I do not understand Equations of state

[Another]

why $\ln(\frac{V}{Nλ^3})$ Equal to $\ln(E^{3/2})$? so $E^{3/2}=\frac{V}{Nλ^3}$ ?
i think $λ∝\frac{N^{1/2}}{E^{1/2}}$ and $E^{3/2}∝\frac{N^{3/2}}{λ^3}$

 

[SchroedingersLion]

Do the following:

Write $\lambda =C*(\frac N E )^{\frac 1 2}$ with some constant C.
Now insert it into the first equation and take the derivative w.r.t E.

 

[vanhees71]

The correct Sackur-Tetrode formula for an ideal monatomic gas reads
$$S=k_{\text{B}} N \left \{ \ln \left [\frac{V}{N} \left (\frac{4 \pi m}{3h^2} \frac{E}{N} \right )^{3/2} \right]-\frac{5}{2} \right \}.$$
Then you have
$$\mathrm{d} E = T \mathrm{d} S-p \mathrm{d} V + \mu \mathrm{d} N \; \Rightarrow \; \mathrm{d} S=\frac{1}{T} \mathrm{d} E + \frac{p}{T} \mathrm{d} V-\frac{\mu}{T} \mathrm{d} N,$$
i.e., the Sackur-Tetrode equation in the above form is already written in the "natural thermodynamical independent variables for entropy", $(U,V,N)$. From the above differential you read off
$$\frac{1}{T}=\frac{\partial S}{\partial U}=\frac{3 k_{\text{B}} N}{2U} \; \Rightarrow \; U= \frac{3 k_{\text{B}} N T}{2}$$
and
$$\frac{p}{T} = \frac{\partial S}{\partial V} = \frac{k_{\text{B}} N}{V} \; \Rightarrow \; p V =k_{\text{B}} N T$$
and
$$\frac{\mu}{T}=-\frac{\partial S}{\partial N} = -k_{\text{B}} \ln \left [V \left (\frac{4 \pi m E}{3 h^2 N} \right)^{3/2} \right].$$