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Homework Statement
Given [itex]dE=TdS-PdV+\mu dN[/itex], [itex]PV=NT[/itex] and [itex]Cv=\frac{3}{2}N[/itex]
Find E
Homework Equations
[itex]dE=TdS-PdV+\mu dN[/itex], [itex]PV=NT[/itex] and [itex]Cv=\frac{3}{2}N[/itex]
The Attempt at a Solution
That part that is confusing me is the chemical potential, not sure what to do with it. Finding the energy for the ideal gas is pretty simple without it
[itex]dE=TdS-PdV+\mu dN[/itex]
[itex]dE=T\frac{\partial S}{\partial T}dT+(T\frac{\partial S}{\partial v}|_T-P)dV+\mu dN[/itex] expanding dS
[itex]dE=C_v dT+(T\frac{\partial P}{\partial T}|_V-P)dV+\mu dN[/itex] by maxwell relation
[itex]dE=C_v dT+(T\frac{N}{V}-P)dV+\mu dN[/itex] by taking derivative of ideal gas law
[itex]dE=C_v dT+(P-P)dV+\mu dN[/itex]
[itex]dE=\frac{3}{2}NdT+\mu dN[/itex]
[itex]E=\int\frac{3}{2}NdT+\int\mu dN[/itex]
Without the chemical potential its simply [itex]E=\frac{3}{2}NT[/itex] using [itex]E(t=0)=0[/itex] and pretty much no source I've looked at makes reference to the chemical potential when calculating the internal energy so I'm confused.
Here are some comments by the person who set homework when someone asked him about N being constant