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

It’s not quite correct. I’ll note internal energy $U$ rather than $E$. It is true for any process: $$ dU=\delta W +\delta Q=-pdV+TdS $$ It is only for reversible processes that you can identify the te …

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 $ …

Yes, actually this reasoning is quite similar with the one for calculating the reflection coefficient of a Fabry-Pérot interferometer. In you case, it is even more simple though. Note that you can sol …

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 …

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. …

You just need to apply Bernoulli’s theorem: $$ \frac{1}{2}v_1^2+h_1= \frac{1}{2}v_2^2+h_2 $$ You therefore need to compare the variation of enthalpy in the compressible and incompressible case. For an …

I'll start with your first question on entropy. The construction is used to explain phase coexistence. Take for example the liquid gas transition and that the $X$ is the total volume. The curve repres …

More commonly, the mechanical/optical analogy is rather between two position/momentum pairs at different instants in time related by a canonical transformation related by the generating function: $$ d …

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 …

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 …

I will add quantitative details to Bob D's answer. Joule expansion There is no work or heat exchange $W=Q=0$ so to internal energy variation $\Delta U=0$ and volume $V$ varies. For a gas, you only hav …

For the exercise 14, assume you have an interacting term $AB$ in $H$, because it is hermitian, you’ll necessarily also have the term $A^\dagger B^\dagger$ as well. You therefore need to reduce $AB+ A^ …

I think that effusivity is more relevant in this context. I don't have the original question, but I guess it's along the line of whether you'd get more easily burnt by touching a hot piece of wood vs …

I'll just write your internal energy $E$ as $U$ and enthalpy $H = U+pV$. To simplify things (less variables) I'll only consider molar quantities and assume extensivity, which gets rid of $N$. The equa …