Chemical potential of an ideal gas

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The discussion focuses on deriving the chemical potential of an ideal gas, expressed as µ(T, V, N) = k_b T(a + 1 + ln(Nv_0/V). It emphasizes that the ideal gas law (pV = NkT) is a fundamental condition, but additional assumptions may be necessary to validate the formula. The relationship between Gibbs free energy and the terms in the proposed formula is explored, particularly linking internal energy and entropy to the chemical potential. The use of the definition of chemical potential and the Sakur-Tetrode equation is suggested as a method for derivation. The conversation highlights the importance of statistical thermodynamics in understanding these concepts.
chrisdk
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hi,
How can I show that the chemical potential of an ideal gas µ(T,V ) can be given by:
<br /> \mu(T , V , N )=k_{b}T\left(a+1+ln(\frac{Nv_{0}}{V})\right)<br />
 
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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.
 
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?
 
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?
 
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.
 

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