# Irreversible Isochoric Process in a Cycle

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• Robert Davidson
Robert Davidson
TL;DR Summary
Consider a reversible ideal gas cycle consisting of: 1. An isochoric heat addition, 2. An isothermal expansion to the initial pressure, and 3. An isobaric compression to the initial volume. What, if any, is the difference in net work done by the gas if the isochoric heat addition was irreversible instead of reversible?
Consider a reversible ideal gas cycle consisting of: 1. An isochoric heat addition, 2. An isothermal expansion to the initial pressure, and 3. An isobaric compression to the initial volume. What, if any, is the difference in net work done by the gas in the cycle if the isochoric heat addition was irreversible instead of reversible?

Since no work is done in the isochoric process,

$$\Delta U_{1-2}=Q=nC_{v}(T_{2}-T_{1})$$

for both the reversible and irreversible process. The change in entropy for both a reversible and irreversible process is, using a reversible path involving an infinite series of thermal reservoirs,

$$\Delta S_{sys}=nC_{v}ln\frac{T_2}{T_1}$$

If we assume the irreversible process involves heat addition from a single thermal reservoir at temperature T2, then I believe the entropy generated would be

$$S_{gen}= nC_{v}ln\frac{T_2}{T_1}-\frac{Q}{T_2}$$

In order to complete the cycle, I would think that this entropy generated would need to be transferred to the surroundings in the form of additional heat and that the only opportunity to do this is during the reversible isobaric compression. Then it would seem that the negative work done in the isobaric process should be greater than for the reversible cycle for an overall positive net work of less. However, that can't be the case if the isobaric process connects the same two volumes for both the reversible cycle and irreversible cycle.

Can someone find the error in my reasoning? Thanks in advance.

## Answers and Replies

Mentor
Hi Bob. Welcome back to PF.

Here is a hint to help you: You don't need to look at the isothermal and isobaric steps of the cycle. The entire answer to your problem lies in the isochoric step.

Chet

Robert Davidson
Hmm. Could it be that since no work is done in the isochoric process, be it reversible or irreversible, then there is no impact of the entropy generated on work? Or, to put it another way, the only time entropy generation affects work is if it occurs in a process that produces work?

Mentor
Hmm. Could it be that since no work is done in the isochoric process, be it reversible or irreversible, then there is no impact of the entropy generated on work? Or, to put it another way, the only time entropy generation affects work is if it occurs in a process that produces work?
I don't know about all that with respect to work. But, in the irreversible process, less entropy is transferred to the system from the surroundings in the form of heat, and this is equivalent to the entropy of the surroundings decreasing less for the irreversible process than the reversible process. And this is equivalent to the generated entropy being transferred to the surroundings.

To be even more straightforward about this, in the irreversible process, less entropy is transferred to the system from the surroundings, so, in order for the change in entropy to be the same for the system, the generated entropy has to make up the difference. So, in reality, the generated entropy stays in the system.

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Robert Davidson
"To be even more straightforward about this, in the irreversible process, less entropy is transferred to the system from the surroundings"

Yes, I calculated that. For the reversible process the entropy transferred to the system from the surroundings is $$\Delta S_{surr}=-nC_{v}ln\frac{T_2}{T_1}$$ For the irreversible path I used (single reservoir T2) it is $$\Delta S_{sur}=\frac{-Q}{T_2}$$ which is less negative than the reversible process for all T2>T1.

"so, in order for the change in entropy to be the same for the system, the generated entropy has to make up the difference".

Yes, that's why I calculated the entropy generated in the system to be $$S_{gen}=nC_{v}ln\frac{T_2}{T_1}-\frac{Q}{T_2}$$

"So, in reality, the generated entropy stays in the system".

OK. So if I the generated entropy stays in the system then it stays in the irreversible cycle I described that includes the irreversible isochoric process. So, are we then to conclude that there is no difference between the cycle with the reversible isochoric heat addition than with the irreversible isochoric heat addition, in terms of the net work produced, cycle efficiency, etc..

In summary, are there any consequences associated with the irreversible isochoric heat addition?

Mentor
"To be even more straightforward about this, in the irreversible process, less entropy is transferred to the system from the surroundings"

Yes, I calculated that. For the reversible process the entropy transferred to the system from the surroundings is $$\Delta S_{surr}=-nC_{v}ln\frac{T_2}{T_1}$$ For the irreversible path I used (single reservoir T2) it is $$\Delta S_{sur}=\frac{-Q}{T_2}$$ which is less negative than the reversible process for all T2>T1.

"so, in order for the change in entropy to be the same for the system, the generated entropy has to make up the difference".

Yes, that's why I calculated the entropy generated in the system to be $$S_{gen}=nC_{v}ln\frac{T_2}{T_1}-\frac{Q}{T_2}$$

"So, in reality, the generated entropy stays in the system".

OK. So if I the generated entropy stays in the system then it stays in the irreversible cycle I described that includes the irreversible isochoric process. So, are we then to conclude that there is no difference between the cycle with the reversible isochoric heat addition than with the irreversible isochoric heat addition, in terms of the net work produced, cycle efficiency, etc..

In summary, are there any consequences associated with the irreversible isochoric heat addition?
Yes. There is no change in the heat and the work, and the changes in entropy and internal energy of the system comprised of the working fluid. The only change is that there is less of a decrease in the entropy of the surroundings in the irreversible path than for the reversible path. So, for the reversible path, there is no change in the entropy of the universe, but, for the irreversible path, there is an increase in the entropy of the universe (equal to the generated entropy). That's all that the 2nd law requires.

Robert Davidson
Interesting. Is this unique to the isochoric process? The increase in entropy of the surroundings is due to less entropy transferred to the system, as opposed to the system ridding itself of the entropy generated by transferring it to the surroundings in the form of heat? Is it also unique to the isochoric process that irreversibility does not effect heat and work in a cycle? Or, as I said previously, in order for there to be less work in a cycle the irreversible process needs to be one where work is performed?

Mentor
Interesting. Is this unique to the isochoric process? The increase in entropy of the surroundings is due to less entropy transferred to the system, as opposed to the system ridding itself of the entropy generated by transferring it to the surroundings in the form of heat? Is it also unique to the isochoric process that irreversibility does not effect heat and work in a cycle? Or, as I said previously, in order for there to be less work in a cycle the irreversible process needs to be one where work is performed?
Sorry Bob,

I don't know the answer to any of these questions. I'm pretty good at applying the fundamentals to solve specific problems, but I never felt it was worth the effort to develop general rules for classes of problems.

Robert Davidson
Chet
No problem. But thank you very much for clearing this one up for me. I confess to a tendency to try and identify general rules, or at least to try and get an in depth feeling for a rule. As usual, I really appreciate your time.

Bob