# 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, 2020 at 14:36
• @FakeMod Sorry couldn't understand what was done there. Apr 2, 2020 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.? Apr 2, 2020 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, 2020 at 15:45

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}

• There is a small catch though. You have proved this for an ideal gas. But what is its not the case and what if thr system is not a gas in the first place? Apr 3, 2020 at 11:03
• @Vilvanesh Well, I don't know of any way of proving it, in general, for a non-ideal gas, liquid or solid. However if you adopt the Boltzmann's approach to entropy, it becomes clear that entropy is a state function for any substance.
– user258881
Apr 3, 2020 at 11:13
• I don't think you can construct a Carnot engine using anything but ideal gas. It is the most efficient engine because you can use ideal gas approximations here only (mainly that your gas won't have any long-range interactions) Feb 15, 2022 at 8:26

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 $$\oint \frac{1}{T} dQ=m\oint \frac{1}{T} C_VdT+\oint \frac{1}{T} PdV=0$$

• Dear ibnAbu. Welcome to Phys.SE. It is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) Aug 14, 2022 at 17:57