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:


*

*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?
 A: 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}
A: It is not obvious from first law of thermodynamics that $dQ,dU,dW$ are differentials for the integration.
It doesnt has to be an ideal gas, all needed is to assume the integrals exist in Riemann sense and certain functions are absolutely continuous.
let $\epsilon , a >0 $,
$T_{n}>T_{n-1}>a,|T_{n}-T_{n-1}| < \epsilon$
Using the first law of thermodynamics,
\begin{align}
\mathrm Q_2-Q_1 & = \mathrm U_2 - U_1 +\mathrm W_2 - W_1\\
\mathrm \int_{Q_{n-1}}^{Q_n}  dQ &= \int_{T_{n-1}}^{T_n} m C_V \mathrm dT + \int_{V_{n-1}}^{V_n}  P \mathrm dV\\
\sum_{n=1}^{n=N} \frac{1}{T_{n-1}}\mathrm  \int_{Q_{n-1}}^{Q_n}  dQ &=\sum_{n=1}^{n=N}\frac{m}{T_{n-1}} \int_{T_{n-1}}^{T_n}  C_V \mathrm dT + \sum_{n=1}^{n=N}\frac{1}{T_{n-1}}\int_{V_{n-1}}^{V_n}  P \mathrm dV\\  
\end{align}
Because of the following inequalities:
$$|\frac{1}{T_{n-1}}\int_{Q_{n-1}}^{Q_n}  dQ-\int_{Q_{n-1}}^{Q_n} \frac{1}{T} dQ| \le \int_{Q_{n-1}}^{Q_n} |\frac{1}{T}-\frac{1}{T_{n-1}}| dQ \le \int_{Q_{n-1}}^{Q_n} \frac{\epsilon}{a^2} dQ $$
$$|\frac{1}{T_{n-1}}\int_{T_{n-1}}^{T_n}  C_VdT-\int_{T_{n-1}}^{T_n} \frac{1}{T} C_VdT| \le \int_{T_{n-1}}^{T_n} |\frac{1}{T}-\frac{1}{T_{n-1}}| C_VdT \le \int_{T_{n-1}}^{T_n} \frac{\epsilon}{a^2} C_VdT $$
$$|\frac{1}{V_{n-1}}\int_{V_{n-1}}^{V_n}  PdV-\int_{V_{n-1}}^{V_n} \frac{1}{T} PdV| \le \int_{V_{n-1}}^{V_n} |\frac{1}{T}-\frac{1}{V_{n-1}}| PdV \le \int_{V_{n-1}}^{V_n} \frac{\epsilon}{a^2} PdV $$
letting $\epsilon \to 0$ we have $$\int \frac{1}{T} dQ=m\int \frac{1}{T} C_VdT+\int \frac{1}{T} PdV$$
$$\frac{df}{dQ}=\frac{1}{T}$$
$$\frac{dg}{dT}=\frac{C_V}{T}$$
$$\frac{dh}{dV}=\frac{P}{T}$$
if $f,g,h$ are absolutely continuous then $$\oint \frac{1}{T} dQ=m\oint \frac{1}{T} C_VdT+\oint \frac{1}{T} PdV=0$$
