So in my second year thermodynamics class we are currently talking about entropy and deriving its definition from the 1st law.

$$\mathrm{d}q_{rev} = dU - dw_{rev}$$ $$= C_{V}(T)dT + PdV$$ $$= C_{V}(T)dT + \frac{nRT}{V}dV$$

And then it just says since the mixed second partial derivatives aren't equal, i.e.

$$(\frac{\delta Cv(T)}{\delta V})_T \neq (\frac{\delta(\frac{nRT}{V})}{\delta T})_V$$ dq is not an exact differential (no proof of this is given). It then stipulates by dividing both sides of the equation by T

$$\frac{\mathrm{d}q_{rev}}{T} = \frac{C_{V}(T)}{T}dT + \frac{nR}{V}dV$$ suddenly it becomes an exact differential because

$$(\frac{\delta Cv(T)/T}{\delta V})_T = (\frac{\delta(\frac{nR}{V})}{\delta T})_V$$ Can anybody give me a proof of this with calculus because my textbook gives 0 proof whatsoever and it irks me because I want to understand the theory underlying this. Thanks!


These statements can be proved in general i.e. irrespective of whether you take your system as ideal gas or not. As you have used in the question $PV = nRT$, you inherently assuming that your system of thermodynamic interest is an ideal gas. So, let's first prove this for an ideal gas and then do the general proof.

$\textbf{Case I : Ideal Gas}$

For an ideal gas, $C_V(T)$ is independent of volume $V$, which means that $\big(\frac{\partial C_V(T)}{\partial V}\big)_T =0$. (We will prove the general relation in the next section, and this statement will be obvious from that proof. ), But on the other hand $\big(\frac{\partial (nR T/V)}{\partial T}\big)_V = \frac{nR}{V}$ by simple partial derivative. So, thus you get the proof that $\big(\frac{\partial C_V(T)}{\partial V}\big)_T \neq \big(\frac{\partial (nR T/V)}{\partial T}\big)_V$. Thus $\delta q_{rev}$ is not an exact differential.

On the other hand if you divide both sides by $T$, you still have, $\big(\frac{\partial (C_V(T)/T)}{\partial V}\big)_T = \frac{1}{T}\big(\frac{\partial C_V(T)}{\partial V}\big)_T =0$. And also on the other hand you have, $\big(\frac{\partial (nR /V)}{\partial T}\big)_V = -\frac{nR}{V^2}\big(\frac{\partial V}{\partial T}\big)_V = 0$ (obviously, as we are taking partial derivative of a constant term. Thus, by the equality of mixed partial derivatives of $\frac{\delta q_{rev}}{T}$ it becomes exact differential.

$\textbf{Case II : General System}$

For general system we have, $\delta q_{rev} = dU +PdV$. But remember we can't write, $dU = C_V(T) dT $ and $PV =nRT$. We have to write $dU = C_V(T)dT + \big(\frac{\partial U}{\partial V}\big)_T \ dV$ .

So, now we have $\delta q_{rev} = C_V(T)dT + \Big(\big(\frac{\partial U}{\partial V}\big)_T + P\Big)dV $.

Also, from the definition of Helmholtz free energy ($A = U - TS $ and $dA = -SdT -PdV$) and by using Maxwell's relation it can be shown that $\Big(\big(\frac{\partial U}{\partial V}\big)_T + P\Big) = T \big(\frac{\partial P}{\partial T}\big)_V$

Now, it can be shown properly that $$\left(\frac{\partial C_V(T)}{\partial V}\right)_T = \frac{\partial^2 U}{\partial T \partial V} = T \left(\frac{\partial^2 P}{\partial T^2}\right)_V \qquad(= 0\text{ if } PV=nRT)\tag{i} $$ $$\left(\frac{\partial (\frac{C_V(T)}{T})}{\partial V}\right)_T = \frac{1}{T} \frac{\partial^2 U}{\partial T \partial V} = \left(\frac{\partial^2 P}{\partial T^2}\right)_V \tag{ii}$$ and for the right hand side $$\frac{\partial}{\partial T}\Bigg(T\Big(\frac{\partial P}{\partial T}\Big)_V\Bigg)_V = T\Big(\frac{\partial^2 P}{\partial T^2}\Big)_V + \left(\frac{\partial P}{\partial T}\right)_V\tag{iii}$$ and if you divide the right side by $T$ and then check the partial derivative criterion, you will ultimately get $\Big(\frac{\partial^2 P}{\partial T^2}\Big)_V$ on the r.h.s also (exactly same as $(ii)$ ) , But note that, $(i) \neq (iii) $. Thus $\frac{\delta q_{rev}}{T}$ will become exact differntial but $\delta q_{rev}$ will remain inexact.


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