# Maxwell Relations

• Feb 11th 2009, 10:02 AM
thelostchild
Maxwell Relations
Hey I'm having a little problem with this question
Quote:

The enthalpy of a gas is defined by $\displaystyle H=U+pV$ where U satisfies $\displaystyle dU=TdS-pdV$ Determine a relationship between the differentials of H, S and p. Hence show that

$\displaystyle \left ( \frac{\partial V}{\partial S} \right )_p= \left ( \frac{\partial T}{\partial p} \right )_S$

By regarding U as a function of p and V show that

$\displaystyle (\frac{\partial S}{\partial V})_p (\frac{\partial T}{\partial p})_V - (\frac{\partial S}{\partial p})_V (\frac{\partial T}{\partial V})_p = 1$

I can do the first one absolutely fine, just take the total derivative of H and substitute in the expression in for dU then use the fact that mixed partials commute.

The second part I'm having more trouble with, any chance of a hint?

• Feb 11th 2009, 03:10 PM
Jester
Quote:

Originally Posted by thelostchild
Hey I'm having a little problem with this question

I can do the first one absolutely fine, just take the total derivative of H and substitute in the expression in for dU then use the fact that mixed partials commute.

The second part I'm having more trouble with, any chance of a hint?

Here's an idea. Write

$\displaystyle \frac{\partial V}{ \partial S} = \frac{\partial T}{ \partial p}$

as (with Jacobian's)

$\displaystyle \frac{\partial(V,p)}{\partial(S,p)} = \frac{\partial(S,T)}{\partial(S,p)}$

then

$\displaystyle \frac{\partial(S,T)}{\partial(V,p)}= 1$