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, and G ). 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's 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

Given information

Maxwell relation is given.

Derive equation for constant volume

We have the thermodynamics identity:

$dU=TdS-PdV+\mu dN$

at constant volume and number of molecules (at which dN=0 and dV=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=0 and dS=0)

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

In the given we have:

$\frac{\partial }{\partial V}\left(\frac{\partial U}{\partial S}\right)=\frac{\partial }{\partial S}\left(\frac{\partial U}{\partial V}\right).......\left(3\right)$

Now substitute (1) and (2) in (3) We get

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

Derive equation for constant pressure

We have following the enthalpy identity as:

$dH=TdS+VdP+\mu dN$

t constant pressure and number of molecules (at which dN=0 and dP=0),

we have

$T={\left(\frac{\partial H}{\partial S}\right)}_P.........\left(4\right)$

again differentiate equation (4) w.r.t. P, we get

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

Then at constant entropy and number of molecules (at which dN=0, dS=0),

we have

$V={\left(\frac{\partial H}{\partial P}\right)}_S.........\left(5\right)$

again differentiate equation (5) w.r.t. V, we get

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

Combine these two we get

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

Derivation continued

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=0 and dP=0)

we have

$S=-{\left(\frac{\partial F}{\partial T}\right)}_P......\left(6\right)$

again differentiate equation (6) 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=0 and dS=0),

we have

$P=-{\left(\frac{\partial F}{\partial V}\right)}_S......\left(7\right)$

again differentiate equation (7) 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 we get

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

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=0 and dP=0)

we have

$S=-{\left(\frac{\partial G}{\partial T}\right)}_P.......\left(8\right)$

again differentiate the equation (8) 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=0 and dT=0)

we have:

$V={\left(\frac{\partial G}{\partial P}\right)}_S.......\left(9\right)$

again differentiate equation (9) 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:

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