Introduction

The free adiabatic expansion of an ideal gas in a vacuum showed that this thermodynamic process cannot be described by the isentropic equations, although these equations were derived for an adiabatic system. The adiabatic expansion against a vacuum is an isothermal process in which the temperature remains constant. Therefore, the question arises under which condition a thermodynamic process in an adiabatic system can actually be described with the isentropic equations.

Reversibility of thermodynamic processes

For this purpose, we will take a closer look at different thermodynamic processes. In doing so, we will see that, from an energetic point of view, all these processes can run in one direction as well as, in principle, in the other direction. For example, in the article Concept of pressure-volume work, we looked at a thermodynamic process in which a gas expands in a cylinder by adding heat and finally lifts a weight via a rack and pinion. From an energetic point of view, this process can be reversed in principle.

To do so, one can imagine the original process as being recorded with a video camera, which is then played backwards. In such a reverse case, the weight lowers by removal of heat (cooling). Here, work is no longer done by the gas on the weight, but conversely work is done on the gas by the weight – the gas is compressed by the weight, so to speak. At the same time, cooling lowers the temperature and with it the internal energy. The energy flows are therefore reversed in such a reversal process, without any contradiction arising from experience.

Isochoric process

Such a reversal process can also be found in an isochoric heating of a gas cylinder. The heat supplied causes an increase in internal energy, i.e. the temperature rises. If this process is considered energetically reversed, one would obtain an isochoric cooling. In such a reversal case, the internal energy would decrease due to the dissipated heat and the temperature would decrease.

Isobaric process

The process explained in the article on the isobaric process, in which a weight is lifted when heat is supplied, can also be reversed in terms of energy. In the original case, the gas performs work on the weight (lifting) by adding heat, while the internal energy increases and the temperature rises. In the reverse case, an isobaric cooling would be obtained, in which the gas is compressed by the weight (lowering), while at the same time heat is dissipated. The internal energy and thus the temperature are reduced.

Isothermal process

The compression of air in an air pump explained in the article on the isothermal process also shows such an energetic reversibility. If the gas in an air pump is compressed very slowly with the valve closed, the temperature (nearly) remains constant due to the heat transfer to the surroundings. Conversely, the gas can also expand isothermally from this compressed state. In this case, the expansion must again be very slow so that the surroundings can transfer enough heat to the expanding gas to prevent a drop in temperature.

Isentropic process

Also, the rapid compression of air in an air pump presented in the article on the isentropic process can be reversed from an energetic point of view. In the original process, a gas is compressed rapidly (work is done on the gas) so that heat cannot be transferred in such a short time (adiabatic system), which leads to an increase in internal energy and causes the temperature to rise. In the reverse case, the gas expands very rapidly starting from the compressed state. In this isentropic reversal process, work is then done by the gas at the expense of the internal energy and the temperature consequently drops.

Reversibility

The processes considered in the previous sections make it clear that they are all reversible from an energetic point of view. The reversibility of thermodynamic processes presupposes, among other things, that there is no friction or, more generally, no dissipation of energy (e.g. when a piston slides in a cylinder).

In thermodynamics, reversible processes are processes which are reversible from an energetic point of view without violating everyday experience!

Furthermore, the gas states must always be in equilibrium, so that there is no temperature difference within the gas itself. Both the assumption that no energy is dissipated by friction and the assumption of the always existing equilibrium exist, of course, only in the ideal conception. Reversible processes therefore do not exist in reality. However, these ideal conditions have been assumed for all thermodynamic processes considered so far.

Adiabatic expansion against a vacuum as a non-reversible process

As far as reversibility is concerned, however, the situation is different for the free adiabatic expansion of an ideal against a vacuum. This thermodynamic process is not reversible from everyday experience! In adiabatic expansion against a vacuum, the gas expands at constant temperature without transfer of work or heat. For the reverse case, this would have to mean that the gas compresses itself without help.

However, this contradicts all experience, because in order to compress a gas in a cylinder, work must be done on the gas – a gas will not do this by itself. The temperature would then also no longer remain constant, but it would increase due to the work transferred to the gas. Consequently, an adiabatic expansion against a vacuum is a non-reversible process from an energetic point of view. Such non-reversible processes are also called irreversible processes.

Basically all equations presented in the respective articles – in particular those of the isochoric, isobaric, isothermal and isentropic process – are valid only for reversible processes. This now explains why the equations of the isentropic process cannot be applied to the adiabatic expansion against a vacuum: the process is irreversible! The equations of the isentropic process may only be used if it is a reversible process of an adiabatic system. Thus, an isentropic process be defined as follows:

An isentropic process is a reversible thermodynamic process of an adiabatic system!

The reversibility of thermodynamic processes is described by the quantity entropy. For a reversible process of an adiabatic system, the entropy remains constant. Therefore, such a process is called an isentropic process (process at constant entropy).

In the article Free expansion of an ideal gas in a vacuum it was explained that the free expansion of a gas against a vacuum taking place in an (ideal) adiabatic system is an isothermal process. For this purpose, a gas-filled cylinder was considered, which is closed with a massless and frictionless sliding piston.

The temperature of the ideal gas remains constant during such an adiabatic expansion. However, this applies only to ideal gases. If, on the other hand, such an experiment is carried out with a real gas, then a slight decrease in temperature would be observed in reality (even though the piston can still be considered to be massless). This phenomenon is due to the attractive forces acting between the gas molecules, which are neglected in the case of ideal gases.

Because when a real gas expands, the distances between the molecules increase due to the increasing volume. The gas must therefore increase its volume against the attractive forces acting between the molecules. This process obviously requires energy. This energy must be delivered by the gas itself and can only be provided at the expense of the kinetic energy of the gas molecules. This means a decrease in temperature, because the temperature is directly linked to the kinetic energy. This phenomenon does not only occur during the expansion against a vacuum, but ultimately whenever a real gas expands against a lower pressure. This effect of temperature decrease is also called Joule-Thomson effect.

The Joule-Thomson effect is the phenomenon of a decrease in temperature of real gases when they expand against a lower pressure!

Note that in many cases the gas temperature decreases during an expansion. This generally has nothing to do with the Joule-Thomson effect. The Joule-Thomson effect only describes the stronger decrease in temperature for real gases than would be the case for ideal gases, since binding energies must be taken into account for real gases. And as already explained, these binding energies cause an additional amount of work during an expansion, which is at the expense of the kinetic energies, which additionally decreases the temperature. Only this additional temperature decrease is called the Joule-Thomson effect.

Note also that even with an adiabatic (!) expansion of a real gas against a vacuum, the internal energy of the gas remains constant. In a real gas, the internal energy is not only composed of the kinetic energy of the molecules as in an ideal gas, but also binding energies are part of the internal energy. In real gases, there is only an “internal” redistribution of energies, namely at the expense of the kinetic energy and in favor of the binding energies. The work done by the gas to increase the molecular distances is not transferred across the system boundary, but only redistributed internally!

The first law of thermodynamics only balances energies that are tranferred accross the system boundary! In an adiabatic expansion against a vacuum, neither such boundary work nor heat is transferred (W=0, Q=0)! Thus, even for real gases, there is no change in the internal energy ΔU:

\begin{align}
\underbrace{W}_{=0} + \underbrace{Q}_{=0} = \Delta U = 0
\end{align}

In the article on the isentropic process it was already indicated that the often used term “adiabatic process” can be misleading in some situations and is generally not correct. To clarify this, the expansion of an ideal gas against a vacuum is considered in the following.

For this purpose, imagine a gas in a cylinder which is closed by a piston with negligible mass. The piston slides frictionless inside the cylinder. The piston is initially held in position by force. The cylinder is located in an evacuated chamber so that there is a vacuum on the outside of the piston. Cylinder and piston are assumed to be perfectly thermally insulated. It is therefore an adiabatic system. Now the piston is released and the gas expands against the vacuum.

However, just because the cylinder and the piston form an adiabatic system, this does not mean that the derived laws of the supposed “adiabatic process” can be applied! According to these laws, the temperature should decrease during an expansion with increasing volume (T⋅V^κ-1=constant). However, this is not observed in the case considered, where the gas expands against the vacuum! The surprising result of this process shows up, if one takes a look at the first law of thermodynamics:

\begin{align}
\underbrace{W}_{=0} + \underbrace{Q}_{=0} = \Delta U = 0
\end{align}

Since it is an adiabatic system, by definition no heat is transferred (Q=0). Furthermore, the gas does not perform any work (W=0), because the gas expands against a vacuum. The gas does not have to exert any force to move the (massless!) piston. There is, so to speak, no counterpressure against which the gas has to push. Thus, the gas does not have to perform any work for the expansion of its volume. However, if no energy is transferred, neither as work nor as heat, the internal energy consequently remains constant (ΔU=0). Since the internal energy of an ideal gas is directly linked to the temperature (and this does not change), the free expansion of an ideal gas against a vacuum is therefore an isothermal process!

When an ideal gas expands in an adiabatic system against a vacuum, the temperature remains constant (isothermal process)!

Note: The fact that the temperature remains constant during expansion against the vacuum applies only to ideal gases. For real gases, on the other hand, a slight decrease in temperature will be observed. This phenomenon is referred to Joule-Thomson effect.

One can explain the isothermal expansion against a vaccum quite clearly. Imagine you are in a vacuum and you hold wildly flying particles in your hand. If you open your hand, the particles fly into the vacuum and are not slowed down by any other particles (since the piston is considered massless in the above example, you can also remove the piston and let the particles fly freely into the vacuum – this is the same situation). So the gas molecules keep their kinetic energy. Since the kinetic energy of the molecules is directly linked to the temperature, this means a constant temperature.

The adiabatic expansion of an ideal gas into a vacuum represents an isothermal process!

This example makes clear that thermodynamic processes in adiabatic systems cannot necessarily be described with the laws of isentropic processes. Therefore, in connection with the isentropic equations derived for this purpose, one should not speak of “adiabatic” processes. Because this suggests falsely that for all adiabatic systems these equations could be used. However, the expansion against a vacuum disproves exactly this.

Of course, the question arises under which condition a thermodynamic process of an adiabatic system can actually be described by the derived isentropic equations. The answer to this question can be found in the article Reversibility of thermodynamic processes (entropy).

Work performed in adiabatic systems

Many thermodynamic processes take place within very short times, such as the expansion of the burned fuel-air mixture in internal combustion engines during the power stroke (approx. 10 ms) or the expansion of hot gases in turbines. Within such short times, a significant heat transfer between gas and surroundings can be neglected in a first approximation (Q=0). The thermodynamic processes are then considered to take place in an adiabatic system. According to the first law of thermodynamics, the work done W by/on the gas is given directly by the change in internal energy ΔU of the gas:

\begin{align}
&\boxed{W+Q=\Delta U} &&~\text{first law of thermodynamics} \\[5px]
&W = \Delta U &&~\text{only applies to adiabatic systems (}Q=0 \text{)}\\[5px]
\end{align}

Since for ideal gases the change in internal energy ΔU is independent of the thermodynamic process and only results from the temperature change, …

\begin{align}
&\Delta U = c_\text{v} \cdot m \cdot \Delta T ~~~~~\text{where}~~~~~\Delta T = T_2-T_1\\[5px]
\end{align}

… the work done by/on the adiabatic system can be determined as follows:

\begin{align}
\label{9067}
&\boxed{W = c_\text{v}~m~\Delta T} ~~~\text{only applies to adiabatic systems (}Q=0 \text{)} \\[5px]
\end{align}

Note that the work denoted by W does not necessarily correspond to the pressure-volume work W_v of the gas. This would only be the case for a non-dissipative process where no friction occurs. Therefore, the work denoted by W describes quite generally the work transferred across the system boundary (boundary work) and thus also includes dissipative processes! More information about this can be found in the article Dissipation of energy in closed systems.

Equation (\ref{9067}) clearly shows that during the expansion of a gas in an adiabatic system, the greater the difference between the initial and final temperature, the more work is transferred from the system to the surroundings. For example, during the power stroke inside a diesel engine, the temperature in the cylinder drops from about 2000 °C to about 500 °C during expansion. However, at a given initial temperature T₁, dissipative processes with friction always lead to higher final temperatures T₂ compared to frictionless processes due to dissipation of energy that benefits the internal energy. The temperature obviously does not decrease as much during the expansion. Therefore, according to equation (\ref{9067}), the work done by the system ist lower!

Dissipative processes in the volume-pressure diagram

The volume-pressure diagram below shows an expansion of a gas in a cylinder which has been closed with a frictionless moving piston (blue curve). In comparison, an expansion is shown in which a constant frictional force between cylinder and piston was assumed over the duration of the expansion (red curve). It was assumed that the entire friction energy was dissipated in internal energy.

Due to the larger temperature difference, the greatest possible work is done by the gas during non-dissipative expansion → see equation (\ref{9067}). For such reversible processes, the laws of the isentropic process (n=κ) apply in adiabatic systems. Dissipative processes, on the other hand, run at a higher pressure level due to the higher temperatures. Such processes can then be approximated by a polytropic process, where in the case of an expansion the polytropic index n has a lower value than the isentropic index κ (n<κ):

\begin{align}
\label{3475}
&\boxed{p_1~V_1^n=p_2~V_2^n} \\[5px]
\label{2070}
&\boxed{T_1~V_1^{n-1}=T_2~V_2^{n-1}} \\[5px]
\label{9579}
&\boxed{T_1^n~p_1^{1-n}=T_2^n~p_2^{1-n}} \\[5px]
\text{ where: }~~ &n=\kappa ~~~\text{for non-dissipative (isentropic) processes} \nonumber \\
&n < \kappa ~~~\text{for dissipative (polytropic) expansion processes} \nonumber \\
&n > \kappa ~~~\text{for dissipative (polytropic) compression processes} \nonumber \\
\end{align}

With the use of these polytropic equations, the work W done by the system at a given initial temperature T₁ can also be determined by the pressure or volume ratio:

\begin{align}
\label{1}
&W=c_\text{v}~m~\left[T_2-T_1 \right] = c_\text{v}~m~T_1~\left[{T_2 \over T_1}-1 \right] \\[5px]
&\boxed{W=c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{n-1}}-1 \right]}~~~\text{only applies to adiabatic systems} \\[5px]
\label{3}
&\boxed{W=c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-n} \over n}}-1 \right]}~~~\text{only applies to adiabatic systems} \\[5px]
\text{ where: }~~ &n=\kappa ~~~\text{for non-dissipative (isentropic) processes} \nonumber \\
&n < \kappa ~~~\text{for dissipative (polytropic) processes} \nonumber \\
\end{align}

At this point, it should be explicitly mentioned that one should refrain from using the term “adiabatic” process. This is because both the isentropic process (n=κ) and the polytropic process (n<κ) take place in an adiabatic system. In this respect, all these processes are “adiabatic”, but they differ, of course, due to the different polytropic index n used to describe the respective processes (see also the article Free expansion of an ideal gas in a vacuum)!

Calculation of the polytropic index for dissipative processes

Which polytropic index n best describes a given dissipative process can be determined on the basis of the initial and final states. In practice, of course, this requires an experimental determination of the respective values. Then either equation (\ref{3475}), equation (\ref{2070}), or equation (\ref{9579}) must be solved for the polytropic index n:

\begin{align}
\label{6526}
\boxed{n
=\frac{\ln{\left(\frac{p_2}{p_1}\right)}} {\ln{\left(\frac{V_1}{V_2}\right)}}
={\frac{\ln{\left(\frac{T_2}{T_1}\right)}} {\ln{\left(\frac{V_1}{V_2}\right)}} +1}
=\frac{\ln{\left(\frac{p_2}{p_1}\right)}} {\ln{\left(\frac{T_1}{T_2}\right)} + \ln{\left(\frac{p_2}{p_1}\right)} }
} \\[5px]
\end{align}

Based on the given initial and final state, a polytropic index of n=1.1689 seems to best reflect the dissipative process shown in the diagram (green curve). The comparison in the diagram below between the approximated polytropic process and the actual process (red curve) shows very good agreement at the beginning and towards the end of the process; however, there are minor deviations in the area in between.

Calculation of dissipated energy

In the article Dissipation of energy in closed systems it has already been explained in detail that the work W transferred across the system boundary (boundary work) is the sum of the pressure-volume work W_v of the gas and the dissipated energy W_diss due to friction:

\begin{align}
\label{ums}
&\boxed{W = W_\text{v} + W_\text{diss}} ~~~\text{work transferred across the system boundary}
\end{align}

In the article Polytropic process in a closed system, it was shown that the pressure-volume work W_v for a polytropic process can be determined using the following formula:

\begin{align}
\label{wv}
&\boxed{W_\text{v} = \left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~\left(T_2-T_1 \right)}
\end{align}

Solving equation (\ref{ums}) for the dissipated W_diss energy and putting in equation (\ref{9067}) and equation (\ref{wv}), gives the dissipated energy:

\begin{alignat}{2}
\label{3440}
&W_\text{diss}= W – W_\text{v} = \underbrace{c_\text{v}~m~(T_2-T_1)}_{W} – \underbrace{\left[\frac{\kappa-1}{n-1}\right]~c_\text{v}~m~(T_2-T_1)}_{W_\text{v}} \\[5px]
\label{5715}
&\boxed{W_\text{diss}= \left[\frac{n-\kappa}{n-1}\right]~c_\text{v}~m~(T_2-T_1)} = \left[\frac{n-\kappa}{n-1}\right]~c_\text{v}~m~T_1~\left[{T_2 \over T_1}-1 \right] \\[5px]
\label{4461}
&\boxed{W_\text{diss}=\left[\frac{n-\kappa}{n-1}\right]~c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{n-1}}-1 \right]} \\[5px]
\label{2394}
&\boxed{W_\text{diss}=\left[\frac{n-\kappa}{n-1}\right]~c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-n} \over n}}-1 \right]} \\[5px]
\end{alignat}

A closer look at these equations shows that they are the same equations that describe the transferred heat in a (non-dissipative) polytropic process (see article Polytropic process in a closed system). However, for the dissipative process at hand, they can no longer be interpreted as transferred heat, since the process takes place in an adiabatic system. However, the identical equations can still be explained from logic, because from an energetic point of view the dissipated energy has the same effect as a heat transfer, namely an increase of the internal energy!

Note

Note that the work W done by/on a closed system according to equations (\ref{1}) to (\ref{3}) depends only on the initial and final state. These equations therefore reflect the actual boundary work, even if the polytropic equations do not accurately describe the actual process (see diagram below). This is due to the fact that the initial and final values agree with reality, provided that the polytropic index was selected according to equation (\ref{6526})!

However, the situation is different for the dissipated work W_diss. For its determination, according to equation (\ref{3440}), the pressure-volume work W_v is relevant, which results as the area under curve in the volume-pressure diagram. However, as can be seen in the diagram, there are deviations between the approximated polytropic process and the actual process. Therefore, equations (\ref{5715}) to (\ref{2394}) will generally not accurately account for the dissipated energy. In the present case, the area under the polytropic curve is larger than for the actual process and thus suggests a somewhat larger dissipated energy than will be the case in reality.

Particular processes shown in a volume-pressure diagram

The figure below shows the course of the pressure as a function of the volume for an isobaric, isochoric, isothermal and isentropic process. All processes take place starting from an initial volume V₁. The pressure curves show both an expansion and a compression. For the isentropic process, an isentropic index of air with κ=1.4 was used.

Although the shown thermodynamic processes appear at first glance to be very different from one another, a closer look reveals common characteristics. These become clear when one takes a closer look at the pressure as a function of the volume:

\begin{alignat}{3}
\text{isobaric process: } ~~~ &p=\text{constant} &&~~~~~~ \Rightarrow ~~~ &&& { p~V^0 =\text{constant}} \\[5px]
\text{isothermal process: } ~~~ &p\sim {1\over V} &&~~~~~~ \Rightarrow ~~~ &&& { p~V^1 =\text{constant}} \\[5px]
\text{isentropic process: } ~~~ &p\sim {1\over V^{\kappa}} &&~~~~~~ \Rightarrow ~~~ &&& { p~V^{\kappa} =\text{constant}} \\[5px]
\text{isochoric process: } ~~~ &V=\text{constant} &&~~~~~~ \Rightarrow ~~~ &&& { p~V^{\infty} =\text{constant}} \\[5px]
\label{8858}
\end{alignat}

From the comparison it is clear that the entire processes shown can be generalized to obey the relation p⋅Vⁿ=constant, where a very specific exponent n can be found for each thermodynamic process:

\begin{align}
\label{3445}
& \boxed{p~V^{n} =\text{constant}} ~~~~~~ {p_1~V_1^n}={p_2~V_2^n} \\[5px]
\end{align}

Note: Why n=∞ holds for the isochoric process becomes clear if one first solves equation (\ref{3445}) with respects to the volume V and then sets n=∞:

\begin{align}
p~V^n=\text{const.} ~~~\Rightarrow ~~~
V ={\left(\text{const.} \over p \right)^{1 \over n}}
={\text{const.} \over p^{1 \over n}}
\overset{n=\infty}={\text{const.} \over \underbrace{p^0}_{=1}}
=\text{constant!} \nonumber \\[5px]
\end{align}

Polytropic equations

Not only the relationship between pressure and volume, but also the relationship between temperature and volume or temperature and pressure can be generalized. These relationships can be derived using the ideal gas law and equation (\ref{3445}). Thus, all special processes (isobaric, isochoric, isothermal and isentropic) treated so far can be reduced to obey the following equations:

\begin{align}
\label{2716}
\boxed{p~V^n=\text{constant}} &~~\Rightarrow~~ \boxed{p_1~V_1^n=p_2~V_2^n} \\[5px]
\label{4375}
\boxed{T~V^{n-1}=\text{constant}} &~~\Rightarrow~~ \boxed{T_1~V_1^{n-1}=T_2~V_2^{n-1}} \\[5px]
\label{7991}
\boxed{T^n~p^{1-n}=\text{constant}} &~~\Rightarrow~~ \boxed{T_1^n~p_1^{1-n}=T_2^n~p_2^{1-n}} \\[5px]
\text{ mit: }~~ &n=0 ~~~\text{isobaric process} \nonumber \\
&n=1 ~~~\text{isothermal process} \nonumber \\
&n=\kappa ~~~\text{isentropic process} \nonumber \\
&n=\infty ~~~\text{isochoric process} \nonumber \\
\end{align}

However, the exponent n is not limited to the special cases such as isobaric, isothermal, isentropic and isothermal. In principle, the exponent can be chosen arbitrarily! The choice depends decisively on which exponent can best describe the thermodynamic process in reality. After all, most thermodynamic processes in reality never run exactly isobaric, isothermal, isentropic or isochoric. The diagram below shows for selected values of n the respective pressure curves. Depending on which curve comes closest to the actual thermodynamic process, the corresponding exponent would then have to be chosen. In principle, negative exponents are also feasible!

Note that ultimately all the curves shown belong to the same set of curves described by the equation (\ref{2716}).

Since equations (\ref{2716}) to (\ref{7991}) can be used to describe a large number of thermodynamic processes (including special cases), the processes behind these equations are commonly referred to as polytropic processes (“poly”=many). The exponent n is therefore also called polytropic index. The isobaric, isothermal, isentropic and isochoric processes are therefore only to be regarded as special cases of a generally polytropic process!

Pressure-volume Work and heat

Besides the equations (\ref{2716}) to (\ref{7991}) describing the state of a gas by state variables, general formulas for calculating the process quantities such as pressure-volume work W_v and transferred heat Q, as well as for the change of the internal energy ΔU can also be found for polytropic processes. Due to their complexity, the derivation of the formulas given below will be discussed in more detail in the linked article. In this article only the formulas are discussed in more detail:

\begin{alignat}{2}
\label{4862}
&\boxed{W_\text{V} = \left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~\left(T_2-T_1 \right)}& &= \left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[\left(T_2\over T_1\right)-1 \right] \\[5px]
&&&= \left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{n-1}}-1 \right] \\[5px]
&&&= \left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-n} \over n}}-1 \right] \\[20px]
\label{5451}
&\boxed{Q = \left[{{n-\kappa} \over {n-1}}\right] ~c_\text{v}~m~\left(T_2-T_1 \right)}& &= \left[{{n-\kappa} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[\left(T_2\over T_1\right)-1 \right] \\[5px]
&&&= \left[{{n-\kappa} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{n-1}}-1 \right] \\[5px]
&&&= \left[{{n-\kappa} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-n} \over n}}-1 \right] \\[20px]
\label{9277}
&\boxed{\Delta U = c_\text{v}~m~\left(T_2-T_1 \right)}& &= c_\text{v}~m~T_1~\left[{T_2 \over T_1}-1 \right] \\[5px]
&&&= c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{n-1}}-1 \right] \\[5px]
&&&=c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-n} \over n}}-1 \right] \\[5px]
\end{alignat}

Note that the isentropic index κ in the above equations is to be understood only as the ratio of the heat capacities (κ=c_p/c_v). Therefore, one should not be deceived into thinking that these equations are valid only for an isentropic process, just because there is an isentropic index κ! In principle, these equations describe any polytropic process, the special cases like isobaric, isochoric, isothermal and isentropic of course included!

However, care must be taken at this point for the special case of the isothermal process with a polytropic index of n=1, since in this case the denominator in the equations would become zero. From a mathematical point of view, there is no solution for such a case. With sufficient accuracy, however, a polytropic index close to one can be chosen as an alternative, for example n=0.999.

Relationship between change in internal energy and work/heat

If we look more closely at equations (\ref{4862}) and (\ref{5451}), we notice that the term c_v⋅m⋅(T₂-T₁) just corresponds to the change in internal energy ΔU. Thus, the pressure-volume work W_v and the transferred heat Q can also be given as a function of the change of the internal energy ΔU:

\begin{align}
\label{6763}
&\boxed{W_V= \left[{{\kappa-1} \over {n-1}}\right] ~\Delta U} \\[5px]
&\boxed{Q= \left[{{n-\kappa} \over {n-1}}\right] ~\Delta U} \\[5px]
\end{align}

It is now obvious that transferred work and heat for a given polytropic process are always in a constant ratio:

\begin{align}
\require{cancel}
\label{3657}
{W_V \over Q} = {{\left[{{\kappa-1} \over \bcancel{n-1}}\right]~\bcancel{\Delta U}} \over {\left[{{n-\kappa} \over \bcancel{n-1}}\right]~\bcancel{\Delta U}}}
= { {\kappa-1} \over {n-\kappa} } \\[5px]
\end{align}

\begin{align}
\label{3285}
\boxed{{ W_V \over Q } = { {\kappa-1} \over {n-\kappa} }} \\[5px]
\end{align}

The diagram below shows for an isentropic index of κ=1.4 the (positive) ratio of work and heat as a function of the polytropic index.

Isobaric process

For any isobaric process with n=0 and κ=1.4 (for air), the quotient of work and heat is (-)0.286. This means that for an isobaric expansion, 28.6% of the supplied heat is converted into (pressure-volume) work. The remaining 71.4% of heat energy benefits the internal energy. This result is independent of the pressure, volume or temperature at which the isobaric process takes place! Only the type of gas influences the efficiency with which the supplied heat is converted into work. For example, the share of the converted pressure-volume work can be increased to 40.1% if helium with an isentropic index of κ=1.67 is used instead of air!

Isentropic process

For a thermodynamic process described with a polytropic index of n<κ, the result of equation (\ref{3285}) is negative. This means that, mathematically speaking, work and heat have opposite signs. Thus, heat must be added to the gas (Q>0) while the gas expands (W_v<0). For the reverse case, heat must be released by the gas (Q<0) when the gas is compressed (W_v>0).

For the special case of a process with an isentropic index n→κ, the denominator in the equation (\ref{3285}) approaches more and more zero. This means that the transferred heat (in relation to the pressure-volume work) decreases and is finally zero at n=κ. For this special case, the isentropic process is obtained, which is exactly characterized by this fact that no heat is transferred.

\begin{align}
{ W_V \over Q } = { {\kappa-1} \over {n-\kappa} } \overset{n\rightarrow\kappa}= \infty
~~~\Rightarrow~~~Q= {W_V\over \infty} = 0
\nonumber \\[5px]
\end{align}

Isochoric process

On the other hand, for a thermodynamic process with an isentropic index n→∞, the denominator in the equation (\ref{3285}) approaches more and more infinity. This means that the transferred work (in relation to the transferred heat) decreases and is finally zero at n=∞. For this special case, the isochoric process is obtained, which is exactly characterized by this fact that no work is transferred.

\begin{align}
{ W_V \over Q } = { {\kappa-1} \over {n-\kappa} } \overset{n\rightarrow\infty}= 0
~~~\Rightarrow~~~W_V= {Q \cdot 0} = 0
\nonumber \\[5px]
\end{align}

Isothermal process

Another special case is shown by the equation (\ref{3285}) for n=1. In this case, the magnitude of the transferred work is as large as the magnitude of the transferred heat (with opposite signs). Thus, the heat absorbed by the gas is completely converted into work and released by the gas. Therefore, the internal energy of the gas is neither reduced nor increased, i.e. it remains constant. For an ideal gas this means that temperature remains constant. The special case n=1 thus corresponds to the isothermal process, which is exactly characterized by this fact that the temperature remains constant.

\begin{align}
{ W_V \over Q } = { {\kappa-1} \over {n-\kappa} } \overset{n\rightarrow 1}= -1
~~~\Rightarrow~~~W_V= -Q
\nonumber \\[5px]
\end{align}

Importance of polytropic processes in real life

As already explained in detail, the polytropic equations (\ref{2716}) to (\ref{7991}) can describe, depending on the polytropic index n, all special cases considered so far, such as the isobaric (n=0), isothermal (n=1), isentropic (n=κ) and isochoric process (n=∞). Also, a variety of other thermodynamic processes can be described depending on the chosen polytropic index. Nevertheless, not all processes obey these polytropic equations (e.g. the expansion process described in this article, in which a weight is lifted by a lever via a rack and pinion).

Nevertheless, polytropic processes are of great technical importance, especially in the range between n=1 (isothermal process) and n=κ (isentropic process). This is because many thermodynamic processes occur between these limiting cases. This becomes clear if one takes a closer look at these thermodynamic processes.

As an example, the compression of air inside an air pump is considered with the outlet valve kept closed. If the compression is carried out (infinitely) slowly, any increase in temperature is immediately compensated by heat dissipation to the surroundings. Thus, the temperature of the gas will not increase during such a slow process. Therefore, an (infinitely) slow thermodynamic process represents an isothermal process – at least the described process for compressing air with the help of an air pump.

However, if the compression of air is performed (infinitely) fast, the temperature increase cannot be compensated, since the system has no time to transfer any significant amount of heat to the surroundings. Therefore, an (infinitely) fast thermodynamic process represents an isentropic process.

In reality, however, a thermodynamic process can be neither infinitely slow nor infinitely fast, so that the actual process will run between these two limiting cases. Consequently, the polytropic index will have a value between 1 (isothermal process) and κ (isentropic process). Often, the term polytropic process is limited to exactly those thermodynamic processes whose polytropic indices lie in the range between n=1 and n=κ. Especially in cases where energy is dissipated, the thermodynamic process is often approximated by such a polytropic process. More information on this can also be found in the article Dissipative thermodynamic processes in adiabatic systems.

Polytropic equations

In the article Polytropic Process in a closed system, the following equations describing polytropic processes were derived and discussed in more detail:

\begin{align}
\label{2716}
\boxed{p \cdot V^n=\text{constant}} &~~\Rightarrow~~ \boxed{p_1 \cdot V_1^n=p_2 \cdot V_2^n} \\[5px]
\label{4375}
\boxed{T \cdot V^{n-1}=\text{constant}} &~~\Rightarrow~~ \boxed{T_1 \cdot V_1^{n-1}=T_2 \cdot V_2^{n-1}} \\[5px]
\label{7991}
\boxed{T^n \cdot p^{1-n}=\text{constant}} &~~\Rightarrow~~ \boxed{T_1^n \cdot p_1^{1-n}=T_2^n \cdot p_2^{1-n}} \\[5px]
\end{align}

The aim of this article is to derive the formulas for calculating work and transferred heat for polytropic thermodynamic processes.

Derivation of the formula for calculating pressure-volume work

The pressure-volume work W_v can generally be calculated by integrating the p(V) function of the polytropic process:

\begin{align}
\label{www}
&\boxed{W_\text{v} = – \int\limits_{V_1}^{V_2} p(V) ~ \text{d}V}\\[5px]
\end{align}

Thus, for calculating the work, the function p(V) must be known. If a polytropic process is mathematically described starting from an initial state 1 with pressure p₁ and volume V₁, equation (\ref{2716}) establishes a relationship to any further state 2 with pressure p and volume V. From this, the pressure function p(V) can be determined:

\begin{align}
\label{9891}
&p_1~V_1^n=p~V^n \\[5px]
\label{1227}
&p = \underbrace{p_1~V_1^n}_{=\text{konstant}} \cdot {1 \over V^n} \\[5px]
\label{9377}
&\boxed{p(V)=p_1~V_1^n \cdot {1 \over V^n}} \\[5px]
\end{align}

To calculate the pressure-volume W_v, the pressure function (\ref{9377}) can now be integrated within the limits V₁ to V₂:

\begin{align}
\label{9445}
W_\text{v} &= – \int\limits_{V_1}^{V_2} p(V) ~ \text{d}V \\[5px]
&= – \int\limits_{V_1}^{V_2} \underbrace{p_1~V_1^n}_{=\text{constant}} \cdot {1 \over V^n} ~ \text{d}V \\[5px]
&= – p_1~V_1^n~\int\limits_{V_1}^{V_2} {V^{-n}} ~ \text{d}V \\[5px]
&= – p_1~V_1^n~ \left[{1 \over {1-n}}~V^{1-n} \right]_{V_1}^{V_2} \\[5px]
&= – {{p_1~V_1^n} \over {1-n}} ~ \left[V_2^{1-n}-V_1^{1-n} \right] \\[5px]
&= {{p_1~V_1^n} \over {n-1}} ~ \left[V_2^{1-n}-V_1^{1-n} \right] ~~~ \text{factoring out } V_1^{1-n} \text{:} \\[5px]
&= {{p_1~V_1^n~V_1^{1-n}} \over {n-1}} ~ \left[{V_2^{1-n} \over V_1^{1-n}} -1 \right] \\[5px]
\end{align}

\begin{align}
\label{8254}
\boxed{W_\text{V}= {{p_1~V_1} \over {n-1}} ~ \left[\left(V_2 \over V_1 \right)^{1-n} -1 \right]}
\end{align}

The pressure-volume work can be determined relatively easily, especially on the basis of the temperatures. For this purpose, equation (\ref{4375}) is combined with equation (\ref{8254}) and the ideal gas law p⋅V=m⋅R_s⋅T is used:

\begin{align}
W_\text{v} &= {\overbrace{p_1~V_1}^{=m~R_\text{s}~T_1}\over {n-1}} ~ \left[\left(V_2 \over V_1 \right)^{1-n} -1 \right] \text{where: } ~{V_2 \over V_1} = \left(T_1 \over T_2 \right)^{1 \over {n-1}}=\left(T_2 \over T_1 \right)^{1 \over {1-n}}~ \text{ : } \\[5px]
&= {{m~R_\text{s}~T_1} \over {n-1}} ~ \left[{T_2 \over T_1} -1 \right] \\[5px]
\label{m}
&= {{R_\text{s} \over {n-1}}} ~m~ \left(T_2 – T_1 \right) ~~~\text{where } \underline{R_\text{s}=c_\text{p}-c_\text{v}} ~~~\text{:} \\[5px]
&= {{{c_\text{p}-c_\text{v}} \over {n-1}}} ~m~ \left(T_2 – T_1 \right) \\[5px]
&= {{{{c_\text{p}\over c_\text{v}}-1} \over {n-1}}} ~c_\text{v}~m~ \left(T_2 – T_1 \right) ~~~\text{where } \underline{\kappa={c_\text{p} \over c_\text{v}}} ~~~\text{:} \\[5px]
\label{eq:9418}
\end{align}

\begin{align}
\boxed{W_\text{V}=\left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~ \left(T_2 – T_1 \right)}
\label{2604}
\end{align}

Derivation of the formula for calculating transferred heat

According to the first law of thermodynamics, the transferred heat results from the difference between the change in internal energy ΔU and the pressure-volume work W_v:

\begin{align}\
&\boxed{ Q + W_\text{v} = \Delta U} ~~~~\text{first law of thermodynamics}\\[5px]
\label{put}
&Q = \Delta U – W_\text{v}\\[5px]
\end{align}

For ideal gases, the change in internal energy ΔU results – regardless of the thermodynamic process (!) – only on the basis of the temperature change ΔT=T₂-T₁:

\begin{align}\
\label{9719}
&\boxed{ \Delta U = c_\text{v}~m~\left(T_2-T_1 \right)} \\[5px]
\end{align}

If the change in internal energy given by equation (\ref{9719}) and the pressure-volume work according to equation (\ref{m}) are put in equation (\ref{put}), the following formula is obtained for calculating the transferred heat Q of a polytropic process:

\begin{align}
Q &= \Delta U – W_\text{V} \\[5px]
&= c_\text{v}~m~\left(T_2-T_1 \right) – {R_\text{S} \over {n-1}} ~m~ \left(T_2 – T_1 \right) \\[5px]
&= \left[c_\text{v}-{R_\text{S} \over {n-1}}\right]~m~\left(T_2-T_1 \right) ~~~\text{where } \underline{R_\text{S}=c_\text{p}-c_\text{v}} ~~~\text{:} \\[5px]
&= \left[c_\text{v}-{{c_\text{p}-c_\text{v}} \over {n-1}}\right]~m~\left(T_2-T_1 \right) \\[5px]
&= \left[{{c_\text{v}~(n-1)}\over {n-1}}-{{c_\text{p}-c_\text{v}} \over {n-1}}\right]~m~\left(T_2-T_1 \right) \\[5px]
&= \left[{{c_\text{v}~n-c_\text{v}-c_\text{p}+c_\text{v}} \over {n-1}}\right]~m~\left(T_2-T_1 \right) \\[5px]
&= \left[{{c_\text{v}~n-c_\text{p}} \over {n-1}}\right]~m~\left(T_2-T_1 \right) \\[5px]
&= \left[{{n-{c_\text{p} \over c_\text{v}}} \over {n-1}}\right]~c_\text{v}~m~\left(T_2-T_1 \right) ~~~\text{where } \underline{\kappa={c_\text{p} \over c_\text{v}}} ~~~\text{:} \\[5px]
\label{eq:7090}
\end{align}

\begin{align}
\boxed{Q= \left[{{n-\kappa} \over {n-1}}\right]~c_\text{v}~m~\left(T_2-T_1 \right)}
\label{eq:8691}
\end{align}

Summary

The derived formulas for the calculation of heat, work and change of internal energy, can also be expressed by the volume and pressure ratios using equation (\ref{4375}) and equation (\ref{7991}), respectively:

\begin{alignat}{2}
\label{6005}
&{W_\text{v} = \left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~\left(T_2-T_1 \right)}& &= \left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[\left(T_2\over T_1\right)-1 \right] \\[5px]
&&&= \left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{n-1}}-1 \right] \\[5px]
&&&= \left[{{\kappa-1} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-n} \over n}}-1 \right] \\[20px]
\label{1733}
&{Q = \left[{{n-\kappa} \over {n-1}}\right] ~c_\text{v}~m~\left(T_2-T_1 \right)}& &= \left[{{n-\kappa} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[\left(T_2\over T_1\right)-1 \right] \\[5px]
&&&= \left[{{n-\kappa} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{n-1}}-1 \right] \\[5px]
&&&= \left[{{n-\kappa} \over {n-1}}\right] ~c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-n} \over n}}-1 \right] \\[20px]
&{\Delta U = c_\text{v}~m~\left(T_2-T_1 \right)}& &= c_\text{v}~m~T_1~\left[{T_2 \over T_1}-1 \right] \\[5px]
&&&= c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{n-1}}-1 \right] \\[5px]
&&&=c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-n} \over n}}-1 \right] \\[5px]
\end{alignat}

If we look more closely at equations (\ref{6005}) and (\ref{1733}), we notice that the term c_v⋅m⋅(T₂-T₁) just corresponds to the change in internal energy ΔU. Thus, the pressure-volume work W_v and the transferred heat Q can also be given as a function of the change in internal energy ΔU:

\begin{align}
\label{6763}
&\boxed{W_\text{v}= \left[{{\kappa-1} \over {n-1}}\right] ~\Delta U} \\[5px]
&\boxed{Q= \left[{{n-\kappa} \over {n-1}}\right] ~\Delta U} \\[5px]
\end{align}

Basic equations

For the derivation of the equations describing the isentropic process, the first law of thermodynamics is used with the restriction that in an isentropic process, by definition, no heat is transferred (Q=0):

\begin{align}
&\boxed{W_\text{V} + Q = \Delta U} ~~~~~&&\text{First Law of Thermodynamics}\\[5px]
\label{wv}
&W_\text{V} = \Delta U ~~~~&&\text{only applies to an isentropic process where }Q=0\\[5px]
\end{align}

The pressure-volume work W_v is determined by integrating the pressure function with respect to the volume (W_v=-∫p⋅dV), whereas the change in internal energy ΔU is determined by the temperature difference (ΔU=cv⋅m⋅ΔT). Thus, the following relationship applies between the change in volume and the change in temperature in an isentropic process:

\begin{align}
&W_\text{v} = \Delta U \\[5px]
\label{eq:5561}
&\boxed{- \int p~\text{d}V =c_\text{v}~m~\Delta T} ~~~~~\text{integral form }\\[5px]
\end{align}

If only infinitesimal changes in volume and temperature are considered at this point (within which the pressure can be considered constant), the pressure-volume work and the change in internal energy can be expressed by the respective differentials (Δ→d):

\begin{align}
\label{eq:9887}
&\text{d}W_\text{V} = \text{d}U \\[5px]
\label{pv}
&\boxed{-p~\text{d}V = c_\text{v}~m~\text{d}T} ~~~~~\text{differential form}\\[5px]
\end{align}

Relationship between volume and temperature

Equation (\ref{pv}) provides the relationship between an infinitesimal volume change dV and the resulting temperature change dT for the isentropic process. However, the pressure p is not independent of the volume V or the temperature T, as can be seen from the ideal gas law:

\begin{align}
\label{0}
&p~V=R_\text{s}~m~T ~~~~~\text{ideal gas law} \\[5px]
\label{p}
&\boxed{p=R_\text{s}~m~{T \over V}} \\[5px]
\end{align}

If the pressure function according to equation (\ref{p}) is used in equation (\ref{pv}), then the variables belonging to their differential can be separated (called separation of variables):

\begin{align}
\require{cancel}
– \overbrace{R_\text{s}~\bcancel{m}~{T \over V}}^{=p}~\text{d}V &= c_\text{v}~\bcancel{m}~\text{d}T \\[5px]
\label{tv}
– R_\text{s}~{1 \over V}~\text{d}V &= c_\text{v}~{1 \over T}~\text{d}T \\[5px]
\end{align}

Now that the variables are separated, this equation can be integrated within the respective limits V₁ to V₂ and T₁ to T₂:

\begin{align}
\label{eq:9552}
– R_\text{s}~\int\limits_{V_1}^{V_2} {1 \over V}~\text{d}V &= c_\text{v}~\int\limits_{V_1}^{V_2} {1 \over T}~\text{d}T \\[5px]
 R_\text{s}~\left[ \ln{(V)} \right]^{V_2}_{V_1} &= c_\text{v}~\left[ \ln{(T)} \right]^{T_2}_{T_1} \\[5px]
– R_\text{s}~\ln \left(V_2 – V_1 \right) &= c_\text{v}~\ln \left(T_2 – T_1 \right) \\[5px]
R_\text{s}~\ln \left(V_1 – V_2 \right) &= c_\text{v}~\ln \left(T_2 – T_1 \right) ~~~\text{where}~~~ \underline{\ln(a)-\ln(b)=\ln \left(a \over b \right)} ~~~\text{:} \\[5px]
R_\text{s}~\ln \left(V_1 \over V_2 \right) &= c_\text{v}~\ln \left(T_2 \over T_1 \right) ~~~\text{where}~~~ \underline{R_\text{s}=c_\text{p}-c_\text{v}} ~~~\text{:} \\[5px]
(c_\text{p}-c_\text{v})~\ln \left(V_1 \over V_2 \right) &= c_\text{v}~\ln \left(T_2 \over T_1 \right) \\[5px]
\label{fre}
\left(\frac{c_\text{p}}{c_\text{v}}-1\right)~\ln \left(V_1 \over V_2 \right) &= \ln \left(T_2 \over T_1 \right) \\[5px]
\end{align}

At this point, the ratio of the specific heat capacities c_p/c_v form a new constant, the so-called adiabatic index κ. This constant serves at this point only the simpler notation!

\begin{align}
&\boxed{\kappa:=\frac{c_\text{p}}{c_\text{v}}} ~~~\text{heat capacity ratio (adiabatic index)}\\[5px]
\end{align}

With this definition of the adiabatic index, equation (\ref{fre}) can be written as follows:

\begin{align}
\label{k}
&(\kappa – 1)~\ln \left(V_1 \over V_2 \right) = \ln \left(T_2 \over T_1 \right) \\[5px]
\end{align}

Equation (\ref{k}) already gives a clear relationship between any initial state 1 (V₁, T₁) and any final state 2 (V₂, T₂) for the isentropic process. With the help of different logarithm rules, however, this equation can be simplified:

\begin{align}
\label{eq:3049}
e^{(\kappa-1)\ln \left(V_1 \over V_2 \right)} &= e^{\ln \left(T_2 \over T_1 \right)} \\[5px]
\left[e^{\ln \left(V_1 \over V_2 \right)}\right]^{\kappa-1} &= e^{\ln \left(T_2 \over T_1 \right)} \\[5px]
\label{w}
\left(V_1 \over V_2 \right)^{\kappa-1} &= {T_2 \over T_1} \\[5px]
\end{align}

\begin{equation}
\boxed{T_1 \cdot V_1^{\kappa-1}= T_2 \cdot V_2^{\kappa-1}} ~~~\text{bzw.}~~~ \boxed{T \cdot V^{\kappa-1}=\text{constant}}
\label{eq:5377}
\end{equation}

Finally, for an isentropic process the product of temperature and the volume exponentiated by κ-1 is constant!

Relationship between temperature and pressure

If the general relationship of two states obtained from the ideal gas law is used at this point, a clear relationship between the temperatures and the pressures can also be found:

\begin{align}
\label{a}
&\boxed{{p_1~V_1 \over T_1}={p_2~V_2 \over T_2} }\\[5px]
&{V_1 \over V_2} = {p_2 \over p_1}~{T_1 \over T_2} \\[5px]
\end{align}

If this ratio of volumes is used in equation (\ref{w}), the following relationship is obtained between the temperatures and the pressures:

\begin{align}
&{\underbrace{\left({p_2 \over p_1}~{T_1 \over T_2}\right)}_{\frac{V_1}{V_2}}}^{\kappa-1} = {T_2 \over T_1} \\[5px]
&\left(p_2 \over p_1\right)^{\kappa-1}~\left(T_1 \over T_2 \right)^{\kappa-1} = {T_2 \over T_1} \\[5px]
&\left(p_2 \over p_1\right)^{\kappa-1} = {T_2 \over T_1}~\left(T_2 \over T_1 \right)^{\kappa-1} \\[5px]
&\left(p_2 \over p_1\right)^{\kappa-1} = \left(T_2 \over T_1 \right)^{\kappa} \\[5px]
&\left(p_1 \over p_2\right)^{1-\kappa} = \left(T_2 \over T_1 \right)^{\kappa} \\[5px]
& \boxed{T_1^\kappa \cdot p_1^{1-\kappa} = T_2^\kappa \cdot p_2^{1-\kappa}} ~~~\text{bzw.}~~~ \boxed{T^\kappa \cdot p^{1-\kappa}=\text{constant}} \\[5px]
\end{align}

Relationship between volume and pressure

In the analogous way, a relation between volumes and pressures can be derived if equation (\ref{a}) is first solved for the ratio of temperatures:

\begin{align}
&\boxed{{p_1~V_1 \over T_1}={p_2~V_2 \over T_2} }\\[5px]
&\frac{T_2}{T_1} = {p_2 \over p_1}~{V_2 \over V_1} \\[5px]
\end{align}

If this ratio of temperatures is used in equation (\ref{w}), the following relationship is obtained between the temperatures and the volumes:

\begin{align}
&\left(V_1 \over V_2 \right)^{\kappa-1} = \underbrace{{p_2 \over p_1}~{V_2 \over V_1}}_{\frac{T_2}{T_1}} \\[5px]
&\left(V_1 \over V_2 \right)~\left(V_1 \over V_2 \right)^{\kappa-1} = {p_2 \over p_1} \\[5px]
&\left(V_1 \over V_2 \right)^{\kappa} = {p_2 \over p_1} \\[5px]
&\boxed{p_1 \cdot V_1^\kappa = p_2 \cdot V_2^\kappa} ~~~\text{bzw.}~~~ \boxed{p \cdot V^\kappa=\text{constant}} \\[5px]
\end{align}

Summary of the equations

Finally, for an isentropic process, all the relationships between the various state variables were derived and are summarized again below:

\begin{alignat}{3}
&p \cdot V^\kappa = \text{constant}& &\Rightarrow&~ &\boxed{ p_1~V_1^\kappa = p_2~V_2^\kappa} \\[5px]
\label{3}
&T \cdot V^{\kappa-1} = \text{constant}& &\Rightarrow&~ &\boxed{ T_1~V_1^{\kappa-1} = T_2~V_2^{\kappa-1}} \\[5px]
\label{4}
&T^\kappa \cdot p^{1-\kappa} = \text{constant}& &\Rightarrow&~ &\boxed{ T_1^\kappa~p_1^{1-\kappa} = T_2^\kappa~p_2^{1-\kappa}} \\[5px]
\label{1}
&&&&&\text{ where } \boxed{\kappa = {c_\text{p} \over c_\text{v}}}>1 \\[5px]
\end{alignat} 

Definition

Whereas in an isochoric process no pressure-volume work is done by the system or on the system (W_v=0) and in an isothermal process no change in internal energy takes place (ΔU=0), in a so-called isentropic process, by definition, no heat is transferred across the system boundary (Q=0). Thus, an isentropic process only takes place in an adiabatic system! For this reason, the isentropic process is also referred to as an adiabatic process.

How to achieve an approximately isentropic process

Basically, no system is perfectly thermally insulated, so that an isentropic process can only be approximated. But even if systems are not particularly well thermally insulated, thermodynamic processes can still be regarded as approximately isentropic under certain conditions. This is the case, for example, when thermodynamic processes occur so rapidly that the system has no time to transfer heat across the system boundary. For this reason, the compression and expansion of the fuel-air mixture in internal combustion engines, for example, which take place within a few milliseconds, are often considered isentropic.

A closed adiabatic system can be realized approximately with a gas-filled cylinder, which is closed with a piston. If the gas is now compressed very quickly, heat transfer to the surroundings can be neglected and an approximate adiabatic system is obtained for the time the compression takes place. Ideally, the walls of the cylinder are still made of a heat-insulating or poorly heat-conducting material.

In practice, the rapid compression of air in an air pump is also approximately an isentropic process when the outlet valve is closed. After compression, however, a drop in pressure is quickly observed, since heat is then transferred from the enclosed air to the inner walls of the air pump. The temperature and thus the pressure drop. The air pump then no longer represents an adiabatic system.

Illustration in the volume-pressure diagram

If a gas is compressed abruptly in an adiabatic system, there is an increase in pressure due to the reduction in volume. However, this is also associated with an increase in temperature. This can be explained by the increased momentum (increased speed) of the gas molecules due to the collisions between the rapidly moving piston and the molecules (see also the article Why do pressure and temperature increase during the compression of a gas?) In diesel engines, the temperature increase during compression, which is considered to be isentropic, is deliberately desired, as this causes the ignition temperature of the diesel fuel to be exceeded and the fuel to ignite (self-ignition).

While in the case of isothermal compression a temperature increase is compensated by heat dissipation, in the case of isentropic compression there is, by definition, no heat transfer. Therefore, the increase in pressure is greater for an isentropic process than for an isothermal process. This is because while in isothermal compression the pressure increase results only from the reduction in volume (the temperature remains constant), in isentropic compression there is an temperature increase. This temperature increase further increases the pressure.

Conversely, the pressure drops to a lower value in an isentropic expansion compared to an isothermal expansion. This is because in order to keep the temperature constant during an isothermal expansion, heat must be supplied to the gas. The molecules obviously maintain their kinetic energy due to the constant temperature (constant internal energy). In an isentropic expansion, on the other hand, no heat is transferred to the gas to keep the temperature constant. Therefore, the temperature decreases as the volume increases. Thus, in isentropic expansion, the decrease in pressure results not only from the increase in volume, but also from the decrease in temperature.

The pressure curve of an isentropic process is steeper in the volume-pressure diagram compared to an isothermal process!

Relationship between volume, pressure and temperature

As long as the gas is considered to be an ideal gas, the ideal gas law still applies, regardless of the exact thermodynamic process. Therefore, for an isentropic process in a closed system, two gas states are related by the following equation:

\begin{align}
&\boxed{p \cdot V = R_\text{s} \cdot m \cdot T} ~~~\text{ideal gas law} \\[5px]
&{p \cdot V \over T}=R_\text{s} \cdot m = \text{constant} \\[5px]
\label{kurz}
\Rightarrow~ &\boxed{{p_1~V_1 \over T_1}={p_2~V_2 \over T_2} }\\[5px]
\end{align}

While at this point for an isobaric process the constant pressure, for an isochoric process the constant volume and for an isothermal process the constant temperature can be cancelled from the equation (\ref{kurz}), now for an isentropic process no constant quantities can be found. However, clear relationships between two quantities can be derived if the ideal gas law and the first law of thermodynamics are taken into account with the restriction that now no heat is transferred (see article Derivation of the formulas of the isentropic “adiabatic” process). This results in the following relationships:

\begin{alignat}{3}
\label{2}
&p \cdot V^\kappa = \text{konstant}& &\Rightarrow&~ &\boxed{ p_1~V_1^\kappa = p_2~V_2^\kappa} \\[5px]
\label{3}
&T \cdot V^{\kappa-1} = \text{konstant}& &\Rightarrow&~ &\boxed{ T_1~V_1^{\kappa-1} = T_2~V_2^{\kappa-1}} \\[5px]
\label{4}
&T^\kappa \cdot p^{1-\kappa} = \text{konstant}& &\Rightarrow&~ &\boxed{ T_1^\kappa~p_1^{1-\kappa} = T_2^\kappa~p_2^{1-\kappa}} \\[5px]
\label{1}
&&&&&\text{ mit } \boxed{\kappa = {c_\text{p} \over c_\text{v}}}>1 \\[5px]
\end{alignat}

Since the ratio of the specific heat capacities c_p and c_v occurs very frequently when deriving these equations, this quotent is conveniently combined to form the so-called heat capacity ratio κ, also known as adiabatic index or Laplace’s coefficient. This ratio is unitless and always greater than 1, since the specific heat capacity c_p is always greater than c_v (see also the article Specific heat capacity of gases).

If we consider the relationship between pressure and volume according to equation (\ref{2}), it is now also mathematically clear that the pressure curve in the volume-pressure diagram is steeper for an isentropic process than for an isothermal process:

\begin{align}
&p \cdot V^\kappa = \text{konstant} \\[5px]
&\boxed{p \sim {1 \over V^\kappa}} ~~~\text{isentropic process}\\[5px]
&\boxed{p \sim {1 \over V}} ~~~\text{isothermal process}\\[5px]
\end{align}

For an isothermal process, the pressure decreases with increasing volume according to the law p~1/V. For an isentropic process, on the other hand, the pressure decreases according to the law p~1/V^κ. Since the adiabatic index κ is always greater than 1, the pressure obviously decreases more rapidly with an isentropic increase in volume than with the isothermal process.

Calculation of the change in internal energy

For ideal gases, the change in internal energy ΔU results independent of the thermodynamic process only on the basis of the temperature change ΔT=T₂-T₁:

\begin{align}
\label{a}
&\boxed{ \Delta U = c_\text{v}~m~\left(T_2-T_1 \right)} \\[5px]
\end{align}

If this equation is solved for the temperature ratio T₂/T₁, the change in internal energy at a given initial temperature T₁ can also be determined by the volume ratio or the pressure ratio by linking with equation (\ref{3}) or equation (\ref{4}):

\begin{align}
& \Delta U = c_\text{v}~m~\left(T_2-T_1 \right) \\[5px]
& \Delta U = c_\text{v}~m~T_1~\left[{T_2 \over T_1}-1 \right] \\[5px]
& \boxed{\Delta U = c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{\kappa-1}}-1 \right]} \\[5px]
\label{c}
& \boxed{\Delta U = c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-\kappa} \over \kappa}}-1 \right]} \\[5px]
\end{align}

Calculation of the pressure-volume work

As in any (reversible) thermodynamic process, the pressure-volume work of the gas is given by the area under the curve in the volume-pressure diagram. This area can basically be determined by integrating the pressure function p(V). However, a closer look at the first law of thermodynamics shows that under the constraint Q=0 the pressure-volume work just corresponds to the change of the internal energy:

\begin{align}
&\boxed{W_\text{v} + Q = \Delta U} ~~~\text{First Law of Thermodynamics}\\[5px]
&\boxed{W_\text{v} = \Delta U} \\[5px]
\end{align}

The fact that the change of the internal energy just corresponds to the pressure-volume work also becomes clear very quickly. Due to the lack of transferred heat, the pressure-volume work done on the gas by compression must completely benefit the internal energy. Conversely, a pressure-volume work done by the gas is carried out completely at the expense of the internal energy, since it obviously cannot originate from a heat transfer. To calculate the pressure-volume work W_v during an isentropic process, the equations (\ref{a}) to (\ref{c}) can be used:

\begin{align}
& W_\text{v} = c_\text{v}~m~\left(T_2-T_1 \right) \\[5px]
&W_\text{v} = c_\text{v}~m~T_1~\left[{\left(V_1 \over V_2 \right)^{\kappa-1}}-1 \right] \\[5px]
&W_\text{v} = c_\text{v}~m~T_1~\left[{\left(p_1 \over p_2 \right)^{{1-\kappa} \over \kappa}}-1 \right] \\[5px]
\end{align}

Important note

Attention: Not all thermodynamic processes that take place in an adiabatic system must necessarily obey the equations explained in this article! Therefore, the term adiabatic process is somewhat misleading. The strict definition of the isentropic process is as follows:

An isentropic process is a reversible change of state of an adiabatic system!

At this point the term reversible is crucial! More information, especially about this topic, can also be found in the article Free expansion of an ideal gas in a vacuum.

A change of state of a gas in which the temperature does not change is also called an isothermal process. This means that the same temperature applies to all the states through which the gas passes between the initial and final state.

How to achieve an isothermal process

An isothermal process in a closed system takes place, for example, when any temperature changes are immediately compensated by absorbing or releasing heat. If an air pump is compressed while the outlet valve is kept closed, the gas inside is compressed. This is normally associated with an increase in temperature. However, if the gas is cooled at the same time, the temperature can be kept constant during compression. Conversely, in an isothermal expansion, heat must be supplied to the gas to counteract the temperature decrease that would otherwise occur.

An isothermal process can approximately be realized by allowing the expansion or compression of the gas to proceed so slowly that any temperature changes that would otherwise occur are very quickly compensated for by the surroundings (through releasing heat by the gas or through absorbing heat by the gas). Just imagine compressing the air of an air pump not within a few seconds, but over several minutes or hours. In this case, there will be no noticeable increase in temperature during compression and the process can be considered quasi-isothermal.

Illustration in the volume-pressure diagram

If a gas is isothermally compressed at a temperature T, the gas volume will decrease from V₁ to V₂. This is associated with an increase in pressure from p₁ to p₂. According to the ideal gas law, pressure and volume behave inversely proportional in an isothermal process. The exact relationship between pressure p and volume V is obtained by solving the ideal gas law with respect to the pressure p:

\begin{align}
\label{eq:6942}
&p~V = R_\text{s} \cdot m \cdot T ~~~\text{ideal gas law}\\[5px]
\label{pv}
&\boxed{p ={ \underbrace{R_\text{s}~m~T}_{=\text{constant}} \cdot {1 \over V}}} ~\Rightarrow~ p \sim {1 \over V}
\end{align}

Note that besides the specific gas constant R_s and the mass m (closed system), now also the temperature T is constant. Thus, of course, the product of all three quantities is also constant (Rs⋅m⋅T = constant). Therefore, the isothermal process is represented as a hyperbola in the p(V) diagram.

Relationship between pressure and volume (law of Boyle-Mariotte)

In an isothermal process, the pressure decreases inversely proportionally with increasing volume. A doubling of the volume (expansion) thus means a halving of the gas pressure. Conversely, a halving of the volume (compression) means a doubling of the pressure. This inverse proportionality is equivalent to statement that the product of volume and pressure is always constant in an isothermal process. This phenomenon is also known as law of Boyle-Mariotte:

\begin{align}
p \sim {1 \over V} ~\Rightarrow~ {p \cdot V} = \text{constant}
\end{align}

Any two states within an isothermal process are consequently linked by the constant product of pressure and volume:

\begin{equation}
\label{3} 
\boxed{ {p_1 \cdot V_1} = {p_2 \cdot V_2} } ~\text{law of Boyle-Mariotte}
\end{equation}

Calculation of the change in internal energy

The change of the internal energy ΔU of the gas is directly linked to the temperature change ΔT=T₂-T₁. However, since the temperature does not change in an isothermal process (T₁=T₂), the internal energy does not change either. This is also already evident from the physical interpretation of temperature as a measure of the mean kinetic energy of a gas molecule. Because if the temperature does not change, then the speeds of the gas molecules do not change either and the internal energy of the gas remains constant.

\begin{equation}
\boxed{ \Delta U = 0}
\label{eq:1203}
\end{equation}

Calculation of the pressure-volume work

The (pressure-volume) work done by the gas during an isothermal compression can no longer be determined by the product of pressure p and volume change ΔV, as was the case with an isobaric process. This is because as the volume decreases, the pressure increases sharply. This means that as compression progresses, an ever greater force and thus more and more work has to be done on the gas to achieve a further reduction in volume. This can also be seen from the p(V) diagram. In this diagram, the area under the curve corresponds to the pressure-volume work done on the gas to change the volume. The area under the curve increases more and more as the volume becomes smaller.

For an isothermal process, the pressure-volume work W_v must be determined as the area under the curve using the integral -∫p(V) dV. For the course of the pressure the function p(V) according to equation (\ref{pv}) has to be used:

\begin{align}
\label{eq:5671}
W_\text{v} &= – \int\limits_{V_1}^{V_2} p(V) ~ \text{d}V ~~~~~ \text{ where } ~~~~~ p(V) = R_\text{s}~m~T \cdot {1 \over V} ~~~~~ \text{ : }  \\[5px]
&= – \int\limits_{V_1}^{V_2} \underbrace{R_\text{s}~m~T}_{=\text{constant}} \cdot {1 \over V} ~ \text{d}V \\[5px]
&= – R_\text{s}~m~T~ \int\limits_{V_1}^{V_2} {1 \over V} ~ \text{d}V \\[5px] &= – R_\text{s}~m~T~ \left[~\ln(V)~\right]^{V_1}_{V_2} \\[5px]
&= – R_\text{s}~m~T~ \left[ \ln(V_2) – \ln(V_1) \right] \\[5px]
&= R_\text{s}~m~T~ \left[ \ln(V_1) – \ln(V_2) \right] ~\text{ where }~ \underline{\ln(V_1) – \ln(V_2) = \ln \left(V_1 \over V_2 \right)} ~\text{ : } \\[5px]
\end{align}

\begin{equation}
\boxed{W_\text{v} = R_\text{s}~m~T \cdot \ln \left(V_1 \over V_2 \right)} = R_\text{s}~m~T \cdot \ln \left(p_2 \over p_1 \right)
\label{eq:4387}
\end{equation}

Note that the pressure-volume work can also be calculated by the quotient of final and initial pressure, since according to equation (\ref{3}) the ratio of volumes (V₁/V₂) is equal to the inverse ratio of pressures (p₂/p₁=V₁/V₂).

Calculation of the transferred heat

The pressure-volume work W_v done on the gas or by the gas obviously does not result in a change of the internal energy due to the isothermal process. Of course, this is only possible if the transferred work is compensated to the same extent by releasing or absorbing heat. Work W_v and heat Q are thus equal in amount, but carry opposite signs. This is also evident from the first law of thermodynamics with the condition ΔU=0:

\begin{align}
&W_\text{v} + Q = \overbrace{\Delta U}^{=0} ~~~~~\text{first law of thermodynamics}  \\[5px]
&\boxed{Q = – W_\text{v}}
\end{align}

A change of state of a gas in which the pressure does not change is also called an isobaric process. This means that the same pressure applies to all the states through which the gas passes between the initial and final state.

How to achieve an isobaric process

An isobaric change of state in a closed system can be realized by a vertically oriented cylinder filled with a gas and closed by a piston. The piston can be loaded with a constant weight to increase the pressure if required. If the gas is now heated, the gas expands and pushes the piston upwards. The gas pressure inside the cylinder results only from the constant ambient pressure and the constant weight forces acting on the piston. This is because, as the gas expands, there must be a balance of forces between the gas pressure acting from the inside of the cylinder on one side of the piston and the forces acting from the outside of the cylinder on the other side of the piston – at least when the piston is raised at a constant speed.

The “external” forces can therefore be used to determine the “internal” force of the gas, and with the piston area A finally the gas pressure p. In the present case, the “external” force is on the one hand the weight force F_G of the piston (including any weights attached to the piston). On the other hand, the force F_amb also acts on the piston, which is due to the ambient pressure! This ambient force can be calculated from the ambient pressure p_amb and the piston area A (F_amb=p_amb⋅A). By setting up the balance of forces, the acting force of the gas F_gas can be calculated:

\begin{equation}
F_\text{gas} \overset{!}{=} F_\text{G} + F_\text{amb} ~~~~~\text{balance of forces} \\[5px]
F_\text{gas} = F_\text{G} + p_\text{amb} \cdot A \\[5px]
\end{equation}

Thus, the gas acts with this force F_gas on the piston surface A, which leads to the following gas pressure p:

\begin{equation}
\label{de}
\underline{p} = {F_\text{gas} \over A} = {{F_\text{G} + p_\text{amb} \cdot A} \over A} = \underline{ {F_\text{G} \over A} + p_\text{amb}}
\end{equation}

Equation (\ref{de}) shows that the gas pressure obviously depends only on quantities which do not change during the entire thermodynamic process. The gas pressure thus always remains constant for the entire change of state. Therefore, the lifting of the piston is an isobaric process.

Illustration in the volume-pressure diagram

If the piston is moved upward while heat is supplied, the gas heats up from T₁ to T₂ and the gas volume increases from V₁ to V₂ at constant pressure. If this process is illustrated in a volume-pressure diagram, a horizontal line results from the initial volume V₁ to the final volume V₂ at constant pressure p.

Relationship between volume and temperature (law of Gay-Lussac)

In an isobaric process, the volume increases proportionally with the temperature. However, this statement only applies if the temperature is given in the unit Kelvin. A doubling of the temperature in an isobaric heating thus also means a doubling of the gas pressure. This proportionality is equivalent to the statement that volume and temperature always have a constant ratio. This phenomenon is also referred to as law of Gay-Lussac:

\begin{align}
&V \sim T ~\Rightarrow~ {V \over T} = \text{constant}
\end{align}

Any two states within an isobaric process are consequently linked by the constant quotient of volume and temperature:

\begin{equation}
\label{p} 
\boxed{ {V_1 \over T_1} = {V_2 \over T_2} } ~ \text{law of Gay-Lussac}
\end{equation}

Calculation of the change in internal energy

For ideal gases, the change in internal energy ΔU results independent of the thermodynamic process only on the basis of the temperature change ΔT=T₂-T₁:

\begin{equation}
\label{du} 
\boxed{ \Delta U = c_\text{v} ~ m ~ \left(T_2-T_1 \right)} = c_\text{v} ~ m ~ T_1 ~ \left({T_2 \over T_1}-1 \right) = c_\text{v} ~ m ~ T_1 ~ \left({V_2 \over V_1}-1 \right)
\end{equation}

In this equation c_v denotes the specific heat capacity and m the mass of the gas. If equation (\ref{p}) is used in equation (\ref{du}) after factoring out the temperature T₁, the change in internal energy can also be expressed in terms of the volume ratio V₂/V₁ (=T₂/T₁).

Calculation of the pressure-volume work

The pressure-volume work done on the gas or by the gas is relatively easy to determine for the isobaric process, since the area under the curve in the p(V) diagram is rectangular (note that the area under the process curve generally represents the pressure-volume change work!). The rectangular area is determined by the “height” p and the “width” ΔV=V₂-V₁. Taking into account the sign convention, the following formula applies to calculate the pressure-volume work W_v of an isobaric process:

\begin{equation}
\label{w}
\boxed{ W_\text{v} = – p \cdot \Delta V = – p \cdot \left(V_2 – V_1 \right) }
\end{equation}

Note: The negative sign results from the sign convention, since with an increase in volume (expansion: ΔV>0) the gas obviously does work on the piston and the pressure-volume work is to be counted negatively (W_v<0). Conversely, a reduction in volume (compression: ΔV<0) is to be counted positively, since work is done on the gas (W_v>0).

The pressure-volume work can be determined not only by the difference in volumes, but also by the difference in temperatures. Some transformations are necessary for this, using the ideal gas law (where R_s denotes the specific gas constant.):

\begin{align}
\label{z}
\boxed{p \cdot V= R_\text{s} \cdot m \cdot T} ~~~\text{ideal gas law}
\end{align}

\begin{align}
\label{www}
W_\text{v} &=- p \cdot \left(V_2 – V_1 \right) \\[5px]
&=- \left( p ~ V_2 – p ~ V_1 \right) ~\text{ where }~ \underbrace{p~V=R_\text{s}~m~T}_{\text{ideal gas law}} ~\text{:}~ \\[5px]
&= – \left( R_\text{s} ~ m ~ T_2 – R_\text{s} ~ m ~ T_1 \right) \\[5px]
&= -R_\text{s} ~ m ~ \left(T_2 – T_1 \right)
\end{align}

The pressure-volume work W_v can therefore also be determined from the difference between the final and initial temperature of the isobaric process:

\begin{equation}
\label{wvv}
\boxed{W_\text{v} = – R_\text{s} ~ m ~ \left(T_2 – T_1 \right)}
\end{equation}

Calculation of the transferred heat

Finally, the first law of thermodynamics can be used to calculate the heat energy Q absorbed or released by the gas for an isobaric process. The transferred heat is determined from the difference between the change in internal energy ΔU and the pressure-volume work W_v:

\begin{align}
&\boxed{W_\text{v} + Q = \Delta U} ~~~ \text{first law of thermodynamics}  \\[5px]
\end{align}

\begin{align}
Q &= \Delta U – W_\text{v} \\[5px]
&= c_\text{v} ~ m ~ \left(T_2-T_1 \right) + R_\text{s} ~ m ~ \left(T_2 – T_1 \right) \\[5px]
&= \underbrace{[c_\text{v}+R_\text{s}]}_{=c_\text{p}} ~ m ~ (T_2-T_1)
\label{eq:9885}
\end{align}

At this point, the constant quantities c_v and R_s can be combined to form a new constant c_p, resulting in the following relationship between the transferred heat Q and the temperature change:

\begin{equation}
\label{q} 
\boxed{ Q = c_\text{p} ~ m ~ (T_2-T_1)} ~\text{where}~ \boxed{c_\text{p}=c_\text{v}+R_\text{s}}
\end{equation}

The reason why these constants are combined to form a new constant becomes apparent when comparing the upper formula with the formula for the transferred heat of an isochoric process:

\begin{align}
\label{eq:6667}
Q_\text{v} &= c_\text{v} ~ m ~ (T_2-T_1) ~~ \text{isochoric process} \\[5px]
Q_\text{p} &= c_\text{p} ~ m ~ (T_2-T_1) ~~ \text{isobaric process}\\[5px]
\end{align}

The constant c_v can be interpreted as the specific heat capacity of the isochoric process and c_p as the specific heat capacity of the isobaric process. In this way, the analogous relationships between the transferred heat and the temperature change now arise in both the isobaric and the isochoric process.

The transferred heat according to equation (\ref{q}) can also be expressed by the ratio of final and initial volumes for a given initial temperature T₁:

\begin{equation}
Q= c_\text{p} ~ m ~ \left(T_2-T_1 \right) = c_\text{p} ~ m ~ T_1 ~ \left({T_2 \over T_1}-1 \right) = c_\text{p} ~ m ~ T_1 ~ \left({V_2 \over V_1}-1 \right)
\label{eq:7052}
\end{equation}

Note on the specific heat capacities

Note that it is evident from equation (\ref{q}) that the isobaric heat capacity c_p is always greater than the isochoric heat capacity by the value of the specific gas constant R_s. This means that more heat has to be transferred in an isobaric process than in an isochoric process if the same temperature change is to be achieved.

The reason for this is that in an isobaric process the supplied heat energy Q does not completely benefit the change in internal energy ΔU and thus the temperature increase. This is because part of the supplied heat is converted into pressure-volume work W_v. For this reason, more heat energy must be supplied so that, minus this pressure-volume work, sufficient energy is still available for the change in internal energy and thus for the temperature increase.

For further information, see also the article Specific heat capacity of gases.