A monatomic gas is slowly and adiabatically compressed from initial conditions T$0$ and P$0$ to twice the original pressure. The equation of state of this gas is P $=$ k$\_$B T$/($v $-$ b$),$ where v is the volume per particle and b is a constant. Note that this is not an ideal gas and therefore you should not make any assumptions as such. What is the final temperature of the gas? How much work per molecule is required in order to accomplish this change of state?

Answer and Tip $($Show all the steps for this solution$)$

$9\mathrm{.}21T_f=2^{\frac{2}{5}}{T}_0,\frac{W}{N}=\frac{3}{2}(2^{\frac{2}{5}}-1)k_B{T}_0$

To find $T_0,$ consider a $2-$step process of pressurizing at constant temperature followed by

heating$/$cooling at constant pressure: $ds=(\frac{\text{d}\text{e}\text{l}\text{s}}{\text{d}\text{e}\text{l}\text{P}}{)}_{T}dP+(\frac{\text{d}\text{e}\text{l}\text{s}}{\text{d}\text{e}\text{l}\text{T}}{)}_{P}dT=0.$ Relate the partial

differential at constant $T$ to measurable properties. Find an expression for the partial

differential at constant $P$ from the entropy obtained from the Gibbs free energy of a

non$-$ideal gas. Find the work obtained by proper integration of $dE.$