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Suppose we compress a gas from $(p,t)$ to $(P',T')$ reversibly & adiabatically (isentropically). If we perform non-isentropic irreversible compression from $(p,t)$ to $(P',T)$, we can say with certainty that $T>T'$.

Now let us consider the same starting point and same final pressure $P'$, but now we fulfill the constant entropy condition in a different way, say during compression the decrease in entropy due to heat lost by gas to surroundings is balanced by the increase in entropy due to friction(irreversible constant entropy compression). Final pressure-temperature coordinates are $(P', T'')$.

Now, will $T''$ be greater than or smaller than $T$?

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  • $\begingroup$ "now we fullfil the constant entropy condition in a different way" do you mean the isentropic condition that the change in the system entropy is 0 or the reversibility condition that the total entropy change of the system + surroundings is 0? $\endgroup$ Aug 3, 2018 at 7:20
  • $\begingroup$ I only mean entropy change of system is zero. Since i am only concerned with the working fluid. That's why i have labelled it as irreversible constant entropy compression. $\endgroup$ Aug 3, 2018 at 7:28
  • $\begingroup$ So you are saying that, in the second case, the process is not carried out adiabatically and not reversibly, but the heat transfer is controlled in such a way that the entropy change is zero? $\endgroup$ Aug 3, 2018 at 11:43
  • $\begingroup$ @Chester Miller : Yes, indeed. Such a process is possible, at least in theory. To be clear only entropy change of gas is zero, since when it rejects heat to surroundings, entropy of surroundings is bound to increase. $\endgroup$ Aug 3, 2018 at 11:45
  • $\begingroup$ Is the final volume for both cases the same? $\endgroup$
    – Bob D
    Aug 3, 2018 at 16:43

2 Answers 2

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If the final pressure is the same as for the adiabatic reversible path P', and the change in entropy for this non-adiabatic irreversible process is zero, the only possibility is that, irrespective of the details of the process, the final temperature T'' must be equal to T'. What other possibility is there? After all, entropy is a physical property of the gas, and a unique function of state.

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  • $\begingroup$ You're right. I think the same can be inferred from the fact that if we try to draw the 2 paths (p,t,s) to (P',T',s) & (p,t,s) to (P',T'',s), on T-S diagram, then since starting points are same & entropy is constant in both cases, we would need to draw a vertical line. But since final pressure is also same(P'), therefore T'' must be at the same height as T', since constant pressure curves in T-S plane, vary in their value according to the height they occupy. So if T'' were to be higher than T', then it would not have pressure P' but a pressure greater than P'. $\endgroup$ Aug 3, 2018 at 16:33
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Agree with Chester, there is no other possibility irrespective of the path between the two states.

I think you can also answer this by simply using the ideal gas law.

$P_1V_1/T_1 = P_2V_2/T_2$

For the first process:

$pv/t = P’v’/T’$

If the second process has the same final volume as the first

$pv/t = P’v’/T’’$

then obviously T’’ must equal T’

I said if the final volume is the same. But if the answer is T’’=T’ then obviously the final volumes for both processes must be the same.

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