Solving Entropy Expression Homework Statement

[ChrisVer]

Homework Statement

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

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

Homework Equations

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

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?

 

[unscientific]

Homework Statement

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

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

Homework Equations

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

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?

Start with:

dU = TdS - pdV + \mu dN