Relating the entropy of an ideal gas with partial derivatives

[Mayan Fung]

Homework Statement

For an ideal gas, use $dE=TdS-PdV+\mu dN$ to prove
1. $V(\frac{\partial P}{\partial T})_{\mu} = S$
2. $V(\frac{\partial P}{\partial \mu})_T = N$

Relevant Equations

$dE=TdS-PdV+\mu dN$

It looks very easy at first glance. However, the variable S is a variable in the given expression. I have no clue to relate the partial derivatives to entropy and the number of particles.

 

[LCSphysicist]

Using the extensible properties of the system as variables, we know that $E (x \lambda) = \lambda E(X)$ (Homogeneous function of degree a=1), so that we can say

$x * \nabla f = a f$ (Euler's homogeneous theorem), where $x = (x_{1},x_{2},...,x_{n})$ is the vector with the variables.

So that
$$S( \partial E/ \partial S ) + V ( \partial E/ \partial V)+ N (\partial E/ \partial N )= a * E$$
$$ ST - PV + N \mu = E$$

The rest i think you can go on, eventually you will get the Gibbs Duhem equation

 

[etotheipi]

Just to add to what @LCSphysicist wrote, first try to re-write each term on the RHS according to $x\mathrm{d}y = \mathrm{d}(xy) - y\mathrm{d}x$.

 

[Mayan Fung]

Thanks! This makes me recall the fact that $G=\mu N$ and $G=U-TS+\mu N$

 

[etotheipi]

Thanks! This makes me recall the fact that $G=\mu N$ and $G=U-TS+\mu N$

Careful, it's $G := U -TS + pV = \mu N$