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I'm supposed to derive the following thermodynamic identity: $$ - \left( \frac{\partial U}{\partial V} \right)_{T,N} + T \left( \frac{\partial P}{\partial T} \right)_{V,N} = P $$ ...by starting with the Helmholtz free energy $F = U - TS$.

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How do I begin this problem? To me it seems like the only way I would be able to relate $U, T, S$ to the variables $P,V$ would be through the use of some kind of equation of state, but I'm not explicitly given one.

Is it enough to use the first law of thermodynamics? I'm also confused as to why I would need the Helmholtz free energy in the first place? A prod in the right direction would be much appreciated.

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So at the most fundamental we have a definition of chemical potentials and temperatures and pressure as$$dU = T ~dS - P ~dV + \sum_i \mu_i ~dN_i,$$where $U$ is the internal energy, $S$ is the entropy, $V$ is the volume, and $N_i$ is the number of particles of species $i.$ One way to read this is that it tells you for example what $\left(\frac{\partial S}{\partial U}\right)_{V,N_i}$ is, it's $1/T$, since you can just set $dV = 0$ and $dN_i = 0$ when you're evaluating this partial derivative.

What if you're not holding $S$ constant but instead holding $T$ constant? Well then the above equation does not tell you much because there is no $dT$ term that we can set to be equal to zero. Instead we decide that internal energy is not the right state parameter for our situation and instead we want to consider a "free energy", a Legendre transform of $U$, in this case (constant volume and temperature) we want $F = U - T S$ to get $$dF = -S ~dT - P ~dV + \sum_i \mu_i~dN_i.$$Now we look at this expression and we just "see" that $$\left(\frac{\partial F}{\partial V}\right)_{T,N_i} = -P.$$Again, if you don't see it, just set $dT = 0, ~~dN_i = 0,$ ask how $F$ changes with a change in $V$.

So now we expand the $F = U - TS$ equation to get the relation that you've got:$$\left(\frac{\partial F}{\partial V}\right)_{T,N_i} = \left(\frac{\partial U}{\partial V}\right)_{T,N_i} + T \left(\frac{\partial S}{\partial V}\right)_{T,N_i}.$$That first term looks fine, what about the second term? Well, that's a Maxwell relation and it also requires coming back to this $F$ term since that has both $S$ and $P$ out front. Since partial derivatives commute we have that $$\left( \frac{\partial}{\partial T} \left(\frac{\partial F} {\partial V}\right)_{T,N_i}\right)_{V,N_i} = \left( \frac{\partial}{\partial V} \left(\frac{\partial F} {\partial T}\right)_{V,N_i}\right)_{T,N_i},$$and going back to this expression for $dF$ you can see that these are precisely $-S$ and $-P$, hence eliminating the double-negative gives:$$\left( \frac{\partial P}{\partial T} \right)_{V,N_i} = \left( \frac{\partial S}{\partial V} \right)_{T,N_i},$$ which we substitute into the above expression.

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  • $\begingroup$ This is great! My only question is why exactly we can make the switch $\left( \frac{\partial}{\partial T} \left(\frac{\partial F} {\partial V}\right)_{T,N_i}\right)_{V,N_i} = \left( \frac{\partial}{\partial V} \left(\frac{\partial F} {\partial T}\right)_{V,N_i}\right)_{T,N_i}$? The fact that we're keeping the terms constant which we are differentiating is freaking me out a bit. $\endgroup$ – Greg.Paul Jan 25 '17 at 3:59
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    $\begingroup$ @Greg.Paul: That should totally freak you out a bit! It is a mathematical theorem. Don't get too scared, just think in normal 2D space. If we take $(\partial f/\partial x)_y$ we get a new function for all $(x, y)$ and we can certainly then take $(\partial(\partial f/\partial x)_y/\partial y)_x.$ Indeed if we raise the 2D space to a plane in 3D space we can apply the curl theorem to an integral of $\nabla f$ about a small loop $(x\pm\delta x, y\pm\delta y)$ to get $4[\partial_y(\partial_x f)-\partial_x(\partial_y f)]\delta x\delta y$ and you just need to show that this is 0 to second order. $\endgroup$ – CR Drost Jan 25 '17 at 23:08
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    $\begingroup$ Actually hammering away at that a little bit I have not yet seen it work out cleanly. Hm. Googling it up reveals some proofs, though, and it seems to be popular to use the mean value theorem to do the heavy work of "continuous implies ___." $\endgroup$ – CR Drost Jan 25 '17 at 23:25
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Start from:

$$dU = TdS - PdV$$

Now "divide by dV"

$$\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial S}{\partial V}\right)_T - P$$

Now you use one of the Maxwell relations:

$$\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V$$

and you have the result:

$$\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P$$

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