A perfect gas in equilibrium in a volume $V$ has quantum stationary states $|i\rangle$ with energies $E_{i}$. In a Boltzmann distribution, the probability that the system is in state $|i\rangle$ is $\rho_{i}=Z^{-1} e^{-E_{i} / k T}$. The entropy is defined to be $S=-k \sum_{i} \rho_{i} \log \rho_{i}$.

For two nearby states establish the equation

$$d E=T d S-P d V$$

where $E$ and $P$ should be defined.

For reversible changes show that

$$d S=\frac{\delta Q}{T}$$

where $\delta Q$ is the amount of heat transferred in the exchange.

Define $C_{V}$, the heat capacity at constant volume.

A system with constant heat capacity $C_{V}$ initially at temperature $T$ is heated at constant volume to a temperature $\Theta$. Show that the change in entropy is $\Delta S=$ $C_{V} \log (\Theta / T)$.

Explain what is meant by isothermal and adiabatic transitions.

Briefly, describe the Carnot cycle and define its efficiency. Explain briefly why no heat engine can be more efficient than one whose operation is based on a Carnot cycle.

Three identical bodies with constant heat capacity at fixed volume $C_{V}$, are initially at temperatures $T_{1}, T_{2}, T_{3}$, respectively. Heat engines operate between the bodies with no input of work or heat from the outside and the respective temperatures are changed to $\Theta_{1}, \Theta_{2}, \Theta_{3}$, the volume of the bodies remaining constant. Show that, if the heat engines operate on a Carnot cycle, then

$$\Theta_{1} \Theta_{2} \Theta_{3}=A, \quad \Theta_{1}+\Theta_{2}+\Theta_{3}=B$$

where $A=T_{1} T_{2} T_{3}$ and $B=T_{1}+T_{2}+T_{3}$.

Hence show that the maximum temperature to which any one of the bodies can be raised is $\Theta$ where

$$\Theta+2\left(\frac{A}{\Theta}\right)^{1 / 2}=B$$

Show that a solution is $\Theta=T$ if initially $T_{1}=T_{2}=T_{3}=T$. Do you expect there to be any other solutions?

Find $\Theta$ if initially $T_{1}=300 \mathrm{~K}, T_{2}=300 \mathrm{~K}, T_{3}=100 \mathrm{~K}$.

[Hint: Choose to maximize one temperature and impose the constraints above using Lagrange multipliers. ]