- #1

thecommexokid

- 70

- 2

## Homework Statement

(Excerpted from a longer, multipart problem but essentially)

Show that for an ideal gas,

$$ \frac{\partial p}{\partial T}\bigg)_\mu = \frac{S}{V}. $$

## Homework Equations

• The ideal gas law, of course

$$ pV = Nk_{\rm B}T $$

• Pressure, temperature, and chemical potential are related in an ideal gas via

$$ \mu-\mu_0=k_{\rm B}T\ln\frac{p}{p_0} $$

• The Gibbs free energy

$$ G = E - TS + pV = \mu N $$

• Hopefully not the Sackur–Tetrode equation.

## The Attempt at a Solution

*Thoughts regarding left-hand side:*

So I rearranged the ##\mu## equation for ##p##:

$$ p(T,\mu) = p_0e^{(\mu-\mu_0)/k_{\rm B}T}. $$

Computing the partial derivative,

$$ \frac{\partial p}{\partial T}\bigg)_\mu = -\frac{\mu-\mu_0}{k_{\rm B}T^2}p. $$

By the ideal gas law ##p/k_{\rm B}T = N/V##, so

$$ \frac{\partial p}{\partial T}\bigg)_\mu = \frac{(\mu_0-\mu)N}{VT}. $$

*Thoughts regarding the right-hand side:*

From the definition of the Gibbs free energy,

$$ \frac{S}{V} = \frac{E + pV - \mu N}{VT}. $$

These two would be equal if

$$ E + pV = \mu_0N $$

in an ideal gas. But I don't really understand what the definition of the "reference" pressure and chem. potential ##p_0## and ##\mu_0## are. So maybe this is true, but if it is I don't know why!

…Or perhaps there's an altogether different starting point or process that gives the result. I am by no means married to my approach here.