# Calculating entropy change: reversible vs irreversible process

Since the change in internal energy and enthalpy, which are equal to the heats for a constant-volume and constant-pressure process, respectively, are state functions, the heats for a reversible v.s. irreversible process should be equal. Thus, the entropy change of the surroundings which equals the quotient of the heat and temperature are equal for reversible v.s. irreversible processes.

As Engel's Physical Chemistry states it:

Because H and U are state functions, the amount of heat entering the surroundings is independent of the path; q is the same whether the transfer occurs reversibly or irreversibly.

However, when the book proceeds to evaluating the change in entropy for the surroundings for a reversible vs irreersible process, the q used for evaluating the change in entropy of surroundings are not equal for the reversible v.s. irreversible process, which contradicts what the textbook previously stated.

The question:

One mole of an ideal gas at 300. K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath thermal reservoir in the surroundings is very large, T remains essentially constant at 300. K during the process.

Calculation of change in entropy for reversible path:

Calculation for irreversible path:

It seems that the book used the actual heat for the irreversible process. However, I don't understand why the entropy's from the two calculations are not equal. Also, I fail to understand why the change in entropy for the surroundings are not equal for the two process for which the final and initial states are equal --> change in entropy is a state function and should be the same regardless of path.

References

Engel, Thomas, and Philip Reid. Physical Chemistry. San Francisco: Pearson Benjamin Cummings, 2006.

• Update: I am starting to get the feeling that (somehow) the final state of the surroundings after an irreversible process is not the same as the final state of the surroundings after a reversible process. Something (for example, friction between the piston and walls of container) make the final states of surroundings different for irreversible vs reversible --> change in S is not the same. However, since the initial and final T and V are the same for the system in reversible and irreversible, the change in entropy is the same for reversible vs irreversible for system. Is this a valid argument? – Ethiopius Apr 3 at 17:09
• Entropy change of your system will be the same for both the reversible and irreversible path. However, the entropy of the surroundings will not be the same , as you have seen. The system goes from the same state A to the same state B for both the reversible and irreversible paths, the surroundings are not in the same state after an irreversible process as they would be after a reversible one. @Ethiopius – Tyberius Apr 3 at 17:31
• @Tyberius What you mean is that the surroundings are not in the same state after an irreversible process imposed on the system as they would be after a reversible process imposed on the system between its same two end states. – Chet Miller Apr 3 at 23:01
• @ChetMiller That is a clearer description. The way I had phrased I was open to misinterpretation. – Tyberius Apr 3 at 23:03