I am trying to figure out where my reasoning falls apart in this thought experiment:

To determine if a process "A" is reversible (or at the very least internally reversible), I try to picture a reversible process "B" that involves only heat transfer and links the same two endpoints that bound process "A".

In this particular case, "A" is the reversible adiabatic compression of a fixed mass of any one ideal gas confined in a frictionless piston-cylinder arrangement. Given that a reversible process by definition does not lead to entropy generation, the gas undergoes isentropic compression and receives an amount of energy equal to the PV work done by the piston upon the gas. State 1, then, can be characterized by entropy s1 = s, temperature T1 and volume V1. State 2 can be correspondingly be defined by s2 = s and temperature T2 != T1 (different), and volume V2.

"B" is as follows: first, heat is removed reversibly from the system while the volume is kept constant until the pressure and temperature are only infinitesimally above absolute zero. At this point, the piston is allowed to move, changing the volume from V1 to V2. Once again the volume is kept fixed by some means and heat input is effected reversibly. Since the internal energy of an ideal gas is dependent only on its temperature, an amount of heat different from that which was removed at the beginning of the process must flow inward so as to attain temperature T2, which is different from T1. It follows that the amount of entropy transfer must also be different, which would entail different entropy values for the initial and final states, which is unverifiable in process "A". However, the volume V2 and temperature T2 are the same as those which lie at the end of process "A", which is the same as saying that both states are identical.

My question is: where have I made a mistake? Is it because the ideal gas hypothesis falls apart over this range of different temperatures/pressures? Or is it the assumption of constant specfic heat? If so, how does the underlying mathematics "know" this and reflect accordingly?

Thanks in advance.

  • $\begingroup$ I did not read the rest after "Given that a reversible process by definition does not lead to entropy generation, the gas undergoes isentropic compression ..." This is incorrect. An adiabatic (no heat exchanged) and reversible process is isentropic, but this is not the only way to be reversible. Any reversible process is such that the total entropy is unchanged, adiabatic or diathermal; here total entropy means the sum of the entropies of the system under consideration and that of its environment, and not just of the system. $\endgroup$
    – hyportnex
    Sep 9 '18 at 21:33
  • $\begingroup$ Your reasoning is faulty in the first two statements of your presentation. First of all, it is not true that a “reversible process does not lead to entropy generation”. For example, a reversible isothermal expansion generates entropy. It’s a reversible cycle that does not generate entropy. Next, if process A is a reversible adiabatic (isentropic) compression, then T2 can’t equal T1. T2 has to be higher than T1. I don’t think anyone can address the rest of your question until your correct the above. I hope this is of help. $\endgroup$
    – Bob D
    Sep 9 '18 at 21:37
  • $\begingroup$ @BobD This is not correct. There are two ways that entropy can change within a system: (a) by irreversible generation of entropy within the system and (b) by entropy exchange with the surroundings across the boundaries of the system. In a reversible process, only the 2nd mechanism is operative. $\endgroup$ Sep 9 '18 at 23:56
  • $\begingroup$ I did state T2 is different from T1 ("State 2 can be correspondingly be defined by s2 = s and temperature T2 != T1..."). @hyportnex That is correct. I left out that the compression is effected without heat exchange. In that case, if it's reversible, then it has to be isentropic. I'll amend it. $\endgroup$
    – jvf
    Sep 10 '18 at 0:47
  • $\begingroup$ @BobD If a process involves no irreversibilites, internal and external, then no entropy is generated. If that weren't the case, a reversible cycle as you described it would involve a process that would "generate" negative entropy. $\endgroup$
    – jvf
    Sep 10 '18 at 0:54

I didn't read about process B, but your results for process A are definitely incorrect. For the isothermal reversible compression you described, the entropy of the system definitely decreases. When you compress the gas, the amount of heat you remove from the system must equal to work you do in compressing the gas, if the temperature is to remain constant. So the entropy change is equal to the heat removed from the system (or the work done on the system) divided by the constant absolute temperature. Even though entropy is not generated within the system during this reversible process, entropy is nonetheless exchanged with the surroundings (by the removal of heat).


The change in entropy of the initial cooling step in process B is $$\Delta S=nC_v\ln{(T_{L1}/T_1)}$$where $T_{L1}$ is the low temperature at the end of the step. When the gas is adiabatically compressed during the second step, we have to control the motion of the piston such that process is reversible and the change in entropy is zero. This will require the temperature to rise from $T_{L1}$ to $T_{L2}$ such that $$nC_V\ln{(T_{L2}/T_{L1})}+nR\ln{(V_2/V_1)}=0$$Finally, when we reheat the gas back up to $T_2$, the change in entropy will be $$\Delta S=nC_v\ln{(T_{2}/T_{L2})}$$When we add these three changes in entropy together for process B, we obtain for the overall change:$$\Delta S =nC_V\ln{(T_{2}/T_{1})}+nR\ln{(V_2/V_1)}=0$$This is the same as the entropy change for process A.

  • $\begingroup$ It isn't a reversible isothermal compression, it's isentropic: " State 2 can be correspondingly be defined by s2 = s and temperature T2 != T1, and volume V2." T2 is different from T1 $\endgroup$
    – jvf
    Sep 10 '18 at 0:55
  • $\begingroup$ You mean != is supposed to mean $\neq$? I didn't get that. Maybe you should learn how to use LaTex? If that's the case, then your assessment of process A is indeed correct. $\endgroup$ Sep 10 '18 at 1:35
  • $\begingroup$ I'll be back to analyze process B in a little while. $\endgroup$ Sep 10 '18 at 1:43
  • $\begingroup$ Yes, != indicates difference. Maybe I should get around to learning LaTex... I'm less than a beginner when it comes to using it. Could you give consideration to the remainder of my question? $\endgroup$
    – jvf
    Sep 10 '18 at 1:43
  • $\begingroup$ See my analysis of process B. $\endgroup$ Sep 10 '18 at 2:10

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