# Extra 5/2 tau term in chemical potential of a monoatomic ideal gas? [closed]

The chemical potential of an ideal monoatomic gas should be: $\mu = \tau ln \frac {n}{n_Q}$ http://web.mit.edu/ndhillon/www/Teaching/Physics/bookse5.html

I get this result if I derive it using the definition of Gibbs Free Energy: G=U+PV-TS

However, if I use the differential equation of enthalpy, H: dH=TdS+VdP+$\mu$dN

I get the result: $\mu = \tau ln (\frac {n}{n_Q}) - \frac 5 2 \tau$

This is the proof, which is where I'm trying to figure where the error is:

dH=$\frac 5 2 Nk_BdT$ = TdS+VdP+$\mu$dN

At constant T and P, -TdS =$\mu$dN

$\mu$ = -T$\frac{dS}{dN}$

For a monoatomic gas (Sackur Tetrode), S = $Nk_Bln(\frac{V}{N\lambda^3}) + \frac{5}{2}Nk_B$

Since $\frac{P}{k_BT} = \frac{N}{V}$

$\mu$=-T$\frac{dS}{dN}$ = -$k_BTln(\frac{k_BT}{P\lambda^3}) - \frac{5}{2}k_BT$ at constant P and T.

Rearrange, $\mu = \tau ln (\frac {n}{n_Q}) - \frac 5 2 \tau$

Just trying to figure out what I am doing wrong with this proof. Shouldn't the answers be the same if I start from G vs. H?

## closed as off-topic by Kyle Kanos, Daniel Griscom, JamalS, Sebastian Riese, GertJan 23 '16 at 1:36

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