Question

Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,

$\frac{\partial }{\partial V}\left(\frac{\partial U}{\partial S}\right)=\frac{\partial }{\partial S}\left(\frac{\partial U}{\partial V}\right)$

where each $\partial /\partial V$is taken with $S$fixed, each $\partial /\partial S$is taken with $V$fixed, and $N$is always held fixed. From the thermodynamic identity (for$U$) you can evaluate the partial derivatives in parentheses to obtain

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

a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for $H,F,andG$ ). Hold $N$ fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to $N$, but after you've done four of them the novelty begins to wear off. For applications of these Maxwell relations, see the next four problems.

Short answer

Maxwell relations are:

$$\begin{gathered} {\left(\frac{\partial T}{\partial V}\right)}_S=-{\left(\frac{\partial P}{\partial S}\right)}_V \\ {\left(\frac{\partial T}{\partial P}\right)}_S={\left(\frac{\partial V}{\partial S}\right)}_P \\ {\left(\frac{\partial T}{\partial P}\right)}_S={\left(\frac{\partial P}{\partial T}\right)}_V \\ {\left(\frac{\partial S}{\partial P}\right)}_T=-{\left(\frac{\partial V}{\partial T}\right)}_P \end{gathered}$$

Step-by-step solution

To find

Four Maxwell relations.

Keeping N as constant derive the given equation.

We have the thermodynamics identity:

$dU=TdS-PdV+\mu dN$

at constant volume and number of molecules (at which $dN=0anddV=0$)

we have:

$T={\left(\frac{\partial U}{\partial S}\right)}_V............\left(1\right)$

and at constant entropy and number of molecules (at which $dN=0anddS=0$)

we have:

$P=-{\left(\frac{\partial U}{\partial V}\right)}_S............\left(2\right)$

In the given we have: role="math" localid="1648414830374" $\frac{\partial }{\partial V}\left(\frac{\partial U}{\partial S}\right)=\frac{\partial }{\partial S}\left(\frac{\partial U}{\partial V}\right).......\left(3\right)$

Now substitute equation (1) and (2) in (3)

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

continuing derivation

We have following the enthalpy identity as:

$dH=TdS+VdP+\mu dN$

at constant pressure and number of molecules (at which $dN=0anddP=0$) we have:

role="math" localid="1648416938771" $T={\left(\frac{\partial H}{\partial S}\right)}_P.........\left(3\right)$

again differentiate equation (3) w.r.t. P

${\left(\frac{\partial T}{\partial P}\right)}_S=\frac{\partial H}{\partial P\partial S}$

Then at constant entropy and number of molecules (at which $dN=0anddS=0$) we have:

role="math" localid="1648416952292" $V={\left(\frac{\partial H}{\partial P}\right)}_S.........\left(4\right)$

again differentiate equation (4) w.r.t. V

${\left(\frac{\partial V}{\partial S}\right)}_P=\frac{\partial H}{\partial P\partial S}$

combine these two equations together to get the following result:

${\left(\frac{\partial T}{\partial P}\right)}_S={\left(\frac{\partial V}{\partial S}\right)}_P$

continuing derivation 

We have following the Helmholtz free energy is given by:

$dF=-SdT-PdV+\mu dN$

at constant pressure and number of molecules (at which $dN=0anddP=0$) we have:

role="math" localid="1648418905850" $S=-{\left(\frac{\partial F}{\partial T}\right)}_P......\left(5\right)$

again differentiate equation (5) w.r.t. V

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

and at constant entropy and number of molecules (at which $dN=0anddS=0$) we have:

role="math" localid="1648418958034" $P=-{\left(\frac{\partial F}{\partial V}\right)}_S......\left(6\right)$

again differentiate equation (6) w.r.t. T

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

combine these two equations together to get the following result:

${\left(\frac{\partial T}{\partial P}\right)}_S={\left(\frac{\partial P}{\partial T}\right)}_V$

continuing derivation 

We have following the Gibbs free energy is given by:

$dG=-SdT+VdP+\mu dN$

at constant pressure and number of molecules (at which $dN=0anddP=0$) we have:

role="math" localid="1648419302437" $S=-{\left(\frac{\partial G}{\partial T}\right)}_P.......\left(7\right)$

again differentiate the equation (7) w.r.t. P

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

and at constant temperature and number of molecules (at which $dN=0anddT=0$) we have:

role="math" localid="1648419374743" $V={\left(\frac{\partial G}{\partial P}\right)}_S.......\left(8\right)$

again differentiate equation (8) w.r.t. T

${\left(\frac{\partial V}{\partial T}\right)}_P=\frac{\partial G}{\partial P\partial T}$

combine these two equations together to get the following result:

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

Final answer

Maxwell relations are:

$$\begin{gathered} {\left(\frac{\partial T}{\partial V}\right)}_S=-{\left(\frac{\partial P}{\partial S}\right)}_V \\ {\left(\frac{\partial T}{\partial P}\right)}_S={\left(\frac{\partial V}{\partial S}\right)}_P \\ {\left(\frac{\partial T}{\partial P}\right)}_S={\left(\frac{\partial P}{\partial T}\right)}_V \\ {\left(\frac{\partial S}{\partial P}\right)}_T=-{\left(\frac{\partial V}{\partial T}\right)}_P \end{gathered}$$