Proof for $\oint \frac{dQ}{T}=0 $ in a reversible process

Question

I'm actually trying to prove that Entropy is a state function. I get struck at the point where I need to prove that $\oint \frac{dQ}{T}=0 $ for a reversible process. Clausius in his book The Mechanical Theory of Heat proved this by considering any process to be a combination of small isothermal and adiabatic process. This will break any reversible process into carnot cycles, for which the result is well establisted. The problem is I'm not really sure whether such a break-up will actually converge to the required process. If someone can prove that even that is good enough.

Else I'm looking for any proof where one can mathematically (or by any logical means) prove it. I have already tried the following answers:

• How does Fermi jump to this conclusion in Clausius inequality?

• Can anyone prove this overstated-but-almost-never-justified fact from thermodynamics?

• How do we know that entropy is a state function given by $\delta Q_{\mathrm{rev}}/T$ for any arbitrary (not just reversible process)?

• How is entropy a state function?

Answer 1

Using the first law of thermodynamics,

\begin{align} \mathrm dQ & = \mathrm dU +\mathrm dW\\ \mathrm dQ &= n C_V \mathrm dT + P \mathrm dV\\ \frac{\mathrm dQ}{T}&=n C_V \frac{\mathrm dT}{T} + \frac{P}{T} \mathrm dV \end{align}

Since the gas under consideration is an ideal gas, we can apply the equation of state, $PV=nRT$, to replace $P/T=nR/V$. Substituting this in the above equations, \begin{align} \frac{\mathrm dQ}{T}&=n C_V \frac{\mathrm dT}{T} + \frac{nR}{V} \mathrm dV\\ \oint \frac{\mathrm dQ}{T}&=n C_V \oint \frac{\mathrm dT}{T} + nR \oint\frac{\mathrm d V}{V}\\ \oint \frac{\mathrm dQ}{T}&=nC_V \ln T \biggr |_T ^T + nR \ln V \biggr |_V ^V\\ \oint \frac{\mathrm d Q }{T} &=0 \end{align}