- #1
ChrisVer
Gold Member
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Homework Statement
From the 2nd TD law:
[itex] TdS= d(\rho V) + P dV - \mu d(nV) [/itex]
find that:
[itex] S= \frac{V}{T} (\rho+P- \mu n)[/itex]
Homework Equations
[itex] \frac{dP}{dT}= \frac{P+\rho - \mu n}{T}[/itex]
The Attempt at a Solution
[itex] TdS= d(\rho V) + P dV - \mu d(nV) [/itex]
[itex] TdS= d[(\rho+ P- \mu n) V] - V dP + nV d \mu [/itex]
or I can write:
[itex] dS=\frac{1}{T} d[(\rho+ P- \mu n) V] - \frac{V}{T} dP + \frac{nV}{T} d \mu [/itex]
Now I write that (using the given formula):
[itex] dP= \frac{P+\rho - \mu n}{T} dT [/itex]
[itex] dS=\frac{1}{T} d[(\rho+ P- \mu n) V] - V (P+\rho - \mu n) \frac{dT}{T^{2}} + \frac{nV}{T} d \mu [/itex]
My problem is that this result gives the correct formula I'm looking for the entropy, except for the last [itex] \frac{nV}{T} d \mu [/itex]
For the last I tried to take:
[itex] \frac{dS}{dT}= \frac{dS}{d \mu} \frac{d \mu}{dT}= \frac{nV}{T} \frac{d \mu}{dT} [/itex]
[itex] \frac{d^{2}S}{d(nV)dT}= \frac{1}{T} \frac{d \mu}{dT} [/itex]
Also:
[itex]\frac{dS}{d(nV)}= -\frac{ \mu }{T}[/itex]
[itex]\frac{d^{2}S}{dT d(nV)}= \frac{ \mu }{T^{2}}[/itex]
So that I have:
[itex] \frac{1}{T} \frac{d \mu}{dT} =\frac{ \mu }{T^{2}}[/itex]
That means:
[itex] d \mu = \frac{ \mu }{T} dT [/itex]
Inserting in the expression for the entropy at last:
[itex] 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}} [/itex]
Which can't be written as:
[itex] dS= d[(\rho+ P- \mu n) \frac{V}{T}] [/itex]
due to the last term...
Any help?