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

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:

• I would advise you to take a look at this answer: physics.stackexchange.com/a/511590/258881 – user258881 Apr 2 '20 at 14:36
• @FakeMod Sorry couldn't understand what was done there. – Vilvanesh Apr 2 '20 at 15:42
• The answer might be there in the links I've shared above. I tried reading every single one of them. But couldn't get much. I guess I'm dumb.? – Vilvanesh Apr 2 '20 at 15:44
• No that doesn't mean you are dumb, but honestly, even I don't know anything more than the stuff you have linked to. – user258881 Apr 2 '20 at 15:45

\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}