Chemical potential of an ideal gas

[chrisdk]

hi,
How can I show that the chemical potential of an ideal gas µ(T,V ) can be given by:
$$\mu(T , V , N )=k_{b}T\left(a+1+ln(\frac{Nv_{0}}{V})\right)$$

 

[DrDu]

The only condition for an ideal gas is that it fulfills pV=NkT.
There are no restrictions to its internal energy and it doesn't even need to be extensive, so I don't think you cann show that formula for mu without further assumptions.

 

[Bill_K]

The Gibbs free energy is G = μN = U + PV -TS. It has three terms, and in your formula there are also three terms. It must be that they correspond: U = akT and S = -k ln(Nv₀/V). Can you derive either of these?

 

[chrisdk]

Does it have to do something with Stirling's approximation (the latter one)? Concepts of statistical thermodynamics were never my strong side... :( Bill, can you help me out with this?

 

[Curl]

I don't know wtf u got in the OP but the way of deriving the chemical potential of an ideal gas is to use the definition μ= -T (∂S/∂N)|U,V

Use this on the Sakur-Tetrode equation. Use mathematica is you suck at differentiation.