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As a consequence of the 3rd law of thermodynamics or more generally a finite value of residual entropy, you also need to have a vanishing of the heat capacity at low temperature. …

It depends on what you are interested in. It is often more physically relevant to think at fixed number of particles rather than fixed $\mu$, so $\mu$ picks up an implicit temperature dependence. Take …

While you can choose any two independent variables as your basis function to express any quantity, you need to be careful in the choice of your variables when defining new quantities as derivatives. D …

Neglecting pressure and volume, your energy is: $$ dU=TdS+HdM $$ You can therefore apply the same method by formally substituting $P\to-H$ and $V\to M$. The analogue for enthalpy is: $$ \mathcal H = U …

Actually, thermodynamic entropy depends on the definition of your system in some sense, just as the statistical entropy depends on the micro-states. The classic example is Gibbs’ paradox. If you have …

On a side note, you can avoid the Trotter formula by using instead the standard interaction picture (if you are already familiar with it from previous QM courses). I will write $\partial_{x_i} = \part …

This will rather lead to the consideration of free enthalpy $G = F+pV$ which should be more familiar if you've done chemistry or thermodynamics. The discussion is essentially the same though. …

Quasistatic processes encompass reversible processes. The inclusion is strict, because there are some irreversible quasistatic transformations. A typical example of a quasi static process would be a s …

Question 1 Yes, you can derive this from the first law of thermodynamics. For a cyclic process $W=Q$, heat equals work. …

In the following, I'll set $k_B=1$. Usually, the fluctuations of energy are rather captured by heat capacity. This is captured by the following formula saying that heat capacity is proportional to the …

Actually, your question has little to do with statistical mechanics, but more about classical thermodynamics. … It typically holds in the thermodynamic limit, which is why for most applications classical thermodynamics applies. …

In general, the two distributions have little to do with each other. After all, $\rho_e$ is fixed but $\rho_{mc}$ depends on the variable energy $E$. I will therefore reformulate your question as: can …

Yes, it works without any problems. I’ll just add that you can do it all in one go by modifying the expression of your entropy: $$ S =nc_v\ln(T/T_0)+nR\ln\left(\frac{V/n}{ V_0/n_0}\right)+ns_0 $$ with …

Assuming your system is isolated you want to maximize $S$ with fixed $U$ according to the second law. More concretely, you want to maximize $S=S_A+S_B$ by only varying $U_A,U_B$ under the constraint $ …

The fact that the variables are intensive/extensive is irrelevant. These partial derivatives just come from differential calculus. Slight notation change, I’ll use $M$ for the magnetic moment instead …