gas can not only release energy into the environment in terms of heat but can also perform mechanical work. Think, for example, of a piston in a cylinder of a combustion engine, which is pushed downwards by the gas pressure with a certain force during this work cycle. The gas obviously releases energy in terms of mechanical work.

In principle, a gas will always perform mechanical work when its volume increases. This is because an increase in volume always means that the gas increases its volume along a distance with a certain force (caused by the gas pressure at the system boundary). After all, a force acting along a distance is precisely the definition of mechanical work. Therefore, a change in volume of a gas is always associated with mechanical work. In this context, one often speaks of a pressure-volume work.

Pressure-volume work (PV work) refers to the mechanical energy supplied to a gas during compression or the mechanical energy released by the gas during expansion!

For the equation (\ref{qv}) to be valid, the gas volume must therefore be kept constant during the heat supply, otherwise this equation loses its validity due to the mechanical work still to be taken into account (for this reason, the vessel considered is closed with a cork!). Such a type of heat supply under constant volume is also referred to as an isochoric process.

A thermodynamic change of state at a constant volume is also called an isochoric process!

To make it clear that equation (\ref{qv}) only applies to an isochoric process (and Q as net heat amount), the letter v is added to the molar heat capacity (“constant volume”). Note that with an isochoric process, the heat supplied is completely used to increase the internal energy, or the heat dissipated is completely at the expense of the internal energy (see energy flow diagram below).

Important note

With the definition of the molar heat capacity of the isochoric process C_m,v the change of the internal energy ΔU can also be expressed with this quantity [see equation (\ref{du})]:

\begin{align}
\label{du2}
&\boxed{\Delta U = C_{m,v} n ~ \Delta T~} ~~~~~\text{applies to any thermodynamic process of ideal gases} \\[5px]
\end{align}

In contrast to equation (\ref{qv}), equation (\ref{du2}) applies to any thermodynamic process of an ideal gas, even if it contains the molar heat capacity of the isochoric process! For the change of the internal energy the use of this quantity is only a purely formal way of writing!

Molar heat capacity of isobaric processes

During the considered isochoric heat supply of the closed vessel, the pressure will also increase as a result of the temperature increase. If, on the other hand, the cork is removed, the gas can expand under constant (ambient) pressure during the heat supply, i.e. increase its volume. In contrast to an isochoric process, a thermodynamic change of state under constant pressure is referred to as an isobaric process.

A thermodynamic change of state at constant pressure is also called an isobaric process!

Even for an isobaric process, a molar heat capacity can be defined as a function of the degrees of freedom. This will be shown in the following.

As an example of an isobaric process, a gas in a vertical cylinder is considered which is closed by a plate (“piston”) sliding frictionless in the cylinder. Under the weight of the plate, the gas is first compressed and the pressure increases until a state of equilibrium is reached.

In this state, the gas pressure p exerts the force F_p=p⋅A from inside the cylinder on the plate surface A. From the outside, the weight F_g=m⋅g acts on the plate surface (more precisely: on the interface to the gas, which is also called the system boundary) and the ambient pressure p₀ as well (!), which additionally exerts a force F₀=p₀⋅A on the plate surface or on the system boundary.

In the state of equilibrium, the external forces F_g and F₀ are in equilibrium with the internal force caused by the gas F_p. The gas pressure p can be determined from this equilibrium analysis as follows:

\begin{align}
&F_p \overset{!}{=} F_g + F_0 \\[5px]
&p \cdot A = mg + p_0A\\[5px]
&\boxed{p = \frac{mg}{A}+p_0} \\[5px]
\end{align}

The gas pressure is obviously only influenced by mass of the plate (and its surface area) and by the ambient pressure. The weight of the plate and the ambient pressure forces the pressure of the gas. Even if the gas is now heated, this situation will not change in principle. The pressure will always be determined by the weight of the plate and by the ambient pressure (just imagine your hand being the gas; then you must apply the same force in each position to keep the plate in position against its weight and the ambient pressure). Thus one obtains a heat supply with which the pressure does not change!

One could also argue as follows: If the gas is heated, then a possible increase in pressure is compensated by a rising of the plate upwards (expansion), so that the pressure always remains constant.

From an energetic point of view, the gas is supplied with energy in terms heat on the one hand, while on the other hand the gas releases energy in terms of mechanical work by lifting the weight of the plate during expansion. The gas thus increases its internal energy by the amount of heat supplied Q and simultaneously reduces it by the amount of work done W, so that the internal energy effectively increases by the difference:

\begin{align}
\label{ees}
&\Delta U = Q – W \\[5px]
\end{align}

While the change of the internal energy is already given by equation (\ref{du}), the work done by the gas W can be determined from the constant force F_p=p⋅A with which the gas raises the plate by a distance Δh:

\begin{align}
&W = F_p \cdot \Delta h = p \cdot \underbrace{A \cdot \Delta h}_{\Delta V} = p \cdot \Delta V\\[5px]
&W = p \cdot \Delta V \\[5px]
\end{align}

In the equation above it was used that the product of the plates’s surface area A and lifting distance Δh corresponds to the change in volume ΔV of the gas. The work done as product of pressure and volume change can finally be expressed by the temperature change ΔT using the ideal gas law:

\begin{align}
&p V =R_m \cdot n \cdot T ~~~~~\text{ideal gas law} \\[5px]
&p \cdot \Delta V = R_m \cdot n \cdot \Delta T \\[5px]
\end{align}

The following relationship therefore applies between the work performed by the gas and the resulting temperature change of the gas:

\begin{align}
\label{w}
&W = p\cdot \Delta V= R_m \cdot n \cdot \Delta T \\[5px]
\end{align}

If the equations (\ref{du}) and (\ref{w}) are used in equation (\ref{ees}), the following relationship between the supplied heat and the resulting temperature change becomes apparent for an isobaric process:

\begin{align}
\Delta U &= Q – W \\[5px]
\frac{f}{2} R_m n ~ \Delta T &= Q – R_m \cdot n ~ \Delta T \\[5px]
Q &= \frac{f}{2} R_m n ~ \Delta T + R_m \cdot n ~ \Delta T \\[5px]
Q &= \underbrace{\left(\frac{f}{2} + 1\right)R_m}_{C_{m,p}} \cdot n ~ \Delta T \\[5px]
\end{align}

The expression marked above the underbrace can now be interpreted as the molar heat capacity of the isobaric process C_m,p, which describes the relationship between a heat supply at constant pressure and the resulting temperature change:

\begin{align}
\label{qp}
&\boxed{Q = C_{m,p} \cdot n \cdot \Delta T~} ~~~~~\text{and}~~~~~ \boxed{C_{m,p} := \left(\frac{f}{2}+1\right)R_m} \\[5px]
\end{align}

Relationship between the molar heat capacities

A closer look at the molar heat capacities shows that the molar heat capacity of the isobaric process is always greater by the value of the molar gas constant than the molar heat capacity for the isochoric process:

\begin{align}
\require{cancel}
& C_{m,p} – C_{m,v} = \left(\tfrac{f}{2}+1\right)R_m – \tfrac{f}{2}R_m = \bcancel{\tfrac{f}{2}R_m} + R_m – \bcancel{\tfrac{f}{2} R_m} = R_m \\[5px]
\label{cc}
&\boxed{C_{m,p} = C_{m,v} + R_m}
\end{align}

The always greater molar heat capacity C_m,p compared to C_m,v means that for an isobaric process obviously more heat has to be supplied in order to achieve the same change in temperature. This also becomes clear, because with an isobaric process the supplied heat is not completely used to increase the internal energy and thus the temperature, but a part of the heat is converted into mechanical work (note that a temperature change is directly linked to a change in the internal energy according to equation (\ref{du})!) However, if the same temperature change were to be caused, an amount of heat greater by the amount of work done by the gas would have to be added.

Adiabatic processes

Other important processes in thermodynamics are state changes that occur so quickly that in principle there is no time for heat supply or heat dissipation. Due to the high speeds in combustion engines, this is approximately the case, for example, with the expansion and compression processes that take place there. Such processes without heat transfer are also referred to as adiabatic processes. The experiment already described, in which a cotton pad was ignited by rapid compression of a gas, can be regarded as approximately adiabatic.

The heat capacity ratio plays an important role in describing these processes. This ratio is also called adiabatic exponent or isentropic exponent κ:

\begin{align}
\require{cancel}
& \kappa:=\frac{C_{m,p}}{C_{m,v}} = \frac{\left(\tfrac{f}{2}+1\right)\bcancel{R_m}}{\tfrac{f}{2}\bcancel{R_m}}= 1+ \frac{2}{f} \\[5px]
&\boxed{\kappa = 1 + \frac{2}{f}}
\end{align}

Molar heat capacities of selected gases

The table below shows the molar heat capacities and the isentropic exponents at room temperature for ideal gases and for selected real gases. It can be seen that helium comes very close to a monatomic ideal gas with three degrees of freedom. Similarly, nitrogen and air come very close to a diatomic ideal gas with seven degrees of freedom.

*) Important: diatomic gases can theoretically have 7 degrees of freedom, but in practice the two degrees of freedom of vibration are usually frozen at room temperature – more about this in the next section.

Note that in practice, the molar heat capacity is first determined from the temperature change at a given heat input according to equation (\ref{c}). Then the number of degrees of freedom can be deduced. Thus, it can be seen that the microscopic state in terms of the degrees of freedom of a gas can be inferred from macroscopically measurable quantities!

Dependence of molar heat capacities on temperature

If one considers the equations for determining the molar heat capacities, then these obviously depend only on the number of degrees of freedom:

\begin{align}
&C_{m,v} := \frac{f}{2}R_m \\[5px]
&C_{m,p} := \left(\tfrac{f}{2}+1\right)R_m \\[5px]
\end{align}

One could assume that the molar heat capacities are independent of the temperature. In practice, however, it is found that the heat capacities are largely constant only within certain temperature ranges. If, for example, the molar heat capacity C_m,v is plotted against the temperature in a diagram, then it can be seen that the molar heat capacity also decreases with decreasing temperature.

The figure above shows the course of the molar heat capacity of the isochoric process as a function of temperature (an ideal diatomic gas is assumed). Since the molar gas constant is a physical constant, the change in heat capacity must be due to the number of degrees of freedom. Obviously the number of degrees of freedom decreases with decreasing temperature.

The reduction of the degrees of freedom can only be explained by quantum mechanics, whose basic idea is based on the fact that energies cannot assume arbitrary values, but only discrete values. In order to excite a certain energy state of a molecule, certain minimum energies are necessary. For example, the oscillation of a diatomic molecule along its molecular axis requires a certain minimum energy. If this minimum energy is not present, then the vibration cannot be excited. If no vibration can be made possible, then this degree of freedom can not be used for energy storage. Due to insufficient energy of the molecules (equivalent to too low temperatures) this degree of freedom is so-called frozen.

A degree of freedom is called “frozen” if due to too low energy (i.e. too low temperature) this degree of freedom cannot be excited for quantum mechanical reasons and is therefore not present!

The situation can be illustrated with two balls (“atoms”) connected by an elastic spring (“binding forces”), but additionally by a fragile rod. At low kinetic energies, the rod will thus prevent the balls from vibrating. Thus, no energy can be stored in terms of vibrational energy. If, on the other hand, the kinetic energy is high (in the figurative sense this corresponds to a high temperature), then the rod can break in a collision with another particle. Now the balls are able to vibrate and can store vibrational energy accordingly.

The reason why the molar heat capacity drops within a temperature range and not suddenly is due to the characteristic speed distribution in an ideal gas. At a certain temperature, not all particles have the same speed. Some particles are slower and some faster. If the temperature drops, the degree of freedom is not frozen for all molecules at the same time, but only for those particles whose kinetic energies are too low.

With a diatomic gas, the degree of freedom of vibration is first frozen with decreasing temperature. Then the degree of freedom of rotation takes place.

Even if a gas is regarded as ideal, degrees of freedom are frozen in practice and the heat capacities change as a result. If such changes in the heat capacities are neglected for the sake of simplification, then one also speaks of perfect gases.

A perfect gas is an ideal gas whose heat capacities are regarded as constant, i.e. in particular no degrees of freedom are frozen with decreasing temperature!

Molar heat capacity of solids

A distinction between an isobaric and an isochoric process does not have to be made for liquids or solids in general. The reason for this is that solids and liquids are incompressible, i.e. they do not change their volume under pressure or with a change in temperature. The low thermal expansion is negligible! These substances are therefore not able to perform significant pressure-volume work. For this reason, a heat supply always resembles an isochoric change of state.

Empirical research has shown that in many cases the considerations about the degrees of freedom of ideal gases can also be applied to solids. In contrast to gases, molecules in solids cannot move freely but are bound to certain locations. In these cases, rotation of the molecules is also omitted. The movements of the molecules in solids are thus limited to vibrations around an equilibrium position, whereby the oscillations can take place in all three spatial directions.

For the molecules in solids there are three degrees of freedom of vibration, which have to be counted twice due to the potential and kinetic energy of an vibration. So with effectively six degrees of freedom there is a molar heat capacity for solids which theoretically corresponds to three times the molar gas constant:

\begin{align}
&C_{m} = \frac{f}{2}R_m = \frac{6}{2}R_m = 3 R_m \\[5px]
& \boxed{C_m = 25~\tfrac{\text{J}}{\text{mol}\cdot\text{K}} } ~~~~~\text{Dulong-Petit law} \\[5px]
\end{align}

And indeed, empirical research shows that this simple approach applies to many solids in good approximation. This connection is also known as the Dulong-Petit law. The term “law” is not really appropriate at this point, since it is rather an estimation.

The Dulong-Petite Law states that the molar heat capacity of solids is about three times the molar gas constant!

The table below shows the molar heat capacity at room temperature for selected solids. However, it can also be seen that the estimation according to the Dulong-Petite Law is a good approximation for many solids.

Note that even in the case of solids with lower temperatures, more and more degrees of freedom are frozen. This is especially true in the vicinity of absolute zero, so that all degrees of freedom of vibration are frozen at this point at the latest. In principle, this applies to all substances, since no energy is present at absolute zero, but each degree of freedom in principle requires a certain minimum energy in order to be excited at all.

The heat capacity of substances decreases to zero as the temperature approaches absolute zero!

Specific heat capacity

As already explained, the molar heat capacity indicates how much heat has to be supplied to raise the temperature of a substance with one mole of particles by one Kelvin. If one wants to determine the temperature change of a substance with a heat supply, the amount of substance n must obviously be known:

\begin{align}
& Q = C_m n \Delta T \\[5px]
\end{align}

In practice, however, one usually does not measure the amount of substance, but the mass. Therefore, it is often usual to express the heat capacity not in relation to the amount of substance but in relation to the mass. The amount of substance and mass are connected by the molar mass. The molar mass M is defined as the quotient of the mass m and the amount of substance n and thus indicates how much mass per mole a substance has:

\begin{align}
& \boxed{M := \frac{m}{n}} \\[5px]
\label{n}
& n = \frac{m}{M} \\[5px]
\end{align}

If the amount of substance is expressed by the mass and the molar mass, then the following relationship between heat and temperature change applies:

\begin{align}
& Q = C_m \cdot \frac{m}{M} \cdot \Delta T = \underbrace{\frac{C_m}{M}}_{=c} \cdot m \cdot \Delta T =c \cdot m \cdot \Delta T \\[5px]
&\boxed{Q= c \cdot m \cdot \Delta T} \\[5px]
\end{align}

In the equation above, the quotient of molar heat capacity and molar mass (substance-dependent!) was combined to the specific heat capacity c:

\begin{align}
& \boxed{c = \frac{C_m}{M}}=\frac{Q}{m \cdot \Delta T} ~~~~~[c]=\frac{\text{J}}{\text{kg} \cdot \text{K} } \\[5px]
\end{align}

The specific heat capacity is a substance-specific quantity and indicates how much heat is needed to raise the temperature of a substance with a mass of one kilogram by one Kelvin!

The specific heat capacity of gases must also be differentiated between an isochoric and an isobaric heat supply. Accordingly, a distinction is made between the specific heat capacity of the isochoric process c_v and the specific heat capacity of the isobaric process c_p.

The relationship between the molar heat capacities C_m,v and C_m,p according to the equation (\ref{cc}) can now be used to create a relationship between the corresponding specific heat capacities:

\begin{align}
&C_{m,p} = C_{m,v} + R_m \\[5px]
&\frac{C_{m,p}}{M} = \frac{C_{m,v}}{M} + \underbrace{\frac{R_m}{M}}_{=R_s} \\[5px]
&\boxed{c_p=c_v+R_s} \\[5px]
\end{align}

The substance-dependent quotient of the molar gas constant R_m and the molar mass is combined to the so-called specific gas constant R_s:

\begin{align}
&\boxed{R_s=\frac{R_m}{M}} \\[5px]
\end{align}

Instead of using the molar heat capacity C_m,v to calculate the change in internal energy, the specific heat capacity c_v can now also be used:

\begin{align}
&\boxed{\Delta U = c_{v} m ~ \Delta T~} \\[5px]
\end{align}

Introduction

In the article Maxwell-Boltzmann distribution it was shown that the mean speed (average speed) of the particles of an ideal gas can be determined with the following formula:

\begin{align}
\label{v}
&\boxed{ \bar{v} = \sqrt{\frac{8 k_B T}{\pi m}}  } \\[5px]
\end{align}

This formula can be used, for example, to estimate the mean speed of air particles. Since 78 % of air consists of nitrogen, the average speed of the nitrogen molecules (N₂) is to be calculated. Such a nitrogen particle has a mass of 4.65⋅10^-26 kg. At a temperature of 20° C (293 K) this results in a mean speed of about 470 m/s.

On average, the air particles move at supersonic speed. Due to the statistical distribution of the velocities, however, significantly higher speeds are also present. About 1 % of the molecules have a speed of more than 1000 m/s. One molecule out of one billion even reaches a speed of 2000 m/s.

If gas molecules generally have such high speeds, why not immediately perceive the smell of an open perfume bottle at the other end of a room, as one would expect at speeds of several hundred meters per second. Experience shows that it obviously takes some time for the fragrance to be noticed.

The apparent contradiction lies in the fact that the gas particles do not have a free path when moving. The molecules will permanently collide with other particles and change their direction of motion in a random way. The distance a molecule can travel on average without colliding with other molecules is called mean free path. In the present case, the relatively small mean free path of the fragrance particles prevents the perfume from being perceived immediately.

The mean free path is the average distance a particle travels without colliding with other particles!

The fact that the fragrance is nevertheless perceived relatively quickly is mainly due to air currents (convections), which carry the particles over greater distances. Note that convection is no longer a completely random motion. In this case, the molecules are moved over macroscopic distances in a certain direction.

Calculation of the mean free path

In order to determine the mean free path of a particle, a gas consisting of only one type of molecule is considered. The molecules are assumed to be spheres with a diameter d. If one now follows a particle (shown in red in the figure below), it will collide with other particles at irregular intervals (shown in black). The average distance travelled between two successive collisions corresponds to the mean free path λ.

For the sake of simplicity, all particles (with the exception of the particle followed in thought) are considered to be completely at rest. The particle traced thus moves through an “ocean” of resting particles and changes direction after each free path.

A collision between the moving particle and a stationary particle will occur when the surfaces of the spherical particles touch each other. This will be the case if the distance of the centers of gravity is smaller than twice the particle radius.

Around the center of gravity of the moving molecule a circular collision cross section σ with the radius R=2r=d can thus be defined perpendicular to the direction of motion. The center of gravity of a stationary particle must lie within this cross-sectional area in order to collide with the moving particle:

\begin{align}
&\sigma = \pi R^2 = \pi d^2 ~~~~~\text{collision cross section} \\[5px]
\end{align}

In the direction of motion, one will obtain an imaginary collision cylinder (collision volume). If the center of gravity of a particle is within this cylinder, a collision will occur. The length of the collision cylinder corresponds to the free path λ.

The (mean) volume of the collision cylinder V_c results from the product of the collision cross section σ and the length of the (mean) free path λ:

\begin{align}
& V_c = \sigma \cdot \lambda = \pi d^2 \cdot \lambda~~~~~\text{collision cylinder} \\[5px]
\end{align}

In every cylindrical collision volume V_c=σ⋅λ there is by definition just one particle, namely the particle with which the moving particle collides. Then the moving particle will change its direction and define a new collision volume, again with a “target” particle with which it will collide.

The statement that there is only one single molecule within a collision cylinder ultimately refers to the centre of gravity of the particles. In general, there will be several particles whose surface will reach into the collision volume, but only if the centre of gravity is inside this cylinder will there actually be a collision. This will only be the case for one molecule, because from then on a new collision volume will be defined, until finally a new collision will occur. This consideration of the center of gravity also makes sense insofar as the particles can still be imagined as mass points surrounded by a spherical “collision shell”. With this assumption as mass points it becomes clear that there is actually only one particle in the collision volume (see figure below).

How many molecules are in a certain volume can also be determined by the particle density n in the gas. This particle density n is calculated by the quotient of the total number of particles N and the entire volume of the gas V:

\begin{align}
& n = \frac{N}{V} ~~~~~\text{particle density}\\[5px]
\end{align}

The particle density indicates the number of particles per unit volume. If one multiplies this particle density n by any volume V_c, then one obtains the average number of particles N_c inside this volume. For the collision volume, this number is of course one, so that the mean free path λ can be determined by using this condition (!):

\begin{align}
& N_c = n \cdot V_c =n \cdot 2 \pi d^2 \lambda \overset{!}{=} 1\\[5px]
& \underline{\lambda = \frac{1}{n \pi d^2}}
\end{align}

The length of the mean free path is only dependent on the particle density and the particle diameter! However, this consideration was based on stationary particles. In fact, however, the individual molecules will move relative to each other. It is to be assumed that this will then lead to increased collisions per time and therefore the mean free path will be shortened. Taking into account the Maxwell-Boltzmann speed distribution, the mean free path is then shortened by the factor 1/√2 (the derivation of this factor is shown in the last section):

\begin{align}
\label{l}
& \boxed{\lambda = \frac{1}{\sqrt{2}n \pi d^2}} ~~~~~\text{}n=\frac{N}{V}
\end{align}

For an ideal gas, the particle density n=N/V can also be expressed by the thermodynamic temperature T and the pressure p according to the ideal gas law:

\begin{align}
&pV=N k_B T ~~~~~\text{ideal gas law}\\[5px]
\label{n}
&n=\frac{N}{V}=\frac{p}{k_B T}\\[5px]
\end{align}

If equation (\ref{n}) is used in equation (\ref{l}), then the length of the mean free path can also be determined as follows:

\begin{align}
\label{lam}
& \boxed{\lambda = \frac{k_B T}{\sqrt{2} p \pi d^2}}
\end{align}

If a nitrogen molecule with a diameter of about d = 370 pm (this corresponds to a collision cross section of σ = 4.3·10^-19 m²) and a temperature of T=293 K and a pressure of p=1 bar is considered, a mean free path of λ = 67 nm results. In this case, the mean free path is about 10 times less than the wavelength of visible light!

Calculation of the collision frequency

If, in addition to the length of the (mean) free path λ, the (mean) speed \(\overline{v}\) of the molecules is also known, then the (mean) time period τ between two collisions can be determined:

\begin{align}
& \text{speed } \overline{v}= \frac{\text{distance }\lambda}{\text{time } \tau} \\[5px]
\label{t}
& \boxed{\tau = \frac{\lambda}{\overline{v}}} ~~~~~\text{duration between two collisions}
\end{align}

This mean time between two collision has the meaning of a period τ, since it indicates the repetitive time intervals in which on average collisions take place. Therefore, the reciprocal of the time τ can be understood as the collision frequency f, which indicates the number of collisions per unit time. This collision frequency is often denoted by the letter Z.

\begin{align}
& Z = f = \frac{1}{\tau} \\[5px]
\label{zz}
& \boxed{Z = \frac{\overline{v}}{\lambda}} ~~~~~\text{collision frequency}
\end{align}

The number of collisions per unit time between a particle and its “target” particles is referred to as the collision frequency!

If equation (\ref{v}) for the mean speed and equation (\ref{lam}) for the mean free path are used in the formula for the collision frequency, the following formula results:

\begin{align}
&Z = \frac{\overline{v}}{\lambda}
= \frac{\sqrt{\frac{8 k_B T}{\pi m}} }{\frac{k_B T}{\sqrt{2} p \pi d^2} }
= \sqrt{\frac{8 k_B T}{\pi m}} \frac{\sqrt{2} p \pi d^2 }{k_B T}
= \sqrt{\frac{8 k_B T}{\pi m}} \sqrt{\left(\frac{\sqrt{2} p \pi d^2 }{k_B T}\right)^2} \\[5px]
&= \sqrt{\frac{8 k_B T}{\pi m} \left(\frac{\sqrt{2} p \pi d^2 }{k_B T}\right)^2}
= \sqrt{\frac{8 k_B T}{\pi m} \frac{2 p^2 \pi^2 d^4 }{k_B^2 T^2}}
= \sqrt{\frac{16 \pi p^2 d^4}{k_B T m}} \\[5px]
\label{z}
&\boxed{Z=\sqrt{\frac{16 \pi p^2 d^4}{k_B T m}} } \\[5px]
\end{align}

For the nitrogen molecule already considered at a temperature of 293 K and at a pressure of 1 bar, a collision frequency of 7·10⁹ 1/s results, i.e. within one second a single nitrogen molecule will collide on average with 7 billion other molecules!

To obtain the total number of collisions per unit volume and time, the collision frequency Z only has to be multiplied by the particle density n (“number of particles per unit volume”). It must be noted that two particles each will collide, so that a factor ½ must still be taken into account. If the particle density n is expressed by temperature and pressure according to the equation (\ref{n}), then the total number of collisions per unit volume and time is determined as follows:

\begin{align}
&z=\frac{1}{2} \cdot n \cdot Z = \frac{1}{2} \cdot \frac{p}{k_BT} \cdot \sqrt{\frac{16 \pi p^2 d^4}{k_B T m}} = \sqrt{\frac{4 \pi p^4 d^4}{k_B^3 T^3 m}} \\[5px]
&\boxed{z= \sqrt{\frac{4 \pi p^4 d^4}{k_B^3 T^3 m}} } ~~~~~\text{total number of collisions per unit volume and time}
\end{align}

For nitrogen, a total number of z = 8.7·10³⁴ 1/sm³ is obtained, i.e. 8.7·10³⁴ collisions occur within one second in a volume of one cubic metre (a hell of a lot of collisions in a second!).

Derivation of the factor 1/√2

In this section the question shall be clarified how exactly the factor 1/√2 in the equation (\ref{lam}) for the mean free path comes about. The initial situation was the assumption that the “target” molecules of a moving molecule are all at rest.

Model conception

In analogy one can imagine a market place with many people, which one would like to cross on foot. However, all persons remain at rest for the time being. Now one starts walking straight ahead. Every time one meets a person, one randomly changes direction and then walks straight ahead again. The distance that one can walk on average without colliding with a person corresponds in a figurative sense to the mean free path.

One can determine the free path on the market place as follows. For this one has to measure the time between two collisions and on the other hand one needs the speed with which one moves. From the product of (mean) time τ₀ and (mean) speed \(\overline{v}\) the (mean) free path λ results:

\begin{align}
& \lambda_0 = \overline{v} \cdot \tau_0 ~~~~~\text{mean free path for stationary “targets”}
\end{align}

Now we assume that the people on the marketplace move themselves in a chaotic way when one crosses it. The duration τ between two collisions can again be determined by a simple measurement. Although one still walks across the marketplace at the same speed, one will notice that the mean time between two collisions is shortened (τ<τ₀). This also shortens the mean free path (λ<λ₀):

\begin{align}
& \lambda = \overline{v} \cdot \tau ~~~~~\text{mean free path for moving “targets”}
\end{align}

The shortening of the time (or the mean free path) is ultimately due to the fact that not only the own speed determines the time interval between two collisions, but also the relative speed with which one moves towards a target. Thus, the speed of the surrounding persons is also relevant. The relative speed is basically also decisive in the case of stationary “targets”, but since these are considered to be stationary, the relative speed corresponds to the speed at which one moves oneself!

Transfering the model conception to gases

If this model conception is transferred to a gas, then the mean speed \(\overline{v}\) of a particle may not be used as the basis for the mean time between two collisions, but rather the mean relative speed \(\overline{v_{rel}}\) with which the particles approach each other on average. From statistical considerations, a relationship can be deduced between these two speeds.

For this purpose, two particles are considered which will approach each other and collide with each other. The velocity vectors of the two particles can be arranged in any way (see figure below). From the point of view of particle 1, the relative velocity at which particle 2 moves results from the difference between the velocity vectors:

\begin{align}
\label{rel1}
& \vec{v_{rel}} = \vec{v_2} – \vec{v_1} \\[5px]
\end{align}

Since later anyway only the magnitude of the (mean) relative velocity vector is relevant for the (mean) free path, it would have made no difference at this point in which order the velocity vectors would have been subtracted from each other. One could have exchanged the indices in the equation above. In this way one would have described the approach of the particles from the point of view of particle 2.

Basically it makes no difference from a mathematical point of view whether you square the magnitude of a vector (v²) or whether you square the vector itself (\(\vec{v}^2\)). In both cases you get the same scalar result:

\begin{align}
\label{mat}
& v^2 = \vec{v}^2\\[5px]
\end{align}

Note: For the sake of clarity, the magnitude of a velocity ist denoted only by the symbol v without any further indication. If, on the other hand, the velocity vector \(\vec{v}\) is meant, then an arrow is explicitly placed above the symbol.

The magnitude of the relative velocity v_rel (relative speed) can thus be expressed as follows by the corresponding vector \(\vec{v_{rel}}\):

\begin{align}
\label{rel2}
& v_{rel}^2 = \vec{v_{rel}}^2\\[5px]
\end{align}

If equation (\ref{rel1}) is now applied in equation (\ref{rel2}), then the following relationship results between the relative speed and the speeds of the individual particles:

\begin{align}
& v_{rel}^2= \vec{v_{rel}}^2 =\left(\vec{v_2} – \vec{v_1} \right)^2 =\vec{v_2}^2 + \vec{v_1}^2 – 2\vec{v_1}\vec{v_2} \\[5px]
\end{align}

To simplify this equation, the square of the velocity vectors can be replaced by the square of the respective speeds (see equation (\ref{mat})):

\begin{align}
& v_{rel}^2= v_2^2 + v_1^2 – 2\vec{v_1}\vec{v_2} \\[5px]
\end{align}

With this equation, only the relative speed of individual collisions can be determined so far. A statement about the mean relative speed is only possible if all potential collisions (with their respective velocity vectors) are considered and the average relative speed ist calculated:

\begin{align}
& \overline{v_{rel}^2}= \overline{v_2^2 + v_1^2 – 2\vec{v_1}\vec{v_2}} \\[5px]
\end{align}

Since the right side of the equation is the mean value of a sum, the mean value of the individual summands can be calculated instead:

\begin{align}
\label{term}
& \overline{v_{rel}^2}= \underbrace{\overline{v_2^2}}_{\text{term 1}} +
\underbrace{\overline{v_1^2}}_{\text{term 2}} –
\underbrace{\overline{2\vec{v_1}\vec{v_2}}}_{\text{term 3}} \\[5px]
\end{align}

The first term, which contains the squared speed of particle 2, does not differ on average from the second term, which contains the squared speed of particle 1. Due to the random motions, the mean speed of the two particles – or rahter all particles – is identical. It has already been mentioned that the indices could have been interchanged. This is only a question of whether the collision is observed from one particle or the other. The mean speeds v₁ and v₂ will thus be identical and correspond to the mean speed \(\overline{v}\) of any particle:

\begin{align}
& \overline{v_{1}^2}=\overline{v_{2}^2}=\overline{v^2} \\[5px]
\end{align}

The third term in the equation (\ref{term}) contains the scalar product of the velocity vectors. As is usual with a scalar product, the individual velocity components are multiplied by each other and then summed up. The individual components can be both negative and positive. However, since this is a completely statistical distribution of the components, positive and negative values for the scalar product are obtained to the same extent. When considering a sufficient number of particles or collisions (this is of course the case with a collision rate in the order of 10³⁴ collisions per second and cubic meter!), the positive values of the scalar product will on average compensate the equally negative scalar products. On statistical average the third term in equation (\ref{term}) will be zero:

\begin{align}
& \overline{2\vec{v_1}\vec{v_2}} = 0 \\[5px]
\end{align}

For the mean of the squared relativ speed the following formula applies:

\begin{align}
& \overline{v_{rel}^2}= \underbrace{\overline{v_2^2}}_{\overline{v^2}} +\underbrace{\overline{v_1^2}}_{\overline{v^2}} –
\underbrace{\overline{2\vec{v_1}\vec{v_2}}}_{=0} \\[5px]
\label{vrel}
& \overline{v_{rel}^2}= 2 \cdot \overline{v^2} \\[5px]
\end{align}

According to the Maxwell-Boltzmann speed distribution, the mean of the squared speeds \(\overline{v^2}\) is in a constant ratio to the square of the mean speed (this also applies to the relative speed, because a gas does not change its temperature just because it is described from the point of view of a gas molecule):

\begin{align}
\boxed{\frac{ \sqrt{\overline{v^2}} }{\overline{v}}=\sqrt{\frac{3\pi}{8}}}\\[5px]
\end{align}

\begin{align}
\frac{ \overline{v^2} }{\overline{v}^2}&=\frac{3\pi}{8}\\[5px]
\overline{v^2} &=\frac{3\pi}{8} \cdot \overline{v}^2 \\[5px]
\end{align}

Therefore, equation (\ref{vrel}) can also be expressed by the square of the mean speed. Hence, the speeds are related by the factor √2:

\begin{align}
\require{cancel}
\overline{v_{rel}^2}&= 2 \cdot \overline{v^2} \\[5px]
\cancel{\frac{3\pi}{8}} \cdot \overline{v_{rel}}^2 &= 2 \cdot \cancel{\frac{3\pi}{8}} \cdot \overline{v}^2 \\[5px]
\end{align}

\begin{align}
&\boxed{\overline{v_{rel}}= \sqrt{2} \cdot \overline{v}} \\[5px]
\end{align}

Conclusion

The relative speed of the molecules in a gas is thus on average greater by the factor √2 than the mean speeds. The actual approach speed of two particles is therefore greater by the factor √2 than if only their own speed were taken as a basis for the approach (in principle this would correspond to the case that all other particles were at rest).

The higher speeds when taking into account the relative motions therefore shortens the time between two collisions by the factor 1/√2. Compared to the situation in which all particles are at rest, the mean free path will therefore also be smaller by the factor 1/√2! For this reason the factor 1/√2 is introduced in equation (\ref{l}).

Introduction

For ideal gases, the distribution function f(v) of the speeds has already been explained in detail in the article Maxwell-Boltzmann distribution. The figure below shows the distribution function for different temperatures.

\begin{align}
\label{p}
&\boxed{  f(v) = \left(  \sqrt{\frac{m}{2 \pi k_B T}}  \right)^{3}  \cdot 4 \pi v^2 \cdot \exp{\left(- \frac{m \cdot v^2}{2 k_B \cdot T} \right)} } ~~~\text{Maxwell-Boltzmann distribution function} \\[5px]
\end{align}

The purpose of this article is to derive this distribution function.

Barometric formula

The barometric formula describes the course of atmospheric pressure p or air density ϱ as a function of altitude z above a reference level (e.g. sea level):

\begin{align}
\label{bar}
&\boxed{p(z) = p_0 \cdot \exp{\left(-\dfrac{\rho_0 g z}{p_0}\right)}} ~~~\text{barometric formula} \\[5px]
\end{align}

The barometric formula is not limited to air. It can be applied to any (ideal) gas exposed to a gravitational field. The ideal gas law shows the following relationship between the pressure p, the volume V, the number of particles N and the thermodynamic temperature T (k_B denotes the Boltzmann constant):

\begin{align}
&p V = N k_B T \\[5px]
\end{align}

The total mass of the gas m_gas (which occupies the volume V) can be determined from the product of particle mass m and number of particles N (m_gas=N⋅m). Thus the number of particles results from the quotient of gas mass and particle mass (N=m_gas/m). If this is taken into account in the equation above, then the gas density ϱ=m_gas/V can be calculated as follows:

\begin{align}
&p V = \frac{m_{gas}}{m} \cdot k_B T \\[5px]
&p = \frac{m_{gas}}{V} \cdot \frac{k_B T}{m} \\[5px]
\label{density}
&\boxed{p = \rho \cdot \frac{k_B T}{m}} ~~~\text{or}~~~\boxed{p_0 = \rho_0 \cdot \frac{k_B T}{m}} \\[5px]
\end{align}

In equation (\ref{density}), ϱ₀ denotes the gas density at the reference level and ϱ is the density associated with a pressure p. If equation (\ref{density}) is now applied in the barometric formula (\ref{bar}), then the following relationship results between the density at the reference level ϱ₀ and the density ϱ at an arbitrary height z:

\begin{align}
\require{cancel}
&p(z) = p_0 \cdot \exp{\left(-\dfrac{\rho_0 g z}{p_0}\right)} \\[5px]
&\rho(z) \cdot \bcancel{\frac{k_B T}{m}}= \rho_0 \cdot \bcancel{\frac{k_B T}{m}} \cdot \exp{\left(-\dfrac{\bcancel{\rho_0} g z}{ \bcancel{\rho_0} \cdot \frac{k_B T}{m} }\right)} \\[5px]
\label{rho}
&\rho(z)= \rho_0 \cdot \exp{\left(-\dfrac{m g z}{k_BT}\right)} \\[5px]
\end{align}

Since the mass m in equation (\ref{rho}) refers to a single molecule, the expression m⋅g⋅z can be interpreted as the potential energy W_pot) of a molecule at the height z:

\begin{align}
\label{rhoz}
&\rho(W_{pot})= \rho_0 \cdot \exp{\left(-\dfrac{W_{pot}}{k_BT}\right)} \\[5px]
\end{align}

At this point the question arises how the molecules get to their potential energy. First imagine the gas molecules at absolute zero. Due to the lack of brownian motion, all particles will be on the ground because of the influence of gravity (reference level).

Now one slowly increases the temperature by heating the gas, so that the motion of the molecules will increase. Sooner or later the molecules will collide with each other. As a result of the permanent collisions, some particles will catapult each other higher and higher. From an energetic point of view, however, this is nothing more than a supply of kinetic energy at the reference level which is subsequently converted into potential energy. We can conclude that the molecules reach their height due to their kinetic energy at the reference level (supplied by heat).

Model conception

The following model can be used to describe such a behavior of the gas molecules under the influence of gravity. One puts many balls into a vertical tube, which stands on a vibrating plate. The balls will move more or less strongly depending on the strength of the vibration, just as the gas particles move more or less strongly depending on the temperature. The balls are catapulted into the air, just as the gas molecules in reality do due to their collisions.

However, one will notice that there is no uniform distribution of the balls over the altitude of the tube, as is also the case with a real gas. This is due to the fact that not all balls are supplied with the same kinetic energy at the bottom of the tube. The balls collide with each other and therefore some are slowed down and others are accelerated. If e.g. an upward flying ball is hit by a faster ball, then the kinetic energy of the pushed ball will increase. This ball will now be able to reach higher heights. If it is pushed by chance by another ball, it will be able to fly even higher. For a few balls this might happen one or two more times. Such an ideal catapult effect will only be limited to a few balls. That’s why with increasing height less and less balls are to find.

With increasing altitude, the ball density (“number of balls per unit volume”) will decrease, because it becomes more and more improbable that a ball on the ground accidentally gets such a high kinetic energy in order to reach this height (whether there are collisions with other balls on the way up or not plays no role from an energetic point of view). It should be noted that in principle it is not possible to assign a certain number of balls to a certain height. For example, you will not find a single ball at a height of exactly 10 cm, since no ball will ever reach an exact height of 10 cm up to the “last” decimal place (possibly only 10.0003 cm or 9.9998 cm).

For a concrete number of balls you have to divide the height into small intervals Δz and determine the balls occurring in them. As an example, 1000 balls are considered. Each ball has a mass of 10 g. If now at a height between 10 cm and 11 cm 5 balls are registered on statistical average, then obviously 0.5% of the total balls have a potential energy between 10 mJ and 11 mJ (the particles come to a standstill at this height and the former kinetic energy at ground level has been completely converted into potential energy). This means that 0.5 % of the balls have a kinetic energy between 10 mJ and 11 mJ at the reference level. The probability that a randomly picked ball has a kinetic energy between 10 mJ and 11 mJ is therefore also 0.5 %. The frequency with which certain energies are present can therefore also be interpreted as a probability!

The frequency with which balls are detected within a certain altitude range corresponds to the probability with which certain kinetic energy ranges are present at reference level!

Instead of the interval of Δz = 1 cm, one could have chosen an interval of Δz = 2 cm (see figure below). In this case, balls at a height between 10 cm and 12 cm are considered. The probability would be high that double the number of balls would be found. After all, we are now looking at a space that is twice as large. Strictly speaking, this proportionality only applies to (infinitely) small interval widths, since the ball density changes with height. However, this example shows that for the determination of the frequency not only the ball density at a certain height is relevant, but also the interval width that is looked at!

The ball density in combination with the height interval observed at a certain height is a measure of the frequency or probability with which the corresponding kinetic energy ranges are present at the reference level!

What does “ball density in combination with the height interval” mean in concrete terms? As already explained, for small intervals there is a proportional relationship to the number of balls contained in that interval (usually expressed in percent as a relative frequency or probability):

\begin{align}
&\text{Frequency} ~\sim \text{“interval width”} \\[5px]
\end{align}

In addition, the ball density is anyway proportional to the number of balls, because a density twice as high means by definition a number of balls twice as high (per unit volume):

\begin{align}
&\text{Frequency} ~\sim \text{“ball density”} \\[5px]
\end{align}

The frequency with which the balls are found at a certain height z and within a certain interval is thus proportional to the product of the ball density and the interval width:

\begin{align}
\label{hau}
&\boxed{\text{Frequency} ~\sim \text{“ball density”} \times \text{“interval width”} } \\[5px]
\end{align}

This model conception shows that it is possible to deduce the distribution of the kinetic energies at the reference level from the density of the balls at different heights and the interval width. The speed distribution can then be determined from this. Since only the velocity component in the z-direction is relevant for reaching a certain height, the kinetic energies refer only to the speeds related to the z-direction:

\begin{align}
\label{wpot}
&W_{pot}=W_{kin,z}=\frac{1}{2}mv_z^2 \\[5px]
\end{align}

Transfer of the model concept to gases

The ball model described above can now be transferred to gas molecules in the Earth’s gravitational field and thus to the barometric formula. The gas density decreases with increasing height because fewer and fewer particles had sufficiently high kinetic energies on the reference level to reach the corresponding height.

The mass density at a certain height is a measure of the frequency with which certain speeds are present at the reference level!

If, for this purpose, equation (\ref{wpot}) is used in equation (\ref{rhoz}), the following relationship results:

\begin{align}
&\rho(W_{pot})= \rho_0 \cdot \exp{\left(-\dfrac{W_{pot}}{k_BT}\right)} ~~~\text{and}~~~ W_{pot}=W_{kin,z}=\tfrac{1}{2}mv_z^2~~~\text{:}\\[5px]
\label{rhov}
&\boxed{\rho(v_z)= \rho_0 \cdot \exp{\left(-\dfrac{mv_z^2}{2k_BT}\right)}} \\[5px]
\end{align}

In the same way as the density of a gas in the Earth’s gravitational field decreases exponentially with increasing altitude according to the barometric formula (\ref{bar}), the frequency (probability) of certain speeds decreases exponentially with the square of the speed according to equation (\ref{rhov}):

\begin{align}
&\text{Frequency} \sim \underbrace{\exp{\left(-\dfrac{mv_z^2}{2k_BT}\right)}}_{\text{Boltzmann factor}} \\[5px]
\end{align}

This factor, which describes the exponential decrease of the frequency with increasing speed (more generally: increasing energy), is also called the Boltzmann factor and plays a central role in statistical physics!

As the model of the balls made clear, a certain interval width must always be allowed for a concrete frequency. In the same way as one would not find a single ball at an exact given height, one would not find a single molecule with an exactly specified speed. One can therefore only ask how many molecules have a speed within a certain range Δv_z. As long as this speed interval Δv_z is chosen small enough, there is a proportional correlation to the frequency. If one allows an interval twice as large, then potentially twice as many particles can now lie within this doubled interval.

\begin{align}
&\text{Frequency} \sim \Delta v_z \\[5px]
\end{align}

Altogether, the frequency (probability) with which a certain speed range is present is thus proportional to the Boltzmann factor exp(-mv_z²/2k_BT) and to the interval width Δv_z:

\begin{align}
&\text{Frequency} \sim \underbrace{\exp{\left(-\dfrac{mv_z^2}{2k_BT}\right)}}_{\text{“Boltzmann-Faktor”}} \cdot \underbrace{~~~\Delta v_z~~~}_{\text{Intervallbreite}} \\[5px]
\end{align}

In order to obtain a formula for the actual calculation of the frequency, a formal proportionality factor f₀ can be introduced, so that the following equation applies, which describes the frequency of a speed in the range between v_z and v_z+Δv_z:

\begin{align}
\label{z}
&\boxed{\text{Frequency} = \underbrace{f_0 \cdot \exp{\left(-\dfrac{mv_z^2}{2k_BT}\right)}}_{\text{frequency density}} \cdot \underbrace{\Delta v_z}_{\text{speed interval}}} \\[5px]
\end{align}

The figure below graphically shows which velocity vectors are to be considered if a speed v_z and an interval Δz are given. All vectors whose arrowheads lie within the green marked area and thus with their z-components between v_z and v_z+Δz would be considered for the calculation of the frequency.

The figure below shows the qualitative course of the term f₀⋅exp(-mv_z²/2k_BT). This expression corresponds in the figurative sense to the particle density or ball density in the model conception. This function represents in this case a measure for the frequency with which certain speeds v_z are present. It is therefore called frequency density function or probability density function. The term “density” means the frequency or probability of a speed related to the corresponding speed interval:

\begin{align}
\label{norm}
&\text{frequency density} f(v_z) = \frac{\text{frequency}}{\text{speed interval}} = f_0 \cdot \exp{\left(-\dfrac{mv_z^2}{2k_BT}\right)} \\[5px]
\end{align}

Normalization of the function

Now it is necessary to determine the introduced proportionality factor in the frequency density function. This is achieved by a so-called normalization of the function. This normalization will be discussed in more detail in the following.

If equation (\ref{z}) is to be a relativ frequency (= probability!) with which a certain speed range between v_z and v_z+Δv_z is present, then the sum of the probabilities over all possible speed ranges must result in 100 %. Because if you finally sum up all occurring frequencies (probabilities) in the speed range between -∞ and +∞, then you will finally catch all molecules to 100 % (probability = 1), because any molecule must have a speed after all. Note that the molecules can have not only a positive velocity component in the z-direction, but also a negative one!

The calculation of the overall frequency (if all speeds in the range between plus and minus infinite are summed up) can be represented as follows:

\begin{align}
&\text{Overall frequency} = \sum_{-\infty}^{+\infty} f_0 \cdot \exp{\left(-\dfrac{mv_z^2}{2k_BT}\right)} \cdot \Delta v_z \overset{!}{=} 1 \\[5px]
\end{align}

Mathematically, this summation corresponds to integrating when infinitesimal speed ranges dv_z are considered instead of finite speed intervals Δv_z. With the constraint that the result must equal 1, the proportionality factor f₀ can finally be determined:

\begin{align}
&\text{Overall frequency} = \int_{-\infty}^{+\infty} \underbrace{f_0 \cdot \exp{\left(-\dfrac{mv_z^2}{2k_BT}\right)}}_{\text{frequency density function}} \cdot \text{d}v_z = \underline{f_0 \cdot \sqrt{\frac{2\pi k_B T}{m}} \overset{!}{=} 1} \\[5px]
&\boxed{f_0 = \sqrt{\frac{m}{2\pi k_B T}} }
\end{align}

The determination of the integral corresponds to the area under the graph of the frequency density function. This area was finally normalized to 1. This clearly shows that the area under the curve of the density function has the dimension of a frequency or probability!

With the factor f₀, the function for calculating the frequency is now finally normalized, i.e. fixed to the value 1, when integrating over all speeds! Thus, the frequency or probability of a speed in the range between v_z and v_z+dv_z can be calculated as follows:

\begin{align}
\label{dic}
&\boxed{\text{Frequency} = \underbrace{\sqrt{\frac{m}{2\pi k_B T}} \cdot \exp{\left(-\dfrac{mv_z^2}{2k_BT}\right)}}_{\text{frequency density} f(v_z)} \cdot \underbrace{~~~\text{d}v_z~~~}_{\text{speed interval}}} \\[5px]
\end{align}

Note that instead of macroscopic speed ranges Δv_z, infinitesimal speed intervals dv_z are now considered.

According to equation (\ref{norm}), the following function f finally applies to the frequency density or probability density:

\begin{align}
\label{f}
&\boxed{f(v_z) =\sqrt{\frac{m}{2\pi k_B T}} \cdot \exp{\left(-\dfrac{mv_z^2}{2k_BT}\right)} } ~~~\text{Frequency density function}
\end{align}

To calculate a concrete frequency F with which a speed occurs in the range between v_z1 and v_z2, the frequency density function f(v_z) must be integrated within these limits:

\begin{align}
& \boxed{\text{Frequency } F=\int_{v_{z1}}^{v_{z2}} f(v_z) ~~ \text{d} v_z} \\[5px]
\end{align}

The figure below shows the frequency density f(v_z) as a function of velocity v_z for different temperatures. Note that an area under the curve as the integral of the density function represents the frequency or probability with which the corresponding speed range is present!

Frequency density function in three dimensions

So far, the speed distribution is limited only to the velocity component of the particles in z-direction. Much more interesting however, is the distribution of the overall speeds of the molecules. For this, the frequency density function (\ref{f}) must be extended to three dimensions.

In the following, an ideal gas is considered that behaves equally in all three spatial directions. This means that the gas molecules have no direction in which they prefer to move. In particular the influence of gravity is neglected! Such a gas, that behaves the same in all directions, is also called an isotropic gas.

The velocity vector

Generally, gas molecules have velocity components in all three directions (v_x, v_y and v_z). The velocity vector v of a gas particle can be represented by these three components:

\begin{align}
\vec{v} &= \begin{pmatrix}
v_{x} \\
v_{y} \\
v_{z}
\end{pmatrix}
\end{align}

For the magnitude of the velocity vector |v| or for the square |v|² applies:

\begin{align}
&|\vec{v}| = \sqrt{v_x^2+v_y^2+v_z^2} \\[5px]
\label{v22}
&|\vec{v}|^2 = v_x^2+v_y^2+v_z^2\\[5px]
\end{align}

Note that when squaring the magnitude of an vector (as done in equation (\ref{v22})), the bars that indicate the magnitude can be omitted. This leads to the same result:

\begin{align}
&\vec{v}^2 = \begin{pmatrix}
v_{x} \\
v_{y} \\
v_{z}
\end{pmatrix}
\begin{pmatrix}
v_{x} \\
v_{y} \\
v_{z}
\end{pmatrix} = v_x^2+v_y^2+v_z^2 = |\vec{v}|^2 \\[5px]
\label{v2}
&\boxed{\vec{v}^2 = v_x^2+v_y^2+v_z^2}
\end{align}

The square of a vector is therefore a scalar. We can conclude from this that by squaring a vector the information about the direction is lost. This fact will become important later!

Frequency of occurrence of a certain velocity vector

Equation (\ref{dic}) describes the frequency with which a velocity component in z-direction is present within an interval v_z and v_z+dv_z. Since an isotropic gas is assumed, this frequency distribution applies equally to each velocity component i=x,y,z:

\begin{align}
\label{dicc}
&\text{Frequency}(v_i) = \sqrt{\frac{m}{2\pi k_B T}} \cdot \exp{\left(-\dfrac{mv_i^2}{2k_BT}\right)} \cdot \text{d}v_i ~~~~~\text{for } i=x,y,z\\[5px]
\end{align}

For the velocity distribution, the question arises how often a certain vector v with given components v_x, v_y and v_z occurs within given speed intervals dv_x, dv_y and dv_z.

A simple example shows the basic procedure. In this context, it makes more sense to interpret the frequency of the occurrence of a velocity as the probability of the occurrence. We now assume that among all particles the x-component of a given velocity range occurs with a probability of 10 % (0.1) and the y-component with a probability of 5 % (0.05) and the z-component with a probability of 20 % (0.2). The overall probability that a particle has all three given velocity ranges at the same time results then from the product of the individual probabilities. In this case, the total probability is 0.1 % (0.001 = 0.1 × 0.05 × 0.2).

For the overall probability or the frequency of a certain velocity vector v (within the intervals dv_x, dv_y and dv_z equation (\ref{dicc}) must be multiplied for all three velocity components i=x,y,z:

\begin{align}
&\text{Frequency}(\vec{v}) =\text{Frequency}(v_x) \cdot \text{Frequency}(v_y) \cdot \text{Frequency}(v_z) \\[5px]
&\text{Frequency}(\vec{v}) = \left(\sqrt{\frac{m}{2\pi k_B T}}\right)^3 \cdot \exp{\left(-\dfrac{m(v_x^2+v_y^2+v_z^2)}{2k_BT}\right)} \cdot \text{d}v_x \cdot \text{d}v_x \cdot \text{d}v_x \\[5px]
\end{align}

Note that when multiplying exponential functions, the individual exponents can be summated. Therefore, the exponent now contains the sum of the squared velocity components! According to the equation (\ref{v2}), this corresponds to the square of the velocity vector:

\begin{align}
&\text{Frequency}(\vec{v}) = \left(\sqrt{\frac{m}{2\pi k_B T}}\right)^3 \cdot \exp{\left(-\dfrac{m\vec{v}^2}{2k_BT}\right)} \cdot \text{d}v_x \cdot \text{d}v_x \cdot \text{d}v_x \\[5px]
\end{align}

Frequency of occurrence of a certain speed

How is the equation above to be interpreted? If one gives a velocity vector v and for each velocity component a certain range dv_i within which the individual components can vary, then the frequency (probability) with which all the possible velocity vectors are present is determined by that equation. The heads of the velocity vectors to which the frequency is related all lie within the “volume” spanned by dv_x, dv_y and dv_z (see Figure: Interpretation of the speed interval as a spherical shell).

However, the equation also shows that the information about the direction of a velocity vector is lost on the right side of the equation. Only the square of the velocity vector appears and – as already mentioned – this is a scalar! All velocity vectors that have the same magnitude thus occur with the same frequency (since all these vectors have the same scalar value v²).

This is a good thing, because in the end it is assumed that the direction of the velocity does not play a role in the frequency distribution! Finally, an isotropic gas was assumed, in which the probabilities for the occurrence of certain velocities should not depend on the direction! For the frequency distribution the direction of the velocity is not decisive but only the speed!

The arrow symbol that usually denotes a vector can therefore be omitted in the equation above, so that v only refers to the magnitude of the velocity (molecular speed of a molecule):

\begin{align}
\label{g}
&\text{Frequencty }v = \underbrace{\left(\sqrt{\frac{m}{2\pi k_B T}}\right)^3 \cdot \exp{\left(-\dfrac{mv^2}{2k_BT}\right)}}_{\text{frequency density}} \cdot \underbrace{\text{d}v_x \cdot \text{d}v_x \cdot \text{d}v_x}_{\text{three-dimensional speed interval}} \\[5px]
\end{align}

Instead of specifying certain intervals of dv_x, dv_y and dv_z for each velocity component, it would make sense to specify a single interval dv for the speed. For this purpose a relationship between these individual intervals must be found.

Therefore equation (\ref{g}) is interpreted graphically. If a speed v is given and intervals dv_x, dv_y and dv_z are allowed, within which the components of all possible velocity vectors can vary, then a three-dimensional spherical shell results. The thickness of this spherical shell corresponds to the interval dv within which the given speed v can vary.

Thus, the three-dimensional speed interval in equation (\ref{g}) (defined by dv_x, dv_y and dv_z) can be replaced by a spherical shell with radius v and thickness dv. This shell is limited by two concentric spheres. Due to the infinitesimal distance between the two spheres, the volume of the sphere (volume in a velocity space!) can be determined from the product of the area of the sphere 4π⋅v² and the distance dv between the spherical shells:

\begin{align}
&\text{three-dimensional speed interval} = 4\pi v^2 \cdot \text{d}v \\[5px]
\end{align}

If this three-dimensional speed interval is used in equation (\ref{g}), then the following equation applies:

\begin{align}
&\text{Frequency} =\left(\sqrt{\frac{m}{2\pi k_B T}}\right)^3 \cdot \exp{\left(-\dfrac{mv^2}{2k_BT}\right)} \cdot \overbrace{\text{d}v_x \cdot \text{d}v_x \cdot \text{d}v_x}^{4\pi v^2 \cdot \text{d}v } \\[5px]
&\boxed{\text{Frequency} = \underbrace{\left(\sqrt{\frac{m}{2\pi k_B T}}\right)^3 \cdot 4\pi v^2 \cdot \exp{\left(-\dfrac{mv^2}{2k_BT}\right)}}_{\text{frequency density }f(v)} \cdot \underbrace{\text{d}v}_{\text{speed interval}}} \\[5px]
\end{align}

This equation finally describes the frequency or probability with which a certain speed v is present within the speed interval dv. As already explained for equation (\ref{z}), a density function f(v) can be defined at this point, which represents a measure for the frequency of an existing speed related to the speed interval dv. This distribution function is finally called Maxwell-Boltzmann distribution:

\begin{align}
\label{max}
&\boxed{f(v) = \left(\sqrt{\frac{m}{2\pi k_B T}}\right)^3 \cdot 4\pi v^2 \cdot \exp{\left(-\dfrac{mv^2}{2k_BT}\right)}}~~~~~\text{Maxwell-Boltzmann distribution} \\[5px]
\end{align}

For the calculation of a specific frequency F with which a speed occurs in the range between v₁ and v₂, the frequency density function f(v) must be integrated within these limits:

\begin{align}
& \boxed{\text{Frequency } F=\int_{v_{1}}^{v_{2}} f(v) ~~ \text{d} v} \\[5px]
\end{align}

The figure below shows the course of the Maxwell-Boltzmann distribution f(v) as a function of speed v for different temperatures. Note that the area under the curve (as the integral of the density function) represents the frequency or probability with which the corresponding speed range is present!

More detailed explanations on the statement of this distribution function can be found in the article “Maxwell-Boltzmann distribution“.

The apparent contradiction

If one compares the Maxwell-Boltzmann distribution with the distribution of the velocity components, an apparent contradiction appears at first glance. How can the frequency density of the individual components have a maximum at v=0, while in the Maxwell-Boltzmann distribution there is a minimum at v=0 (more precisely, the frequency density is zero). If there is a high probability that the velocity components are zero, shouldn’t the probability for a speed of zero also be quite high?

At this point it is important to interpret the diagrams correctly. The diagrams do not show the frequency or probability but the frequency density or probability density. For an interpretation whether a velocity component occurs rarely or frequently, the speed interval must also be considered (i.e. the area under the curve!).

The following two-dimensional example is intended to illustrate this and resolve the apparent contradiction. A relatively low speed v (see figure on the left) and a relatively high speed v (see figure on the right) are considered. For both cases an equally large speed interval dv is given, within which the speed can vary. This interval is graphically represented by a spherical shell around the velocity vector v.

Question: In which speed range can more particles be found; in the one with the low speed (left side of the figure) or in the one with the higher speed (right side of the figure)?

According to the argumentation with the frequency distribution of the velocity components, one would probably conclude that low velocity components are more likely than large ones. Therefore, there should be more particles with lower speeds than particles with higher speeds (in this argument the frequency density is mistakenly considered as frequency).

However, if one looks at the graphic representation of the two cases, the opposite becomes apparent. In order for the given speeds v to lie within the permissible ranges dv, the x-components v_x can vary to a certain degree. Graphically, a velocity vector v+dv whose arrowhead is located on the outer spherical shell would just be permissible (shown as a green arrow). The corresponding velocity component v_x is represented by a red arrow.

As the comparison shows, at a high speed v the x-component can obviously vary within a much larger range without exceeding the speed range dv. This means that considerably more x-components are possible. The frequency distribution of the velocity components shows how often these occur in concrete terms. However, the area under the curve must be considered and not the frequency density itself (note that the area under the curve indicates the concrete frequency)! So you have to interpret the diagrams correctly and above all you must not equate the frequency density with a frequency!

In this case, the higher speed v is therefore present much more frequently (since the permissible x-components are represented more frequently) than the lower speed. From a mathematical point of view, this is due to the quadratic influence of the speed in the Maxwell-Boltzmann distribution function (which is caused by the “spherical shell” 4π⋅v²)!

Now it also becomes clear why an extremely low speed of v≈0 is almost non-existent according to the Maxwell-Boltzmann distribution. This would mean that the velocity components would also vary only in an extremely small range. The frequency that such a small velocity range is present is extremely low (very small area under the frequency density function of the components).

At this point, the question arises why the Maxwell-Boltzmann distribution flattens out at all if, as argued above, higher speeds should be present more often than lower ones. This argumentation is of course only valid to a certain degree. As the frequency distribution of the individual components also shows, the frequency density decreases with increasing speed. This means that too high speeds are not strongly represented anyway due to the limited frequency. Mathematically, this is due to the exponential influence of the speed in the Maxwell-Boltzmann distribution function (see red dotted line in the figure above)!

Equipartition theorem

In the article Pressure and temperature the following equation was derived to calculate the pressure p exerted by an ideal gas:

\begin{align}
\label{druck}
& \boxed{p = 2 \cdot \frac{N}{V} \cdot \overline{W_{kin,x}}} \\[5px]
\end{align}

In this equation, N denotes the number of particles in a gas volume V (e.g. in a cylinder). W_kin,x refers to the mean kinetic energy with which the molecules collide with a boundary surface and thereby exert the pressure p (e.g. on a piston). The kinetic energy refers only to those velocity components with which the particles actually hit the boundary surface (here: the velocity components along the x-direction). Speed components directed perpendicular to the surface (in y- and z-direction) do not lead to an impulse on the piston and are therefore not relevant for the pressure on the piston.

At first glance, the pressure given by the equation (\ref{druck}) refers only to the pressure exerted by the gas in the x-direction. For the other spatial directions a pressure can be defined analogously, which is connected with the kinetic energies in the corresponding directions:

\begin{align}
\label{pi}
& p_i = 2 \cdot \frac{N}{V} \cdot \overline{W_{kin,i}} ~~~~~i=x,y,z \\[5px]
\end{align}

However, practice shows that the pressure in gases is equally present in all spatial directions. To verify this, the cylinder sealed with a piston can be equipped with three pressure gauges. These are mounted at right angles to each other on the cylinder, so that the pressure exerted by the gas in the various spatial directions can be measured separately. In fact, you will find that the pressure in all three directions is identical.

Now the gas in the cylinder is compressed by suddenly pushing the piston into the cylinder. The moving piston then hits the oncoming molecules in the gas with great force. On a microscopic level, this process is similar to hitting an oncoming ball with a racket. The moving piston thus increases the speed of the molecules in the (negative) x-direction. This also increases the mean kinetic energy that is related to the x-direction (which also explains the increase in temperature during compression).

Consequently, the compression should only increase the pressure in the x-direction and the pressures in the y- and z-direction should remain unaffected. However, this will not be found in practice! Experience shows that even with such a compression the pressure will increase equally in all spatial directions. According to the equation (\ref{pi}), the kinetic energy must have increased equally along the three spatial directions. Obviously the supplied energy during compression must have been divided equally among all three spatial directions. This phenomenon is also called equipartition theorem.

The equipartition theorem in connection with ideal gases states that the kinetic energy of the particles (or the gas) is equally divided along all three spatial directions and thus the pressure in all spatial directions is identical!

The equipartition theorem is a direct consequence of the random, statistical motion of the particles. These “chaotic” movements of the particles leads to continuous collisions in which the particles subsequently move in different directions. In this way, the originally ordered motions quickly become disordered motions in which no direction is preferred and thus an equal distribution of speed and energy takes place.

The equipartition theorem is not limited to translational motion but applies to all forms of energy at the atomic level. The next section deals with this in more detail.

Distribution of energy among degrees of freedom

In the article “Pressure and temperature” it was shown that the mean kinetic energy of a particle Wkin (now no longer limited to one direction but to be understood as “total” kinetic energy of a molecule!) is related to the thermodynamic temperature T as follows:

\begin{align}
\label{kin}
& \boxed{\overline{W_{kin}} = \frac{3}{2} k_B \cdot T} \\[5px]
\end{align}

For the overall energy of the ideal gas with a total of N particles, the following internal energy applies:

\begin{align}
\label{u}
& \boxed{U = \frac{3}{2} N k_B T} \\[5px]
\end{align}

Translation

The three dimensions in which the molecules of an ideal gas can move ultimately stand for the possibilities the gas has to store energy. A particle can put its energy into the motion along the x-direction as well as into the motion in y- or z-direction. These possibilities to store energy are also called degrees of freedom.

In thermodynamics, degrees of freedom are the number of possibilities to store energy at the atomic level!

A molecule of an ideal gas thus has a total of three degrees of freedom – for each spatial direction a possibility to store energy in the form of a translational motion. For each degree of freedom f (i.e. for each spatial direction) an energy of ½⋅k_B⋅T is assigned for a particle or for the entire gas an (internal) energy of ½⋅N⋅k_B⋅T per degree of freedom.

\begin{align}
\label{frei}
&\boxed{W = \frac{f}{2} k_BT} ~~\text{energy of a particle}\\[5px]
\label{inn}
&\boxed{U = \frac{f}{2}Nk_BT} ~~\text{internal energy of the gas}\\[5px]
\end{align}

Rotation

Whereas only three degrees of freedom in terms of the three spatial dimensions are possible for translational motion in monatomic gases, rotational motion can also occur in molecules with two atoms (or more generally: linear molecules). This theoretically results in three further possibilities to store energy: in the form of rotation around the x-axis, the y-axis and the z-axis. Theoretically in this case you get a total of f = 6 degrees of freedom.

One will also notice an equal distribution among these individual possibilities to store energy, regardless of whether it is a translational or rotational motion! Imagine that the linear molecules in a gas are initially only given a translational motion in one direction, but then sooner or later the molecules will begin to rotate due to the collisions between the molecules. Due to the random motion, the originally purely translational kinetic energy will gradually be converted into rotational energy and divided equally among all degrees of freedom. For example, the translational kinetic energy in the x-direction will then be the same as the angular kinetic energy around the y-axis.

The equipartition theorem states that the total energy is divided equally among each degree of freedom!

Note that the temperature is determined exclusively by the translational kinetic energy of the particles, regardless of whether the molecules rotate or not! Thus, at a given temperature, the same amount of energy will always be attributed to the translational motion, which is determined by equation (\ref{kin}). According to the equipartition theorem, the same amount of energy must then be in the rotational motion, as long as the molecules can rotate! Therefore, equation (\ref{frei}) also applies to each degree of freedom of a rotational motion. Analogously this applies to equation (\ref{inn}). A gas with linear molecules with a total of 6 degrees of freedom at the same temperature thus contains twice the (internal) energy compared to a monatomic gas with only three degrees of freedom (assuming the same number of particles).

Note that, for quantum mechanical reasons, not all theoretically possible degrees of freedom can actually be used. For example, with molecular hydrogen (dihydrogen), a very high energy is required for the hydrogen molecule to rotate around the molecular axis (rotation around the z-axis in the figure above). At “normal” temperatures, the energy stored in the gas is usually not sufficient to cause the molecule to rotate around this axis. The use of this rotational energy is therefore only theoretical for the time being. One also speaks of a so-called “frozen” degree of freedom, since this is only actually present at high thermal energies or temperatures. The hydrogen molecule, like many other diatomic molecules, therefore only effectively has five degrees of freedom.

Note: In many cases, the atoms are assumed to be mass points, including the individual atoms of linear molecules. In this case, no energy can be stored around the molecular axis anyway, since effectively no mass is in rotation (no moment of inertia)! In this model, such a two-atom molecule then has only two degrees of freedom for rotation anyway! Therefore in most cases we speak of two degrees of freedom for rotation and not of three. The consideration as mass points also makes sense insofar as one would otherwise have had to take rotational energies into account even for monatomic gases (after all, spherical atoms can also rotate and thus store energy). In addition, practice shows that in most cases only two degrees of freedom are effectively available anyway for the rotational movement of diatomic molecules.

Vibration

Diatomic molecules (dumbbell molecules)

In the case of diatomic molecules, it must be noted that there are generally also binding energies between the particles, which act similarly to an elastic spring. Due to these elastic bonding forces, the molecule can also oscillate (vibrate) along the molecular axis. One might think that this is only one degree of freedom. In reality, however, vibrations contain both potential energies and kinetic energies. For this reason, two degrees of freedom always apply to possible directions of oscillation.

Triatomic linear molecules

Triatomic linear molecules, for example, can vibrate in four different ways, so that eight degrees of freedom effectively result. Such vibrational degrees of freedom are also referred to as normal modes of vibration.

One normal mode of vibration results from the symmetrical stretching of the two outer atoms (red) along the molecular axis around the centrally located atom (blue). In addition, the molecule can also perform an asymmetric stretching motion by all three atoms vibrating along the molecular axis. The third and fourth vibrational mode result from a so-called bending mode (also referred to as scissoring). A distinction is made in which plane (vertical or horizontal) the vibrational bending motion takes place.

Triatomic non-linear molecules

For triatomic non-linear molecules, there are generally three vibrational degrees of freedom. At the one hand, again there is a symmetrical stretching of the outer atoms (red) along their molecular axis. At the other hand, again exists an asymmetrical stretching, where atoms vibrate phase-shifted. The third normal mode of vibration is again a bending (scissoring) mode, where the outer atoms vibrate in a kind of scissoring motion within the molecular plane (i.e. the bond angle oscillates around a resting position).

Determination of the number of normal modes of vibration

To determine the position of an atom within a molecule, one needs three coordinates (x-, y- and z-coordinates). A molecule with n atoms thus needs 3⋅n coordinates in total, by which it is then unambiguously defined at a certain point in time. This number of parameters corresponds to the total number of degrees of freedom f_tot that a molecule has in principle:

\begin{align}
&\boxed{f_{tot} = 3 n} \\[5px]
\end{align}

These degrees of freedom can be divided into translation (f_trans), rotation (f_rot) and vibration (f_vib):

\begin{align}
&f_{tot} = 3 n = f_{trans} + f_{rot} + f_{vib} \\[5px]
\end{align}

If the number of degrees of freedom of translation and the number of degrees of freedom of rotation are known, then the degrees of freedom of vibration (normal modes of vibration) can be determined as follows:

\begin{align}
\label{xx}
&\boxed{f_{vib} = 3 n – f_{trans} – f_{rot}} \\[5px]
\end{align}

Summary

It must be noted that for the number of degrees of freedom f that are relevant for the internal energy, the total degrees of freedom f_tot are not essential, since from an energetic point of view the degrees of freedom of vibration must be counted twice!

If one divides the degrees of freedom between the translational motion (f_trans), the rotational motion (f_rot) and the vibrational motion (f_vib), then the total number of degrees of freedom relevant for the internal energy is:

\begin{align}
\label{yy}
&\boxed{f = f_{trans} + f_{rot} + 2 \cdot f_{vib}} \\[5px]
\end{align}

If at this point equation (\ref{xx}) is used in equation (\ref{yy}), then the “energetic” degrees of freedom f can also be determined by the number of atoms n of a molecule:

\begin{align}
&f = f_{trans} + f_{rot} +2 \cdot \overbrace{ \left(3 n – f_{trans} – f_{rot} \right)}^{f_{vib} } \\[5px]
&\boxed{f =6n- f_{trans} – f_{rot}} \\[5px]
\end{align}

The table below shows the number of degrees of freedom for monatomic, diatomic and triatomic (non-linear) molecules.

Introduction

As already explained in the article Temperature and particle motion, the temperature of a gas is a measure of the kinetic energy of the particles it contains. Even at a constant temperature, however, not all the particles have the same speed. After all, in a gas at the atomic level there are permanent collisions between the particles. Some particles are slowed down and others are accelerated by the collisions. Thus, particles with different velocities can be found in a gas.

The question arises how the speeds of the particles in a gas can be determined experimentally in order to then make a statement about the velocity distribution. In the following, an experiment will be explained in more detail and its result will be discussed.

Experimental setup

To measure the velocity of particles in a gas, a substance is first evaporated in an effusion oven and heated to a constant temperature. The gas molecules can pass through a hole in the oven in different directions. A particle beam is then generated with the use of two apertures (also called collimator). In this molecular beam, the particles move in a common direction, but at different speeds.

The speed distribution of the particles in the unidirectional stream is representative of the speed distribution of the entire gas, even if same particles were filtered out by the collimator. Ultimately, there is no direction in which the particles prefer to move in the gas. Therefore, the speed distribution in any direction is representative for the entire gas.

The molecular beam is now directed onto a rotating drum. At the circumference of this drum, several helical grooves are milled in axial direction (analogous to the threads of screws). Only particles whose speeds are within a certain range pass through the slotted drum at a given rotational speed. Particles that are too fast will hit the left side of the groove. If particles are too slow, they will collide with the right side of the groove.

The speed of interest can be controlled by varying the rotational speed of the drum. If the speed of the drum is high, only particles with a higher speed will pass through the apparatus. At low rotational speed, however, only particles with low speed will pass through. This experimental setup of the slotted discs thus serves as a velocity selector (speed filter). So that the gas particles to be measured are not influenced in their velocity by collisions with air molecules, the entire apparatus must be in a vacuum.

In order to obtain the distribution of the different speeds, the respective number of particles passing through the speed filter must be determined at different speeds. This is achieved with a particle detector that measures the frequency of impact.

Note: Due to the finite size of the groove width, the speed of the particles passing through the experimental setup may vary within certain limits. Therefore, it is not possible to measure the exact speeds of the gas particles, but only to analyze speed ranges. However, this is quite sufficient for the representation of a speed distribution, as will be shown later.

Due to the measuring method only velocity ranges can be assigned to a concrete number of particles!

Experimental Results

A gas usually contains innumerable particles. To better illustrate the speed distribution, we will therefore assume a number of 1000 gas particles in the following. Helium is assumed to be the gas, at a temperature of 273 K (0 °C). The speed intervals to be investigated are 500 m/s each. In this case, the following statistical distribution would typically result:

• 68 particles would have a speed in the range between 0 m/s and 500 m/s,
• 309 particles would be in the speed range between 500 m/s and 1000 m/s,
• 358 particles would be measured in the interval between 1000 m/s and 1500 m/s,
• 195 particles would have a speed in the interval between 1500 m/s and 2000 m/s,
• 59 particles would be in the speed range between 2000 m/s and 2500 m/s,
• 10 particles would have a speed in the range between 2500 m/s and 3000 m/s,
• 1 particle would show a speed greater than 3000 m/s.

Experimental evaluation

Histogram of the speed distribution

This measurement results can be evaluated in a diagram that lists the speed on the horizontal axis and the number of particles on the vertical axis. In such a diagram, the height of a bar represents the number of particles whose speeds lie within the corresponding width of the bar.

Such a graphical representation of a frequency distribution within so-called bins* is also called a histogram.

*) in the present case, the different speed intervals serve as the bins.

The histogram gives a very clear insight into the velocity distribution, but there is no generalization to any velocity intervals. With this diagram form, for example, it is not possible to deduce the number of molecules in the speed range between 1300 m/s and 1600 m/s. Based on the histogram, another form of representation is therefore used. This will be discussed in more detail in the next section.

Dependence of the histogram on the interval width

For a meaningful and more general representation of the speed distribution, it must be noted that it is in principle impossible to directly assign a specific number of particles to a certain speed. This is because the observed velocity of a particle will never correspond to the given value up to the last decimal place. After all, you would not find a single particle that exactly has this given velocity. Even on the basis of the experimental setup, it is not possible to draw conclusions about the exact speed of the gas particles anyway, but only about speed ranges (due to the finite size of the slots).

For a higher resolution of the speed distribution, the slots in the disks could be reduced in size. This would also reduce the speed range to be filtered. This then gives a more detailed picture of how many molecules are moving in a certain speed range. The figure below shows the histogram at a velocity interval of Δv = 250 m/s and Δv = 125 m/s.

The smaller the speed interval Δv is chosen, the “flatter” the diagram will be, since the particles are divided into smaller speed ranges. For small speed intervals there is a proportional correlation between the speed interval Δv and the number of particles n in it:

\begin{align}
&n \sim \Delta v ~~~~~~~~~\text{(only valid for small speed intervals)} \\[5px]
\end{align}

The height of the bars in the histogram thus decreases by half each time the speed interval is halved. This also becomes clear, because if the interval is halved, only half of the particles in the original interval will have a higher speed and the other half a lower speed.

For small speed intervals, the number of particles within a certain speed interval (“height of the bar”) is proportional to the chosen speed interval (“width of the bar”)!

Absolute frequency distribution

For a more general representation of the speed distribution one can now use exactly this fact that speed interval and the number of particles in it are proportional to each other. This is because the quotient of the number of particles and the speed interval is then constant and no longer dependent on the chosen speed interval itself:

\begin{align}
&\frac{n}{\Delta v } = \text{constant} ~~~~\text{(frequency density)} \\[5px]
\end{align}

This quantity is also called (absolute) frequency density (unit: s/m). The animation below shows the respective histograms for different speed intervals. It becomes clear that for smaller intervals the diagram becomes smoother and smoother. For infinitely small interval widths, the result is a continuous curve that is independent of the speed intervall itself.

So it is not necessary to measure the exact velocities of the individual gas particles, a small enough speed interval is sufficient to determine the speed distribution!

In such a diagram the height of the bar no longer corresponds to the number of particles but to the area under the curve! In the figure below, the area marked in red correspond to a number of 185 particles with a speed between 1300 m/s and 1600 m/s. The area below the entire curve in the speed range between 0 and ∞ would finally correspond to 1000 particles, since all the particles would be recorded.

The area under the frequency distribution corresponds to the number of particles in the respective speed range!

Note that the term density in connection with the frequency density is not related to a volume or an area but to the speed! A frequency density of 5 s/m, for example, means that per 1 m/s speed interval 5 particles are found. It has to be considered that the frequency density changes with the speed. Therefore this statement is only valid for speeds close to this point.

Relative frequency distribution

The representation of the frequency distribution in the figure above is valid in this form only for a total of 1000 particles. However, the qualitative distribution would also be the same for 1 million particles. The diagram would only be stretched in height. For a general representation of the speed distribution it is therefore not practicable to plot the absolute number of particles.

The frequency distribution is usually given in percent to be independent from the total number of particles. In this case one no longer speaks of the absolute frequency distribution but of the relative frequency distribution. The size on the vertical axis is called relative frequency density.

In such a diagram, the area under the graph corresponds to the percentage of molecules in the respective speed range (relative frequency). In the figure above the red marked area would correspond to 18,5 % of molecules with a speed between 1300 m/s and 1600 m/s. The area below the entire curve in the speed range between 0 and ∞ would ultimately correspond to 1 (≙ 100 %), since all the particles would be recorded.

The area under the graph corresponds to the percentage of particles in the respective speed range!

Influence of temperature on speed distribution

The figure below shows the influence of temperature on the speed distribution. For higher temperatures the speed distribution is stretched in length and squeezed in height. This results in a broader speed distribution for high temperatures, with correspondingly higher speed proportions.

As the temperature rises, the proportion of particles with higher speeds increases!

This fact matches the statement already made in the article Temperature and particle motion that the temperature of a substance is a measure of the kinetic energy of the particles contained therein. The higher the temperature, the more energy the particles have and the faster they move. In contrast to temperature, the gas pressure has no influence on the velocity distribution (at least for an ideal gas).

The speed distribution of an ideal gas is independent of the gas pressure!

The physicists James Clerk Maxwell and Ludwig Boltzmann attempted to derive such a speed distribution on the basis of the kinetic theory of gases using statistical methods. In 1860, the two physicists finally succeeded. Therefore, such a speed distribution as shown above is also called Maxwell-Boltzmann speed distribution. The article Maxwell-Boltzmann distribution deals with this in more detail.

Introduction

In order to connect the macroscopically observed state variables of a gas such as temperature, volume and pressure with the microscopic variables such as particle mass and particle velocity, the kinetic theory of gases was developed. With its help it is possible, for example, to deduce the temperature or the pressure of a gas from the mean kinetic energy of the molecules.

For (ideal) gases, the kinetic theory of gases provides important relationships between macroscopically measurable state variables (e.g. temperature, pressure, volume, gas mass, etc.) and microscopic variables (e.g. particle velocity, mean kinetic energy, number of particles, partial mass, etc.)!

Assumptions

In order to develop a model of the behaviour of gas particles, some assumptions must first be made about the properties of gases or the molecules they contain. First, it is assumed that the gases are ideal gases. This means in particular:

1. the gas particles are considered as mass points,
2. the gas particles do not exert any binding forces on each other,
3. collisions between gas particles are completely elastic as well as collisions between molecules and surfaces (i.e. no loss of energy)
4. all gas particles move completely randomly, i.e. they have no preferred direction and are therefore statistically distributed in space (i.e. the influence of gravity on the molecules is neglected).

Microscopic interpretation of the gas pressure

Formation of the gas pressure

In the article “Gas pressure“, the formation of the gas pressure has already been explained in detail using the particle model. The macroscopically measurable gas pressure (“force per unit of surface area”) can be explained at a microscopic level by means of collisions. If the gas particles collide with a surface (e.g. the wall of a container), they exert forces similar to tennis balls thrown against a racket.

The pressure in gases is caused on a microscopic level by collisions of the particles contained therein, which collide with adjacent surfaces and thus exert impact forces!

If, for example, a gas under high pressure is enclosed in a cylinder, the particles contained in it collide constantly with the cylinder wall and exert forces. These impact forces can be clearly felt when, the cylinder is closed with a piston.

The animation below schematically shows the collisions between the gas molecules and the piston surface. In contrast to the animation, due to the large number of particles normally contained in a gas, one will not feel a “hammering” of the particles but will perceive a constant force.

The force F required to keep the piston in position is solely due to collisions between the gas particles and the piston surface A, i.e. the gas pressure! From the definition of pressure as “force per unit surface area”, the gas pressure p can finally be determined as follows:

\begin{align}
\label{p}
&\boxed{p = \frac{F}{A}} \\[5px]
\end{align}

Note: Not only the molecules inside the cylinder collide with the piston surface and thus exert an outward force. The surrounding air outside the cylinder also contains particles. These particles therefore additionally exert an opposite force on the piston, which is directed inwards. Only the difference between the two forces corresponds to the force required to hold the piston in position. For the sake of simplicity, a vacuum outside the cylinder is assumed in the following, so that the force F in the equation (\ref{p}) can be attributed exclusively to the gas particles inside the cylinder.

Variables influencing the gas pressure

The gas pressure in the cylinder will be higher the more particles collide with the piston within a certain time. This depends on the number of particles in the cylinder. This is because the more particles there are in total, the more can collide with the piston surface and will exert (impact) forces. This is already shown by everyday experience when inflating a bicycle tyre: the more air is pumped into the tyre (i.e. the more particles it contains), the greater the pressure!

On the other hand, the gas pressure depends on the speed at which the particles hit the piston surface. The greater the speed, the more intense the collisions and the greater the impact forces or the associated pressure. This, too, is shown by everyday experience: If a bottle filled with air is placed in the sun, the pressure in the bottle will increase as the temperature rises, since the speed of the molecules increases with higher temperature.

Calculation of the gas pressure

In order to determine the gas pressure in the cylinder, the number of particles that collide with the piston within a certain time Δt must first be determined. Only those molecules can hit the piston which on the one hand actually fly towards the piston and on the other hand are close enough to the piston surface.

In general, the velocity of a particle consists of three components (x-, y- and z-components), whereby only the velocity component with which the particle hits the piston surface is relevant for the formation of the pressure, i.e. the speed in the x-direction.

First, it is assumed that the x-component of the velocity is identical for all particles (|v_x|). Since no direction is preferred by the particles, half of all particles will move at this speed |v_x| in the positive x-direction and the other half in the negative x-direction.

However, not all particles moving in positive direction can actually reach the surface of the piston within the given time Δt. With the assumed constant velocity |v_x| the particles can only cover a maximum distance Δs within the time Δt:

\begin{align}
&\Delta s = |v_x| \cdot \Delta t \\[5px]
\end{align}

This means, conversely, that only particles less than Δs away from the piston can hit the surface (see yellow volume V_c in the figure above). All other particles will simply not have the time to reach the piston within the time Δt.

Since the particles are evenly distributed in the cylinder, the particle density in the cylinder N/V (ratio of total number of particles N and total cylinder volume V) can be used to determine the number of particles N_c contained in the pressure relevant collision volume V_c as follows:

\begin{align}
&N_c = \frac{N}{V} \cdot V_c \\[5px]
\end{align}

Now it must be considered that only half of the particles in the collision volume actually move towards the piston in the positive x-direction. For the number of particles N_x in the collision volume that actually flies in the direction of the piston, the following applies:

\begin{align}
&N_x = \frac{1}{2} \cdot N_c = \frac{1}{2} \cdot \frac{N}{V} \cdot V_c \\[5px]
\end{align}

The collision volume V_c relevant for the formation of the pressure results from the product of the piston area A and the distance Δs. Thus, the pressure relevant number of particles N_x in the collision voume can be calculated as follows:

\begin{align}
&N_x = \frac{1}{2} \cdot \frac{N}{V} \cdot \overbrace{A \cdot \Delta s}^{V_c} = \frac{1}{2} \cdot \frac{N}{V} \cdot A \cdot \overbrace{|v_x| \cdot \Delta t}^{\Delta s} \\[5px]
\end{align}

Each of the total N_x particles will change its momentum by a certain amount |Δp| (impuls) when it collides with the piston. The total impuls |Δp_tot| of all particles over the time period considered Δt corresponds to the force |F_tot| that the particles exert on the piston surface:

\begin{align}
\require{cancel}
\label{force}
&|F_{tot}| = \frac{|\Delta p_{tot}|}{\Delta t} = \frac{N_x \cdot |\Delta p|}{\Delta t} = \frac{\frac{1}{2} \cdot \frac{N}{V} \cdot A \cdot |v_x| \cdot \bcancel{\Delta t} \cdot |\Delta p|}{\bcancel{\Delta t}} = \frac{1}{2} \cdot \frac{N}{V} \cdot A \cdot |v_x|\cdot |\Delta p| \\[5px]
\end{align}

Equation (\ref{force}) shows that the force on the piston is not dependent on the time period considered Δt! If this were the case, the pressure would have to change over time. Everyday experience shows, however, that the pressure in gases remains constant as long as no changes such as an increase in temperature or a reduction in volume are made!

The impuls |Δp| is identical for each particle, since it was assumed that all particles move at the same speed |v_x| and that the collisions with the piston are completely elastic without (kinetic) energy loss. A particle that moves towards the piston with the speed |v_x| will move after the collision in the opposite direction with the same speed |v_x|. Since the momentum of a particle before the collision is +m⋅|v_x| and after the collision -m⋅|v_x|, the momentum has changed by double the value:

\begin{align}
\label{momentum}
&|\Delta p| = 2 \cdot m \cdot |v_x| \\[5px]
\end{align}

(Analogy: If the temperature changes from -20 °C to +20 °C, then it has changed by 40 °C in total, i.e. by twice the value).

If equation (\ref{momentum}) is now used in equation (\ref{force}), then the total force F_tot that the gas exerts on the piston can be calculated with the following formula:

\begin{align}
\label{f}
&|F_{tot}| = \frac{1}{2} \cdot \frac{N}{V} \cdot A \cdot |v_x|\cdot \overbrace{2 \cdot m \cdot |v_x|}^{|\Delta p|} =\frac{N}{V} \cdot A \cdot m \cdot |v_x|^2 =\frac{N}{V} \cdot A \cdot m \cdot v_x^2 \\[5px]
\end{align}

Note: Since the speed v_x is squared in equation (\ref{f}), it is no longer necessary to calculate the absolute value, since the square of a negative number is always positive.

Since the force |F_tot| is exerted on the piston surface A, the following gas pressure p is obtained by equation (\ref{p}):

\begin{align}
\require{cancel}
\label{pp}
& p = \frac{|F_{tot}|}{A} =\frac{\frac{N}{V} \cdot \bcancel{A} \cdot m \cdot v_x^2}{\bcancel{A}} = \frac{N}{V} \cdot m \cdot v_x^2 \\[5px]
\end{align}

Now it must be considered that even in ideal gases not all molecules have the same speed in x-direction, but is statistically (randomly) distributed. Therefore, according to the equation (\ref{pp}), the arithmetic mean of the squares of the speeds must be calculated:

\begin{align}
\label{ppp}
& \boxed{p =\frac{N}{V} \cdot m \cdot \overline{v_x^2}} \\[5px]
\end{align}

Not the average speed (arithmetic mean speed) of the particles may be squared but the squares of the individual speeds must be averaged. Therefore one speaks also of the mean square speed.

As an example the table below shows for 5 particles the results, if once the average speed (arithmetic mean speed) is squared and once the squares of the speeds are averaged (mean square speed). So the average speed in this case is 3 m/s and the square is 9 m²/s². If, on the other hand, the average of the speed squares is calculated, a value of 11.8 m²/s² is obtained.

Note the different notation between the square of the average speed \(\overline{v~}^2\) and the average of the square of the speed \(\overline{v^2}\).

Relation between the motion in x-direction and the total motion

Kinematic approach

The speed v of a molecule can basically be determined as follows from its velocity components in the x, y and z direction:

\begin{align}
& v = \sqrt{v_x^2 + v_y^2 + v_z^2}~~~~~\text{} \\[5px]
\end{align}

For the square of the speed v² therefore applies:

\begin{align}
& v^2= v_x^2 + v_y^2 + v_z^2 \\[5px]
\end{align}

The average of the squared speed can then be determined from the mean value of the individual squared velocity components as follows:

\begin{align}
& \overline{v^2}= \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2} \\[5px]
\end{align}

The velocity components of the molecules are statistically distributed and no direction is preferred. The mean value of the speed squares will thus be identical for all velocity components:

\begin{align}
&\overline{v_x^2} = \overline{v_y^2} = \overline {v_z^2} \\[5px]
\end{align}

Thus the following relation between the average of the squared speeds of the particles \(\overline{v^2}\) and the mean value of the squared velocity components in x-direction (\(\overline{v_x^2}\)) is obtained:

\begin{align}
& \overline{v^2}= \overline{v_x^2} + \overline{v_x^2} + \overline{v_x^2} = 3 \cdot \overline{v_x^2} \\[5px]
\end{align}

The average of the squared speeds in the x-direction thus corresponds to one third of the squared speeds of the individual particles:

\begin{align}
\label{x}
& \boxed{\overline{v_x^2} = \frac{\overline{v^2}}{3}} \\[5px]
\end{align}

If equation (\ref{x}) is applied in equation (\ref{ppp}), then the following relationship between the gas pressure p and the average of the speed squares \(\overline{v^2}\) of the individual particles becomes evident:

\begin{align}
\label{yy}
& \boxed{p =\frac{1}{3} \cdot \frac{N}{V} \cdot m \cdot \overline{v^2}} \\[5px]
\end{align}

Note that the speed v is no longer limited to the x-direction but represents the total speed of a particle (to be precise: to the mean of the speed squares of all molecules)!

Energetic approach

The same relations as are expressed in equation (\ref{yy}) can also be obtained by an energetic consideration. Starting point is again equation (\ref{ppp}). If equation (\ref{ppp}) is expanded with factor 2, then the term \(\frac{1}{2}m \overline{v_x^2}\) can be interpreted as the mean kinetic energy \(\overline{W_{kin,x}}\) of the particles that is related to the x-direction:

\begin{align}
& p =\frac{2}{2} \cdot \frac{N}{V} \cdot m \cdot \overline{v_x^2} =
2 \cdot \frac{N}{V} \cdot \overbrace{\frac{1}{2} m \cdot \overline{v_x^2}}^{\overline{W_{kin,x}}} \\[5px]
\label{druck}
& \boxed{p = 2 \cdot \frac{N}{V} \cdot \overline{W_{kin,x}}} \\[5px]
\end{align}

Since the velocities are randomly distributed and no direction is preferred, the same speed distribution is obtained for any direction. This in turn means that the mean kinetic energy of the particles in each direction is identical (i.e. the mean kinetic energy that would result if the motion of the particles were only observed along one spatial direction):

\begin{align}
& \overline{W_{kin,x}} = \overline{W_{kin,y}} = \overline{W_{kin,z}}\\[5px]
\end{align}

However, the particles do not only move in one direction but in three-dimensional space. The mean total kinetic energy \(\overline{W_{kin}}\), which the molecules have during their three-dimensional motion, is then the sum of the kinetic energies along the three spatial directions. Since the mean kinetic energy along the different spatial directions is identical, the total energy corresponds to three times the kinetic energy in one spatial direction (e.g. in the x-direction):

\begin{align}
& \overline{W_{kin}} = \overline{W_{kin,x}} + \overline{W_{kin,y}} + \overline{W_{kin,z}} = 3 \cdot \overline{W_{kin,x}} \\[5px]
\label{kine}
&\overline{W_{kin,x}} = \frac{1}{3}\cdot \overline{W_{kin}} \\[5px]
\end{align}

Thus the pressure is linked as follows to the mean kinetic energy of a particle (insert equation (\ref{kine}) in (\ref{druck})):

\begin{align}
&p = 2 \cdot \frac{N}{V} \cdot \frac{1}{3} \overline{W_{kin}} \\[5px]
\label{y}
& \boxed{p = \frac{2}{3} \cdot \frac{N}{V} \cdot \overline{W_{kin}}} \\[5px]
\end{align}

The pressure of a gas depends beside the number of particles and the volume only on the average kinetic energy of a particle!

Note that the mean kinetic energy of a particle refers only to the translational motion. The particles of a gas generally also have a rotational motion and therefore rotational energy, but this has no effect on the pressure!

Since the mean kinetic energy of the particles is linked by their mass to the mean of the speed squares (\(\overline{W_{kin}}=\frac{1}{2}m\cdot \overline{v^2}\)), whereby the speed v no longer refers only to a spatial direction but to the total speed, then the same relationship as in equation (\ref{yy}) becomes evident:

\begin{align}
& p = \frac{2}{3} \cdot \frac{N}{V} \cdot \overbrace{\overline{W_{kin}}}^{\frac{1}{2}m \overline{v^2}} \\[5px]
& \boxed{p =\frac{1}{3} \cdot \frac{N}{V} \cdot m \cdot \overline{v^2}} \\[5px]
\end{align}

Microscopic interpretation of temperature

In the article “Ideal gas law” the following equation was derived experimentally:

\begin{align}
\label{1}
& p \cdot V = N \cdot k_B \cdot T \\[5px]
\end{align}

In this equation, k_B denotes the so-called Boltzmann constant and T the thermodynamic temperature of the gas. If you now bring the gas volume V in equation (\ref{y}) on the left side…

\begin{align}
\label{2}
& p \cdot V =\frac{2}{3} \cdot N \cdot \overline{W_{kin}} \\[5px]
\end{align}

… and then equating equations (\ref{1}) and (\ref{2}), then the relationship between the mean kinetic energy of a particle and temperature of the gas becomes apparent:

\begin{align}
\require{cancel}
& \frac{2}{3} \cdot \bcancel{N} \cdot \overline{W_{kin}} = \bcancel{N} \cdot k_B \cdot T \\[5px]
\label{kin}
& \boxed{\overline{W_{kin}} = \frac{3}{2} k_B \cdot T} \\[5px]
\end{align}

The average kinetic energy of the particles depends only on the temperature!

Also at this point, the kinetic energy again refers only to the translatory motion and not to the rotational motion. Rotational energies (angular kinetic energy) do not influence the temperature at all!

Internal energy of an ideal gas

Since according to equation (\ref{kin}) each particle carries (on average) a kinetic energy of 3/2⋅k_BT, the total energy of all the particles and thus the energy of the gas can be obtained by multiplying the mean kinetic energy of a particle with the total number of particles N. This total energy inside the gas is also called the internal energy U:

\begin{align}
\label{u}
& \boxed{U = \frac{3}{2} N k_B T} \\[5px]
\end{align}

If the expression N⋅k_BT is replaced by p⋅V according to the equation (\ref{1}), then the pressure can also be determined from the energy density u_v of the gas (in order to avoid confusion with the specific internal energy u as a mass-related quantity, a “v” is added to the volumetric energy density in the index):

\begin{align}
& U = \frac{3}{2} N k_B T = \frac{3}{2} pV\\[5px]
& p = \frac{2}{3} \cdot \underbrace{\frac{U}{V}}_{\text{energy density } u_v} \\[5px]
& \boxed{p = \frac{2}{3} u_v} ~~~~~ u_v=\frac{U}{V}
\end{align}

The pressure of an ideal gas only depends on the energy density, i.e. the internal energy in the gas per unit volume!

Introduction

As already explained in the article Temperature and particle motion, the temperature of a gas is a measure of the kinetic energy of the particles. Even at a constant temperature, however, not all the molecules have the same speed. After all, in a gas there are permanent collisions between the particles. Some particles are slowed down by the collision and others are accelerated by it. Thus, molecules with different velocities can be found.

A gas usually contains a large number of molecules. For ideal gases, therefore, statistical predictions can be made about the frequency with which certain speeds occur. The so-called Maxwell-Boltzmann distribution describes such a speed distribution for the particles of an ideal gas.

The Maxwell-Boltzmann distribution describes the speed distribution of the particles of an ideal gas!

Such a statistical prediction about the speed distribution is only possible if a sufficient number of molecules are present. This is true in most thermodynamic cases! To get an impression of the large number of particles in an ideal gas, imagine a football filled with air. The number of gas particles in such a football would then correspond approximately to the number of 1-liter water bottles that would theoretically be necessary to completely fill the entire volume of the earth with water!

This example clearly shows that in practice a sufficiently large number of particles in a gas can usually be assumed to make reliable statistical predictions.

Maxwell-Boltzmann speed distribution

Using statistical methods, physicists James Clerk Maxwell and Ludwig Boltzmann were able to derive the following formula for the molecular speed distribution in an ideal gas. For this reason this function of the relative frequency density f(v) is also called Maxwell-Boltzmann distribution function:

\begin{align}
\label{p}
&\boxed{  f(v) = \left(  \sqrt{\frac{m}{2 \pi k_B T}}  \right)^{3} 4 \pi v^2 \cdot \exp{\left(- \frac{m v^2}{2 k_B T} \right)} } ~~~\text{Maxwell-Boltzmann distribution function} \\[5px]
\end{align}

In equation (\ref{p}), m denotes the mass of a particle (not the total mass of the gas!). For a helium particle as assumed for the graph above, the mass is m = 6.6465⋅10^-27 kg. The constant k_B is the so-called Boltzmann constant with a value of k_B = 1.38065⋅10^-23 J/kg. The temperature T must be expressed in the unit Kelvin! For the considered case T=273 K (0 °C) applies. Note that the speed distribution is independent of the gas pressure!

The distribution of the molecular speeds for an ideal gas is independent of the gas pressure!

The article Determination of the speed distribution in a gas deals with the experimental determination of such a speed distribution in more detail. For ideal gases, this speed distribution can also be derived mathematically as shown in the article Derivation of the Maxwell-Boltzmann distribution function. In this article, however, the speed distribution will only be explained and interpreted in more detail.

Interpretation of the Maxwell-Boltzmann distribution function

The Maxwell-Boltzmann distribution describes the frequency with which certain molecular speeds occur in an ideal gas. In principle, however, it is not possible to assign a specific number of molecules to a specific speed. One will never find a single molecule with a specified velocity down to the “last” decimal place. One can only assign speed intervals to a concrete number of particles, i.e. the number of particles whose speeds lie within a certain range (e.g. between 1300 m/s and 1600 m/s).

Therefore one chooses a graph in which the area under the curve corresponds to the percentage share of the particles whose speed lies within the considered interval! Therefore one does not speak of the frequency but of the frequency density (“frequency with respect to the speed interval”). As it is not an absolute frequency (i.e. not a concrete number of molecules) but a percentage, it is referred to as the relative frequency density.

The area under the Maxwell-Boltzmann distribution function corresponds to the percentage of particles whose speeds lie within the considered interval!

More detailed information on the derivation and interpretation of such a graph can be found in the article Determination of the speed distribution in a gas!

If, for example, there is an area of 0.2 in the range between 1300 m/s and 1600 m/s (see figure above), this means that 20 % of the particles have a speed within this range. If one randomly selects 100 particles in this gas, then 20 particles would be among them whose speeds lie within this interval. The probability of catching a particle whose speed lies within this interval would thus be 20 %. This example shows that the relative frequency density can also be interpreted as the probability density.

The relative frequency density can also be understood as probability density, i.e. the probability to encounter a particle within a certain speed range!

With this interpretation it also becomes clear that if only a certain speed is considered, the area under the curve becomes zero. Thus, the probability of finding a particle with exactly this velocity is also zero! This means that there is no particle with an exactly given speed.

Influence of temperature on the speed distribution

The effects of ever higher temperatures on the Maxwell-Boltzmann distribution are shown in the figure below. As the temperature rises, the curve maximum shifts to higher and higher speeds. The curves are stretched in length and squeezed in height. This results in a broader speed distribution for high temperatures, with higher speed proportions.

With increasing temperatures, the curve maximum shifts to higher and higher speeds!

However, all curves are basically “open to the right”, i.e. even at very low temperatures, there are a few particles with very high speeds!

Even at very low temperatures, there are gas molecules that have very high speeds!

If this statement is transferred qualitatively to liquids, then it can also be used to explain why liquids gradually enter the gas phase, i.e. evaporate, even far below the boiling point. Even at such low temperatures, there are always particles that have a sufficiently high velocity and thus kinetic energy to escape the attraction forces within the liquid phase. They practically break away from the binding forces and thus escape from the liquid. The liquid gradually becomes less until the molecules are completely in the gas phase. Since according to the speed distribution the number of particles that have such a sufficiently high velocity is relatively small, such an evaporation process takes a relatively long time compared to the vaporisation process. Further information can be found in the article Evaporation of liquids.

Characteristic speeds

To characterize the speed distribution, different speeds are introduced, which are explained in more detail in the following.

Most probable speed

If a particle is randomly picked out of an ideal gas, it is most likely to be in the speed range with the highest proportion. This corresponds to the maximum of the speed distribution and is referred to as the most probable speed \(\hat{v}\).

Mathematically, this most likely speed can be determined by setting the first derivative of the Maxwell-Boltzmann distribution (\ref{p}) to zero [df(v)/dv=0]. The result is the following formula for determining the most probable speed:

\begin{align}
\label{w}
&\boxed{ \hat{v} = \sqrt{\frac{2 k_B T}{m}}  } ~~~\text{most probable speed}\\[5px]
\end{align}

The most probable speed is not only a function of temperature but also depends on the mass of the particles. Thus, the most probable speed cannot be deduced directly from the temperature of a gas. Two different gases (whose particles have different masses) therefore have different most probable speeds despite the same temperature (see figure below). For heavier gas molecules, the most probable speeds will be lower than for lighter molecules.

Arithmetic mean speed

Another characteristic speed of the Maxwell-Boltzmann distribution is the arithmetic mean speed \(\overline{v}\) of a particle (also called average speed or just mean speed).

Compared to the most probable speed, the mean speed will be higher, since the number of particles with a higher speed than the most probable is also higher. This becomes clear when the area to the right and to the left of the maximum is compared. 42.8 % of the particles have a velocity below the most probable speed and 57.2 % have a velocity above the most probable speed.

The average speed is obtained by summing up the speeds of the particles and then dividing the sum by the number of particles. Mathematically, this can be calculated by solving the integral ∫v⋅f(v) dv within the entire speed range from 0 to ∞:

\begin{align}
&\overline{v} = \frac{v_1+v_2+v_3+…+v_N}{N} =  \frac{\sum_{i=1}^{N}{v_i}}{N} = \int_0^\infty \! v~f(v) \, \mathrm{d}v \\[5px]
\label{a}
&\boxed{ \overline{v} = \sqrt{\frac{8 k_B T}{\pi m}}  } ~~~\text{mean speed} \\[5px]
\end{align}

At this point it can be seen again that the temperature is not a direct measure of the average particle speed. The particles in a gas with four times the particle mass are on average only half as fast, even with the same temperature.

To get an impression of the order of magnitude of the mean speed in gases, nitrogen molecules N₂ are considered, which make up the majority of the air with 78 %. With a particle mass of 4.65⋅10^-27 kg, the nitrogen molecules have a mean speed of about 470 m/s at 273 K. Note that the average speed of the nitrogen molecules in the air is thus greater than the speed of sound! In contrast to sound, a nitrogen molecule does not travel several meters or even kilometers in a certain direction. It will permanently collide with other air particles and constantly change direction. Such a distance without a collision is usually only a few nanometers and is called the mean free path!

Root-mean-square speed

However, neither the most probable speed nor the average speed is relevant for determining the kinetic energy of a molecule. This is due to the quadratic influence of velocity on kinetic energy. Higher speeds have a disproportionately strong effect on kinetic energy. Thus, a speed twice as high does not mean a kinetic energy twice as high but four times as high. Higher speeds influence the average kinetic energy more than lower speeds.

Therefore, the mean speed must not be used to characterize the kinetic energy of a particle. Rather, the arithmetic mean of the speed squares must be taken as a basis. In order to obtain the dimension of a speed, the square root of this mean value must be calculated. This speed is then referred to as the root-mean-square speed v_rms:

\begin{align}
&\boxed{v_{rms} = \sqrt{\overline{v^2}}} \\[5px]
\end{align}

For the calculation of the root-mean-square speed from the Maxwell-Boltzmann function, the square root of the integral ∫v²⋅f(v) dv has to be calculated within the range between 0 and ∞:

\begin{align}
& v_{rms} = \sqrt{\overline{v^2}} = \sqrt{\frac{v_1^2+v_2^2+v_3^2+…+v_N^2}{N}} =  \sqrt{\frac{\sum_{i=1}^{N}{v_i^2}}{N}} = \sqrt{\int_0^\infty \! v^2~f(v) \, \mathrm{d}v }\\[5px]
\label{q}
&\boxed{v_{rms} = \sqrt{\frac{3 k_B T}{m}} } ~~~\text{root-mean-square speed} \\[5px]
\end{align}

It now becomes apparent that the root-mean-square speed is decisive for the average kinetic energy of a particle:

\begin{align}
\overline{W_{kin}} &= \tfrac{W_{kin,1}+W_{kin,2}+W_{kin,3} + … + W_{kin,N}}{N} = {\tfrac{\frac{1}{2}m\cdot v_1^2+\frac{1}{2}m\cdot v_2^2+\frac{1}{2}m\cdot v_3^2+…+\frac{1}{2}m\cdot v_N^2}{N}}  \\[5px]
&= \frac{1}{2}m\cdot \underbrace{{\tfrac{v_1^2+v_2^2+v_3^2+…+v_N^2}{N}}}_{v_{rms}^2} = \frac{1}{2} m \cdot  v_{rms}^2  \\[5px]
\end{align}

\begin{align}
\label{e}
\boxed{\overline{W_{kin}} = \frac{1}{2} m \cdot  v_{rms}^2}  \\[5px]
\end{align}

Relationship between the different speeds

The average speed of the particles \(\overline{v}\) is always greater than the most probable speed \(\hat{v}\), and the root-mean-square speed \(v_{rms}\) is greater than the average speed. The different speeds are in a constant ratio to each other, independent of temperature or mass. For the ratio of average speed \(\overline{v}\) to most probable speed \(\hat{v}\) applies:

\begin{align}
\frac{\overline{v}}{\hat{v}}= \frac{\sqrt{\frac{8 k_B T}{\pi m}} }{
\sqrt{\frac{2 k_B T}{m}} } = \sqrt{\frac{4}{\pi}} = 1,128\\[5px]
\end{align}

For the ratio of root-mean-square speed \(v_{rms}\) to average speed \(\overline{v}\) a value of 1.085 is obtained:

\begin{align}
\frac{v_{rms}}{\overline{v}}= \frac{\sqrt{\frac{3 k_B T}{m}}}{\sqrt{\frac{8 k_B T}{\pi m}}} = \sqrt{\frac{3\pi}{8}} = 1,085\\[5px]
\end{align}

The average speed is therefore always 12,8 % higher than the most probable speed and the root-mean-square speed is always 8,5 % higher than the average speed.

Relationship between temperature and kinetic energy

The different speeds (the most probable speed, the average speed and the root-mean-square speed) depend not only on the temperature but also on the particle mass. The frequently heard statement that the temperature is a measure for the speed of the particles is strictly speaking not correct.

For example, such a statement does not apply if two different types of gas are considered. Even at temperatures of 473 K (200 °C) the particle speeds of argon are more than half lower compared to the particle speeds of helium at 273 K (0 °C), because argon has an atomic mass about 40 times larger than helium.

Rather, the statement about the temperature aims at the average kinetic energy of the molecules. This becomes clear when the formula for the root-mean-square speed (\ref{q}) is used in equation for the average kinetic energy (\ref{e}):

\begin{align}
&\overline{W_{kin}} = \frac{1}{2} m \cdot v_{rms}^2 = \frac{1}{2} m \cdot \frac{3 k_B T}{m} \\[5px]
\label{wkint}
&\boxed{\overline{W_{kin}}=\frac{3}{2}~k_B~T } \\[5px]
\end{align}

The average kinetic energy of a particle is directly connected to the temperature and independent of the particle mass! Thus the temperature is directly a measure for the average kinetic energy of the gas particles of an ideal gas. This equation is remarkable because it combines a macroscopically measurable quantity (in the form of temperature) with a microscopic quantity (in the form of the kinetic energy of a particle)!

The temperature of an ideal gas is directly a measure of the average kinetic energy of the particles!

Equation (\ref{wkint}) was already derived in the article Pressure and temperature by kinematic and statistical considerations. It is no coincidence at this point that the Maxwell-Boltzmann distribution comes to the same conclusion. After all, the derivation of the Maxwell-Boltzmann distribution function is based on the assumption that the mean kinetic energy of a particle is linked to the temperature according to the equation (\ref{wkint})!

Maxwell-Boltzmann kinetic energy distribution

Since a certain kinetic energy can be assigned to each speed, the speed distribution can also be converted into an energy distribution. Instead of the distribution function f(v) for the speed, a distribution function for the kinetic energy g(W) is obtained.

As already explained, the (relative) frequency with which a speed exists within the limits between \(v_1\) and \(v_2\) results from the integral of the distribution function f(v):

\begin{align}
& \text{Frequency} = \int_{v_1}^{v_2} \! f(v) \, \mathrm{d}v   \\[5px]
\end{align}

If the speeds v₁ and v₂ are converted into the respective kinetic energies W₁ and W₂, then a distribution function of the energy g(W_kin) within these limits must lead to the same frequency:

\begin{align}
& \text{Frequency} = \int_{W_{1}}^{W_{2}} \! g(W) \, \mathrm{d}W   \\[5px]
\end{align}

If both equations are equated, then the following relationship between the two distribution functions is obtained:

\begin{align}
\int_{W_{1}}^{W_{2}} \! g(W) \, \mathrm{d}W &= \int_{v_1}^{v_2} \! f(v) \, \mathrm{d}v \\[5px]
g(W) \text{d}W &=f(v) ~ \text{d}v \\[5px]
g(W) &= f(v) ~ \frac{\text{d}v}{\text{d}W} \\[5px]
g(W) &= f(v) ~ \dfrac{1}{\frac{\text{d}W}{\text{d}v}} \\[5px]
\end{align}

The expression dW/dv contained in the denominator corresponds mathematically to the speed derivative of the kinetic energy:

\begin{align}
& \frac{\text{d}W(v)}{\text{d}v} = \frac{\text{d}(\tfrac{1}{2}mv^2)}{\text{d}v} = mv \\[5px]
\end{align}

Thus the distribution function of the kinetic energy g(W) can be determined from the distribution function of the speed f(v):

\begin{align}
&g(W) = f(v) ~ \dfrac{1}{\frac{\text{d}W}{\text{d}v}} = f(v) ~ \dfrac{1}{mv} \\[5px]
&\boxed{g(W) = f(v) ~ \dfrac{1}{mv} } \\[5px]
\end{align}

If the speed distribution f(v) is used at this point and the terms are transformed and summarized in such a way that only the kinetic energies can be found in them, then the following distribution function of the kinetic energy becomes apparent:

\begin{align}
\require{cancel}
g(W) &= \left(  \sqrt{\frac{m}{2 \pi k_B T}}  \right)^{3} 4 \pi v^2 \cdot \exp{\left(- \frac{\color{red}{m v^2}}{\color{red}{2} k_B T} \right)} ~ \dfrac{1}{mv} \\[5px]
& = \left(  \sqrt{\frac{m}{2 \pi k_B T}}  \right)^{3} 4 \pi \frac{v^\bcancel{2}}{m\bcancel{v}} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \sqrt{\left(  \frac{m}{2 \pi k_B T}  \right)^{3}} \cdot 4 \pi \frac{v}{m} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \sqrt{\left( \frac{m}{2 \pi k_B T} \right)^{3}} \cdot \sqrt{\left(4 \pi \frac{v}{m} \right)^2} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \sqrt{\left( \frac{m}{2 \pi k_B T} \right)^{3} \cdot \left(4 \pi \frac{v}{m} \right)^2} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \sqrt{ \frac{m^\bcancel{3}}{8 \pi^3 k_B^3 T^3} \cdot 16 \pi \frac{v^2}{\bcancel{m^2}} } \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \sqrt{\frac{4}{\pi^2 k_B^3 T^3} \cdot \color{red}{\tfrac{1}{2}mv^2} } \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \frac{2}{\sqrt{\pi}}\sqrt{\left(\frac{1}{k_B T}\right)^3} \cdot \sqrt{\color{red}{\tfrac{1}{2}mv^2}} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
& = \frac{2}{\sqrt{\pi}}\left(\sqrt{\frac{1}{k_B T}}\right)^3 \cdot \sqrt{\color{red}{\tfrac{1}{2}mv^2}} \cdot \exp{\left(- \frac{\color{red}{\tfrac{1}{2}m v^2}}{k_B T} \right)} \\[5px]
\end{align}

The red marked terms correspond to the kinetic energy, so the distribution function can be expressed solely by the kinetic energy:

\begin{align}
&\boxed{g(W) = \frac{2}{\sqrt{\pi}}\left(\sqrt{\frac{1}{k_B T}}\right)^3 \cdot \sqrt{W} \cdot \exp{\left(- \frac{W}{k_B T} \right)}} \\[5px]
\end{align}

The figure below shows the distribution of the kinetic energies for a temperature of 500 K. Again, the area under the curve corresponds to the percentage of particles with kinetic energy within the corresponding limits.

In the article Derivation of the Maxwell-Boltzmann distribution function, the distribution function of the molecular speeds was derived from the barometric formula. It has already been noted that the exponential term exp(-W/k_BT) describes the frequency or probability with which certain (kinetic) energies are present. This term plays a central role in statistical physics (Boltzmann statistics) and occurs whenever random distributions of energetic states are involved. For example also with Planck’s law of blackbody radiation.

The Boltzmann factor exp(-W/k_BT) describes the probability with which certain energetic states are present!

Maxwell-Boltzmann distribution of ideal gases

The figure below shows the speed distribution according to Maxwell-Boltzmann for the particles of an ideal gas. Put simply, this distribution shows the number of particles (vertical axis) for a certain velocity (horizontal axis).

Temperature has a major influence on this distribution. For higher temperatures, the curves are squeezed in height and stretched in length. This results in a greater distribution with higher speed proportions. This corresponds to the fact that the temperature is a measure of the kinetic energy of the gas particles: the higher the temperature, the greater the kinetic energy and thus the velocity of the particles (see also the article Temperature and particle motion).

With increasing temperatures, the curve maximum shifts to higher speeds!

In principle, all curves are open to the right, i.e. even at such low temperatures, there is a certain probability that particles with extremely high speeds can be found.

Even at very low temperatures, there are gas particles that have very high speeds!

Evaporation

If this fact is now qualitatively transferred from ideal gases to liquids, this means that even below the boiling point particles are always found with sufficiently high velocities. Due to the associated high kinetic energy, these particles can escape the attractive forces of the liquid. Such a process is called evaporation and takes place far below the boiling point.

Particles with sufficiently high speeds are capable of escaping the binding forces of the liquid and thus of transitioning from the liquid phase to the gas phase (evaporation)!

Cooling effect

Another effect can be explained in connection with the evaporation of liquids. Since the evaporation causes the liquid to lack particles with high speeds, the average kinetic energy and thus the temperature in the liquid decreases. The liquid cools down during evaporation! This leads, for example, to the cooling effect when sweating, when the sweat evaporates on the skin.

This phenomenon can be checked relatively easily. If the sensor of a thermometer is put in water and then removed, the temperature will very quickly drop below ambient temperature (see video below). This phenomenon is also the reason why you freeze faster with wet skin than with dry skin.