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Enthapy is defined as $$ H \equiv U + PV $$ This definition is a mathematical equality. It is valid even if the pressure is not constant. But it is more useful under isobaric condition. Because the …

In the answer of Bob, the process is described as an irreveraible precess It is certainly OK if the problem meant to be an irreversible process, an irreversible process without "isothermal". That is f …

partial S}{\partial V}\right)_{E,N};\\ \frac{\mu}{T} =& -\left( \frac{\partial S}{\partial N}\right)_{E,V};\\ \end{align} This structure providing a systematic development from statistical mechanics to thermodynamics …

From the first law of thermodynamics: $\Delta Q = P dV + n C_V dT$ \begin{align} \Delta Q =& P dV + n C_V dT.\,\,\text{Using Eq.(1) to replace }\, dV \,\text{ by } dT\\ \Delta Q =& \left\{ - P\left(\frac …

Given the expresstion if entropy (Imagine that it is derived from microcanonical ensemble statistically): $$ S(U,V,N)=A\,\left(UVN\right)^{1/3}. \tag{1}$$ And the temperature $$ \frac{1}{T} = \frac …

Yes. Both Eq.(1) and Eq.(2) are ok. They simply the calculus chain-rule and change variables. Nothing wrong with that. You don't need a thermodynamic property to validate these equations. It is rigoro …

The Van der Waal's equation try to take into account two effects: 1. the finite volume occupied by moleculars, thus the effective vloume becomes smaller by a parameter $b$; and 2. the attractive forc …

Your are right that the microcanonical ensemble is working with a fixed energy ($E$ as a parameter). But after obataining the entropy with Boltzmann relation $$ S = k \ln \Omega(E). $$ Then your tak …

This connection is linked from both the thermal dynamics and the statistical microcanonocal ensemble. From thermal dynamics, the entropy is defined as $dS =\frac{dQ}{T}$. Using the first law of therma …

In canonical ensemble, the partition funciton $Z(T, V, N)$ is directly related to Helmholtz free energy: $$F = -KT \ln Z(T, V, N) \tag{1}$$, and energy: $$\tag{2} U = \frac{\sum_i E_i e^{-\beta E_i} …

You final result is not correct: \begin{align} \Delta S & = S_f - S_i\\ &= K_B\, ln\left\{B(2V)^{2N} (2E)^{2\frac{3N}{2}}\right\} - \left[K_Bln\left(BV^N E^\frac{3N}{2}\right)+K_Bln\left(BV^NE^\frac{ …

Edit: After posted I found that you were talking about re-loading. That is too weird a condition. You might want to double check of thie "re-loading" condition. I keep the post here in case you need a …

In Clausius theorem for second law of thermodynamics, the temperature in the integrand is the temperature of the reservoir, not the system: $$ \oint \frac{\delta Q}{T_{surr}}. $$ Therefore, there exists …

For an ideal gas $PV = n RT$, the expression of entrop can be calculate: \begin{align} dS =& \frac{dU}{T} + \frac{PdV}{T};\\ =& n C_v \frac{dT}{T} + nR\frac{dV}{V};\\ \end{align} Carry out the integ …

Maximum entropy (maximum configurations) In thermodynamics, the equilibrim of a state is not determined by the maximum of entropy. Then when to apply the maximum entropy principle? …