@UmeshKonduru Are you in India, and are you thinking about going to an IIT or IISER? If so these are a bit different from European universities.

When we consider a non-relativistic bose gas, when $T\rightarrow \infty$ then $\mu \rightarrow - \infty$. In the expression for the average occupation number, of an arbitrary 1 particle state we have $\langle n_i \rangle=\frac{1}{e^{\beta(\epsilon_o - \mu)} -1}$ and we say that for $T\rightarrow \infty$ $\langle n_i \rangle \rightarrow e^{-\beta(\epsilon_i - \mu)}$

But in the exponent $\beta \rightarrow 0$ and $\mu \rightarrow -\infty$, which means we have something of the sort $\frac{-\infty}{0}$, how do approximate this

If $\mu$ goes to infinity when $T$ does it's probably a function of it?

Did you try finding out the limit of $\beta \mu(T)$

don't you mean of $e^{\beta \mu(T)}$ ?

Sure why not

Yes, I was able to prove it, thank you

@Slereah One more thing, in our lecture we took this $\Omega= - PV$ and we simply said that it comes from thermodyn. relations

Which relation exactly ?

I do not know

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@imbAF what's your definition of $\Omega$?

the wikipedia definition for the Landau potential is $\Omega=U-TS-\mu N$ where $U$ is the internal energy. Then $\Omega=-PV$ is equivalent to $\mu N = U-TS+PV$, which is just the Gibbs free energy $G$ for a homogeneous system.

so if $\Omega$ is defined via other thermodynamical potentials, it's basically just coming from $G$ being an intensive quantity.