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I want to show that $\delta q$ is not an exact differential.

Starting from $dE = \delta q - pdV$ and because $E := E(V, T)$ is a state function, which allows to express the exact differential as $$ dE = \left.\frac{\partial E}{\partial V}\right|_T dV +\left.\frac{\partial E}{\partial T}\right|_V dT, $$ the two expressions can be set equal giving after rearrangement $$ \delta q = \left[ \left.\frac{\partial E}{\partial V}\right|_T + p \right]dV + \left.\frac{\partial E}{\partial T}\right|_V dT $$ and therefore also $$ \delta q = \left[ \left.\frac{\partial E}{\partial V}\right|_T + p \right]dV + C_V dT. $$

Now, by definition $\partial E/\partial T|_V = C_V$, and so $$ \left.\frac{\partial C_V}{\partial V}\right|_T =\left[\frac{\partial}{\partial V}\left.\frac{\partial E}{\partial T}\right|_V\right]_T $$ and because $E$ is a state function, the sequence of partial derivatives can be exchanged (according to Schwarz' theorem, while I don't understand how it works), allowing to write $$ \left.\frac{\partial C_V}{\partial V}\right|_T =\left[\frac{\partial}{\partial T}\left.\frac{\partial E}{\partial V}\right|_T\right]_V. (*) $$

Also, multiplying the above expression for $\delta q$ by $1/\partial V$ at constant $T$, I obtain $$ \left.\frac{\delta q}{\partial V}\right|_T = \left[ \left.\frac{\partial E}{\partial V}\right|_T + p \right]\left.\frac{\partial V}{\partial V}\right|_T + C_V \left.\frac{\partial T}{\partial V}\right|_T $$ where $\partial V/\partial V = 1$ and the second term on the right hand side equals $0$ because $\partial T = 0$ at constant temperature. Multiplying the remaining equation by $\partial / \partial T$ at constant $V$ gives $$ \left[\frac{\partial}{\partial T}\left.\frac{\delta q}{\partial V}\right|_T\right]_V = \left[ \frac{\partial}{\partial T} \left(\left.\frac{\partial E}{\partial V}\right|_T + p \right)\right]_V. $$

Now assuming $\delta q$ were exact, again the sequence of partial derivatives would not matter and I could write $$ \left[\frac{\partial}{\partial V}\left.\frac{\delta q}{\partial T}\right|_V\right]_T = \left[ \frac{\partial}{\partial T} \left(\left.\frac{\partial E}{\partial V}\right|_T + p \right)\right]_V $$ and by using $q=E$ since the "inner" differential on the left side is evaluated at constant volume, $$ \left[\frac{\partial}{\partial V}\left.\frac{\partial E}{\partial T}\right|_V\right]_T = \left.\frac{\partial}{\partial V} C_V\right|_T = \left[ \frac{\partial}{\partial T} \left(\left.\frac{\partial E}{\partial V}\right|_T + p \right)\right]_V. (**) $$ From this we find that $(*)$ and $(**)$ are different and thus the assumption must be wrong and therefore $\delta q$ is not an exact differential.

Does this make any sense?


Exact wording from book:

Starting with $dE = \delta q - pV$, show that

a) $\delta q = C_V dT + [P+(\partial E/\partial V)_T] dV$

b) $\left(\frac{\partial C_V}{\partial V}\right)_T = \left[\frac{\partial}{\partial T} \left(\frac{\partial E}{\partial V}\right)_T\right]_V$

c) $\delta q$ is not an exact differential.

For c), the book states

If $\delta q$ were an exact differential, then by solution to a), $(\partial C_V/\partial V)_T$ would have to be equal to $[\partial /\partial T(P+(\partial E/\partial V)_T)]_V$ but it is not according to solution of b), hence $\delta q$ is not exact.

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  • $\begingroup$ I don't know if you simply want to show that $\delta q$ is not an exact differential whatever the means used or if you want to use your method in particular...In my opinion, the simplest way to see that $\delta q$ is not exact i.e. cannot be seen as the differential of a function $q$ is to see that its integral is generally non zero along a closed path and more generally its integral between any two points depends on the path chosen $\endgroup$ – gatsu Mar 10 '14 at 12:06
  • $\begingroup$ My above proof is to follow along the argument in a book. I know $\delta q$ could be shown to be not exact by integrating along a closed path, but I'm not quite sure at the moment how to do that. $\endgroup$ – TMOTTM Mar 10 '14 at 12:45
  • $\begingroup$ for the latter it is a simple proof by contradiction: you just need to find one explicit example where the integral along a path is not zero and your proof is done! $\endgroup$ – gatsu Mar 10 '14 at 14:23
  • $\begingroup$ I have an example for an ideal monoatomic gas (where I know the explicit form of the constant volume heat capacity). But how to do this if there is no explicit $C_V$? $\endgroup$ – TMOTTM Mar 10 '14 at 16:51
  • $\begingroup$ In the general case, you can trust the fact that the thermal efficiency of a (working) engine is never zero (that's an experimental fact). Using the first law of thermodynamics, it implies that the heat exchanged over a cycle is non zero whatever liquid, gas or anything you would put in your engine to make it work. Sometimes, facts are more useful to (easily) prove your point than equations... $\endgroup$ – gatsu Mar 10 '14 at 20:01
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I think it is important to not loose yourself in calculations. The method in your book probably starts from saying that considering an object $\delta Q$ that has the following general form:

$\delta Q = C_vdT + hdV$ say,

then it is an exact differential iif

$\left(\frac{\partial C_v}{\partial V}\right)_T = \left(\frac{\partial h}{\partial T}\right)_V$

This is also equivalent to saying that $C_v$ and $h$ can be thought of as partial derivatives of the same state function with respect to $T$ and $V$ respectively.

Now, from the first law of thermodynamics, we know that

$C_v \equiv \left( \frac{\partial U}{\partial T}\right)_V$

hence the point is then to figure out if $h$ is the partial derivative of $U$ with respect to $V$ at fixed $T$ and obviously it is not since the latter would be $h-p$ where $p$ is the thermodynamic pressure which is not a null function.

That ends the story I believe...but please tell me if you don't trust this argument, I can easily be fooled myself by circular arguments.

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  • $\begingroup$ I added the exercise from the book, you're actually stating exactly what the book asks for, I just had some trouble following it. You're right, the easiest way to show this seems to be that "cross-partial derivatives" are not equal. $\endgroup$ – TMOTTM Mar 11 '14 at 10:59

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