# Entropy change in a cycle with two isochoric and two adiabatic processes [closed]

Prove that the change of Entropy in a cycle with two isochoric and two adiabatic processes is 0. How can I prove that? Thanks!

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## closed as too localized by David Zaslavsky♦Mar 3 '12 at 19:29

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## 2 Answers

We don't give full solutions here. But here is a hint: Write $dS=\frac{q}{T}$ and apply first law.

Or just say: Entropy is a state function, so it does not change in a cyclic process.

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Entropy is a state function, so if your system ends up in the same state as it started and there has been no net heat flow the entropy change must be zero.

Are you allowed to assume it's an ideal gas? If so draw a PV diagram of the cycle. The two isochoric stages are vertical lines, and the two adiabatic stages are the usual curves of PV = nRT. Using this relation you can show the temperature change in the two isochoric stages is the same, and therefore that the $\Delta Q$ is the same (assuming $C_v$ is constant). This means the net entropy change in the cycle is zero. To do the calculation start at one point ($P_1$, $V_1$) and work your way round the cycle using PV = nRT to calculate the pressure at the remain three points.

If you're not allowed to assume it's an ideal gas I'd have to think about it because you can no longer use PV = nRT.

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