# The Carnot Cycle, Reversibility and Entropy

## Abstract

**:**

## 1. Introduction

_{rev}exceeds dW and the sign of work processes relative to change in volume means that the difference will always be positive, hence TdS > dQ.

**does**, but of the work which it

**can do**; and similarly, in the first form of the law, it is not of the resistances which heat overcomes, but of those which it

**can overcome**that mention is made.” Clausius himself emphasised the terms in bold. The resistance that heat can overcome is an external pressure, P, that matches the internal pressure, resulting in work PdV. In Clausius’s view, the effect of such work was to reverse the separation between particles and PdV was therefore a measure of the “separative force of heat”. However, the first law is not concerned with potentialities, but with work that is performed and heat that does flow. PdV is a work term and therefore implies a change in energy, but it does not correspond to work performed, and therefore the changes in energy, in an irreversible process.

## 2. The Thermodynamics of an Ideal Gas Based on Rate Equations

_{p}. The piston is moving with a velocity v

_{p}, which can be either positive or negative. Taking the gas in the left-hand side of the piston only, at any instant the kinetic energy of a single particle can be regarded as the algebraic sum of three terms associated with the component of the motion along each axis. The sum over all particles of each of the three separate terms is just the total kinetic energy. As the gas is ideal, there is no interaction with other particles or with the walls except through collisions and within the piston the internal energy is equal to the kinetic energy: that is

_{a}and velocity u

_{a}, and a piston, with mass m

_{p}and velocity u

_{p}. Conservation of both momentum and kinetic energy leads to well-known expressions for the velocities after the collision, respectively v

_{a}and v

_{p}, given by

_{p}is positive, i.e., moving in the same direction as the impinging atom, then $\left|{v}_{a}\right|<\left|{u}_{a}\right|$. In other words, a gas particle loses energy on each collision. Generally, u

_{a}>> u

_{p}, so the effect is quite small for each individual particle, but as even a small volume contains a very large number of particles, the collective effect is to cool the gas as it works on the piston. On the other hand, if u

_{p}is negative and the piston is moving towards the gas particles, the gas gains energy as work acts on it.

_{a}< u

_{p}, as these do not collide with the piston and therefore do not contribute to the force. Strictly, excluding these atoms would require some renormalisation of the probability distribution, but for simplicity we ignore that and move to a model in which we regard all the atoms as travelling with the same average speed. We can therefore divide the 6N atoms into N atoms each travelling in the positive and negative x, y and z directions. The average speeds in each of the different directions will in general be different and in both the positive and negative x directions there could also be a spatial dependence as atoms that have collided with the piston propagate through the system. Collisions between particles will smooth out these differences as energy is redistributed among the different degrees of freedom, but any differences in the total momentum in the atoms induced by collisions must remain.

_{a}, collide with the piston in time ∆t, then clearly $\Delta t=\frac{\mathsf{\Delta}l}{{u}_{a}}$ and $N\prime =NA{u}_{a}\mathsf{\Delta}t$, where A is the area of the piston.

_{x}

^{+}is the kinetic energy of the atoms associated with motion in the positive x direction, i.e., towards the piston, we have

_{x}

^{+}will be different from the kinetic energy of the atoms moving in the opposite direction, denoted by E

_{x}

^{−}. In a kinetic model, this would automatically be accounted for through knowledge of the individual velocities of the particles as they propagate through space and collide with other particles and the walls of the chamber. In a macroscopic formulation, the loss of information about individual momenta and their changes in both space and time means that it will be very difficult, perhaps impossible, to keep track of the differences between E

_{x}

^{+}and E

_{x}

^{−}in both time and space. Out of necessity, the differences between E

_{x}

^{+}and E

_{x}

^{−}are ignored and instead E

_{x}, the total energy associated with the velocity components in both the positive and negative directions, is used. In effect,

_{ext}, the nett force on the piston is

_{p}, $P\raisebox{1ex}{$dV$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.$, and on or by the external load, ${P}_{ext}\raisebox{1ex}{$dV$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.$.

_{x}at the piston will change the internal energy and the system will be out of equilibrium, in the sense given by Equation (10). The exchange of energy between the different degrees of freedom, say E

_{x}, and, for example, E

_{z}, can be described by the rate equation

_{y}denoted by the process variable, E

_{xy}. The constant, α

_{z}

_{,}in (22) is a variable parameter equivalent to an inverse time constant. Thus, for example, if ${E}_{z}<\frac{U}{3}$ energy is scattered out of the x-direction and into the z-direction. Energy conservation requires

_{x}brought about by scattering are being considered in (23). In addition to scattering, heat can flow in from the external walls and affect the total energy. Consider heat flow from the y-z plane into x, i.e., the wall opposite the piston on the left of Figure 2. This affects E

_{x}. The one-dimensional rate of heat flow can be defined as

_{wall}is the wall temperature and is equal to the starting temperature of the gas, ${\propto}_{h}$ is again a variable parameter equivalent to an inverse time constant, $\frac{N{k}_{B}}{2}$ is a one-dimensional heat capacity and the 1-D equivalent temperature, T

_{x}, can be defined from kinetic theory as

_{x}is the effective temperature associated with motion in the x-direction and is greater than, or less than, the equilibrium temperature of the gas according to whether E

_{x}is greater than, or less than, $\frac{U}{3}$. Heat flow from walls in the y-x and z-x planes into E

_{z}and E

_{y}, respectively, can be expressed in the same manner. Under these assumptions, namely that:

- Heat flows into the system from walls through inelastic collisions;
- The collisions with the piston are perfectly elastic and therefore only the x component of velocity is affected;

_{h}and α

_{z}are properties of the gas and its interaction with its surroundings and are therefore independent of δt, but, in as much as they are unknown, they can be varied at will in the simulations. This does mean, however, that this approach is not suited to an exploration of the dynamical evolution of such systems as the evolution depends not only on the choice of scattering parameters, α

_{h}and α

_{z}, but on knowledge of the spatial variation of both energy and particle densities, as well as the flow of momentum. The more complex computational approaches of irreversible thermodynamics [19] are probably better suited to this kind of in-depth exploration. Here, the emphasis is on the final state and, as will be shown in the next section, the final state depends only on the equality of internal and external pressures and is independent of the dynamics of the system.

## 3. Numerical Solution of the Thermodynamic Rate Equations

#### 3.1. Step Changes in Pressure

^{2}with a piston initially located at x

_{p}= 10 cm. The mass of the piston was calculated by specifying its thickness as a variable and using the density of aluminium as a typical metal from which such a piston might be made.

_{x}according to

_{x}adjusted accordingly. For simplicity, only heat flow from the rear wall in Figure 2 was permitted and the only changes to E

_{y}and E

_{z}considered were those due to scattering, with the additional assumption that E

_{z}= E

_{y}at all times. For adiabatic processes, α

_{h}was set to zero.

_{h}= 0 is an idealization that limits the system under consideration to the gas and the piston alone. The only interaction between the piston and gas is through the exchange of momentum, which changes the temperature of the gas but not the piston. The piston is therefore a mechanical device which acts as a repository for the difference in work performed within the gas and the exterior pressure. Setting α

_{h}> 0 is a device that connects the gas to a heat reservoir and allows the final temperature to equal the initial temperature.

_{x}. Both expansion and compression from this starting state were modelled by setting the external pressure to be a factor (1 ± β) of the internal pressure, where −0.1 ≤ β ≤ 0.1 in steps of 0.01. The time step was set to δt = 10

^{−9}s, which is small enough to give stable results. Figure 3 shows the piston position as a function of time for β = 0.5, corresponding to an external pressure of 106,391.25 Pa, with α

_{h}= 0 and α

_{z}= 10

^{4}, for three different piston thicknesses of 10

^{−5}, 10

^{−4}and 10

^{−3}m. For clarity, the data for the thinner pistons has been offset to 0.12 and 0.10 s, respectively, but the timescales remain the same. It is apparent that the heavier the piston, the longer the decay. It is not just that the piston moves slower, hence the larger period of the motion, but also that there are many more oscillations within the decay. For the thinnest piston, 10

^{−5}m, corresponding to a mass of 2.7 mg, there is essentially only one oscillation and the movement from the starting position to the minimum looks almost like a vertical line on this time scale. All three cases settle at the same final position and, for computational convenience, and unless mentioned otherwise, the piston thickness was set at 10

^{−4}m for all other calculations. It is easily verified that the change in internal energy at the final position of 0.097143 m is identical to the external work performed.

_{z}, with α

_{z}= 10

^{2}having the slowest decay of the oscillation followed by a steady decay to the equilibrium position. The data in Figure 4 has been truncated in time, but the downward trend for the piston position for α

_{z}= 10

^{2}is evident. The explanation for this is provided by Figure 5, which shows that, for α

_{z}δt = 1, corresponding to instantaneous equipartition of energy such that the gas can be considered be in internal equilibrium, the motion is still damped. The cause can be traced to the dependence of the pressure on the speed of the piston in Equation (19). By contrast, omission of this term with α

_{z}δt = 1, so that the effective pressure is just 2E

_{x}and equivalent to the equilibrium pressure given by the equation of state, leads to undamped motion. Indeed, a simulation based on the internal energy and the corresponding equilibrium pressure yields the same indefinite oscillation. For α

_{z}= 10

^{2}then, the piston motion has been damped before full equipartition occurs. The gas is therefore out of equilibrium, but scattering out of E

_{x}into E

_{y}and E

_{z}continues according to Equations (26a)–(26c) and the system very slowly relaxes to equilibrium. As it does so, the effective pressure acting against the piston also reduces and, if left long enough, the system settles at the expected position.

- Both α
_{z}and α_{h}set to zero, corresponding to an adiabatic process with no scattering out of E_{x}into E_{y}and E_{z}; - Both α
_{z}and α_{h}set such that both α_{z}δt = 1and α_{h}δt = 1, corresponding to ideal isothermal; - α
_{z}is set such that α_{z}δt = 1 but α_{h}= 0, corresponding to ideal adiabatic, which is the same situation as shown in Figure 4.

_{x}is always different from both E

_{y}and E

_{z}and the system settles in an artificial state.

_{z}and α

_{h,}respectively, represent processes that have been shown to give rise to damping in kinetic simulations [17,18]. When inter-particle collisions were switched off, undamped oscillations were observed, but in the macroscopic model, as shown in Figure 6, switching off the inter-particle collisions still gives rise to damped motion. In a kinetic simulation, the velocity dependence of the effective pressure should automatically be accounted for through the speed of the particles, so the lack of any damping observed at the microscopic level is contrary to the observation of damping in the macroscopic formulation. It is entirely possible, therefore, that the effect is a consequence of the loss of microscopic information and the implicit averaging that necessarily occurs in the transformation from a microscopic to a macroscopic view. It is for this reason that the simulations reported in this paper rely on one or both of α

_{z}and α

_{h}as sources of damping.

_{z}= 10

^{4}. As expected, the system expands to a larger volume upon reversal of the external pressure. The inset shows this more clearly. Although the difference is small, it is only because the initial change is small, but it is nonetheless real and within the limits of computational accuracy, even for β = 0.05. There is, of course, a limit as to how small a change can be modelled on a computer with a set numerical resolution, so rather than attempt to model very small changes, instead the data from slightly larger changes is extrapolated back. Thus, Figure 8a shows the ratio of work performed in expansion to the work performed in compression plotted against 1 + β. The data appears linear over a small range of β, but is in fact best described by a gentle curve. There is no sense of asymptotic behaviour or the ratio becoming unity as β moves towards zero, such that the work performed in either direction approaches equality and the process becomes effectively reversible. The ratio only approaches unity for β = 0, in which case there is no change in pressure and therefore no change of state. This is exactly what would be expected for even very, very small changes in a system that is intrinsically irreversible. There are no circumstances in which a finite change of external pressure leads to a return to exactly the same state on restoration of the external pressure.

^{−4}m and α

_{z}= 10

^{4}. The axes are plotted this way because in the simulation the pressure is the independent variable. Also shown is the line of best fit that has a slope of −0.6009, which is effectively the inverse of γ, the ratio of specific heats. However, this is an average. Closer examination of the data reveals that there are two slightly different slopes, one corresponding to compression (−0.6149) and one corresponding to expansion (−0.58960). In [16], a mathematical argument was made for intrinsic irreversibility using a binomial expansion of $P{V}^{\gamma}=k$ for a small step change in P and a corresponding step change in V. This argument is open to the criticism that this expression is derived using the notion of reversibility, and whilst the existence of two slopes shows that the relationship linking states in the process of compression is different from that in expansion, on average the relationship holds for irreversible processes over the limited range of expansion and compression for which the binomial expansion would apply.

#### 3.2. The Carnot Cycle

_{z}< 1; heat conduction from a reservoir, as represented by 0 < α

_{h}< 1; piston damping, owing to the dependence of the effective pressure on the speed of the piston. The key to the first two mechanisms is the idea of a delay between work performed at the piston and the consequent redistribution of energy among the degrees of freedom or the exchange of heat with a reservoir. For α

_{z}= 1, energy is instantaneously redistributed and the gas can effectively be described by the ideal gas equation of state. In consequence, the work performed on or by the gas, depending on whether the gas is expanding or being compressed, is always PdV. The work performed on or by the external pressure is different and the difference between the two is, by Equation (20), the kinetic energy of the piston. For α

_{h}= 0 the temperature changes with the work performed, and for α

_{h}= 1 the temperature remains constant. As shown in Figure 5, either of these conditions, in conjunction with the elimination of piston damping, leads to undamped oscillatory motion of the piston and this forms the basis of a dynamic Carnot cycle.

^{−4}m, but the piston damping is idealised away by setting the ratio of u

_{p}:u

_{a}to zero for all piston speeds in simulation of a much heavier piston. It is perhaps not immediately or intuitively obvious, but an appropriate combination of ideal isothermal expansion and ideal adiabatic expansion, as illustrated in Figure 9, will cause the piston to undergo a Carnot cycle, as illustrated in Figure 10.

_{h}= 1) and at some point before the end point of the expansion the heat flow from the reservoir is switched off by setting α

_{h}= 0. The expansion then continues adiabatically.

- The gas is allowed to expand isothermally to some arbitrary point before the maximum of the piston motion when the heat is switched off: that is, α
_{h}is switched from a value of 1 to 0. The two cycles correspond to 0.5 and 0.7 of the maximum isothermal extent of the piston motion, which from Figure 8 corresponds to a piston position of just under 0.27 m. The temperature of the wall is set to the temperature of the gas and the total heat input during the isothermal expansion is summed. This allows for a calculation of the entropy change, $\frac{\delta Q}{{T}_{wall}}$. At the point when the heat flow is switched off, the piston is already in motion and the gas continues to carry out work on the piston, thereby cooling in the process. - At the extremum of the motion, α
_{h}is reset to 1. The temperature of the wall is set to the temperature of the gas and the external pressure is reduced. That is to say, from the entropy calculated during the isothermal stage, the nett heat that must be exchanged with the cold reservoir can be calculated. The difference between the heat in (Q_{H}) and the heat out (Q_{C}) must be equal to the nett work performed. If the external pressure is P_{out}and P_{in}on the outward and inward piston motions, respectively, and the total movement of the piston is δx_{tot}, then

_{h}to 0.

- 3.
- The piston continues its motion, but now without any heat flow the temperature rises. The program continues until the minimum piston position is found. This turns out to be the same as the starting point and both the temperature and internal pressure are restored to their starting values.

## 4. Discussion

## 5. Conclusions

- Although the quasi-static approach to reversible processes is accepted as standard, it is not compatible with the idea that a difference of pressure must exist in order to change the state.
- In the dynamic cycle, on the other hand, the external pressure is different from, and independent of, the internal pressure and the movement of the piston is a consequence of simple mechanical principles.
- The Carnot cycle is intended to represent the ideal engine and the assumption of ideal conditions within the gas to eliminate damping mechanisms is fully compatible with that.
- There is no need to consider whether entropy is a property of a body or not as the entropy change due to the heat flow from the reservoir necessarily corresponds to the state function of thermodynamic phase space.
- Finally, the absence of damping means that the system can map out the reverse trajectory in thermodynamic phase space under the same external conditions and, in consequence, the cycle itself is reversible.

## Funding

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

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**Figure 1.**A 3-D representation of changes in δQ/T according to Equation (2) for (

**a**) the Carnot cycle and (

**b**) the same cycle with the initial isothermal expansion (AB) replaced by a free expansion (EF).

**Figure 3.**The effect of piston mass on the decay time in an adiabatic process. For computational purposes, the mass is determined by the thickness using the density of aluminium (2700 kg m

^{−3}) as a typical metal from which such a piston might be made. The origins of curves (a) and (b) are offset for clarity but the timescale is unchanged.

**Figure 4.**The effect of the energy scattering parameter, α

_{z}, on the decay of the piston motion for an adiabatic process with an external pressure of 1.05 × atmospheric. The solid line, α

_{z}= 10

^{4}, was used for the majority of simulations in conjunction with a time step of 10

^{−9}s.

**Figure 5.**A demonstration of piston damping for an adiabatic process with an external pressure of 1.05 × atmospheric. The solid line is the same decay as in Figure 3 for comparison. Elimination of the dependence of the effective pressure on the speed of the piston gives undamped oscillations.

**Figure 6.**Piston damping for an external pressure of 1.05 × atmospheric and other damping mechanisms eliminated. The curves correspond to an adiabatic process with no equipartition, an adiabatic process with instantaneous equipartition and an isothermal process with instant heat flow and instant equipartition.

**Figure 7.**Two 2-step processes corresponding to a step increase in external pressure followed by a return to the starting external pressure. The starting position of the piston is indicated by the dotted line at a position of 0.1 m. The inset shows clearly that the system does not return to the starting state but, as would be expected, the final volume is greater than the initial volume.

**Figure 8.**The ratio of work performed in expansion to the work performed in compression for a two step process in which compression is followed by expansion (

**a**). For small changes, the ratio of works is approximately linear with the ratio of initial to final pressure in the compression stage and always less than unity. In (

**b**) a plot of log V against log P yields a straight line with a best fit slope of −0.6002, which is effectively the inverse of γ, the ratio of specific heats.

**Figure 9.**Isothermal and adiabatic undamped expansion for a gas initially at 995 K and an external pressure of two atmospheres.

**Figure 10.**Pressure against volume for two Carnot cycles based on the isothermal expansion in Figure 8 plotted linearly (

**a**) and logarithmically (

**b**) to show that the curves correspond to isothermal and adiabatic transitions. Two cycles are shown to illustrate that the point at which heat flow from the reservoir is switched off is unimportant as long as the temperature of the reservoir and external pressure are adjusted at the extreme piston motion.

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Sands, D.
The Carnot Cycle, Reversibility and Entropy. *Entropy* **2021**, *23*, 810.
https://doi.org/10.3390/e23070810

**AMA Style**

Sands D.
The Carnot Cycle, Reversibility and Entropy. *Entropy*. 2021; 23(7):810.
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**Chicago/Turabian Style**

Sands, David.
2021. "The Carnot Cycle, Reversibility and Entropy" *Entropy* 23, no. 7: 810.
https://doi.org/10.3390/e23070810