# Entropy expression

1. May 25, 2014

### ChrisVer

1. The problem statement, all variables and given/known data

From the 2nd TD law:
$TdS= d(\rho V) + P dV - \mu d(nV)$

find that:
$S= \frac{V}{T} (\rho+P- \mu n)$

2. Relevant equations

$\frac{dP}{dT}= \frac{P+\rho - \mu n}{T}$

3. The attempt at a solution
$TdS= d(\rho V) + P dV - \mu d(nV)$

$TdS= d[(\rho+ P- \mu n) V] - V dP + nV d \mu$

or I can write:

$dS=\frac{1}{T} d[(\rho+ P- \mu n) V] - \frac{V}{T} dP + \frac{nV}{T} d \mu$

Now I write that (using the given formula):
$dP= \frac{P+\rho - \mu n}{T} dT$

$dS=\frac{1}{T} d[(\rho+ P- \mu n) V] - V (P+\rho - \mu n) \frac{dT}{T^{2}} + \frac{nV}{T} d \mu$

My problem is that this result gives the correct formula I'm looking for the entropy, except for the last $\frac{nV}{T} d \mu$

For the last I tried to take:

$\frac{dS}{dT}= \frac{dS}{d \mu} \frac{d \mu}{dT}= \frac{nV}{T} \frac{d \mu}{dT}$

$\frac{d^{2}S}{d(nV)dT}= \frac{1}{T} \frac{d \mu}{dT}$

Also:
$\frac{dS}{d(nV)}= -\frac{ \mu }{T}$

$\frac{d^{2}S}{dT d(nV)}= \frac{ \mu }{T^{2}}$

So that I have:

$\frac{1}{T} \frac{d \mu}{dT} =\frac{ \mu }{T^{2}}$
That means:
$d \mu = \frac{ \mu }{T} dT$

Inserting in the expression for the entropy at last:

$dS=\frac{1}{T} d[(\rho+ P- \mu n) V] - V (P+\rho - \mu n) \frac{dT}{T^{2}} + (nV \mu) \frac{dT}{T^{2}}$

Which can't be written as:

$dS= d[(\rho+ P- \mu n) \frac{V}{T}]$
due to the last term...

Any help?

2. May 25, 2014

### unscientific

$$dU = TdS - pdV + \mu dN$$