# Efficiency of a thermodynamic cycle

We are given a thermodynamic cycle of an ideal gas (but of rigid spheres) composed of a single substance which undergoes a Carnot cycle illustrated below. All processes are assumed to be reversible. We have isotherms at $(1,2)$ and $(3,4)$ and adiabatics at $(2,3)$ and $(4,1)$. The gas obeys the following law:

$p(V-B)=NRT, B=const$ An infinitesimal heat transfer is given by $\delta Q=C_VdT+L_VdV$, where $C_V, L_V$ are the specific heat at constant volume and latent heat of expansion respectively.

I am attempting to find the the efficiency $e$ of the cycle (the correct answer is $e= 1-\frac{T_-}{T_+}$ (which is the same as for the Carnot cycle of a normal ideal gas?)

Here is my attempt: as $e$ is defined as $e=-\frac{W}{Q^+}$, I just have to find the total work done ($W=W_{12}+W_{23}+W_{34}+W_{41}$) and the heat extracted from the hot source, $Q^+$. Now, here is where I get stuck:

$W_{12}+W_{34}= -NR(T^+\ln(\frac{V_2-B}{V_1-B})+T^-\ln(\frac{V_4-B}{V_3-B}))$. Apparently $W_{23}+W_{41}=0$, but I do not know how to prove this: (can I still use $dU=cNRdT$ even for this rigid-sphere model?) However, even with this result taken into consideration, I can still not deduce the final result. Is there perhaps a faster approach, using a T-S cycle?

• Let $V^\prime = V-B$. Then $\mathrm{d}V^\prime = \mathrm{d}V$. You should find things look a little familiar. – By Symmetry Jun 26 '17 at 16:20
• @ChesterMiller Actually, in this case, $L_V=p$, which I have proved in the other exercise. Otherwise, it is the temperature times the partial derivative of entropy(in terms of T and V) wrt to volume. I dont think that this is needed though. – Ruslan Mushkaev Jun 26 '17 at 17:49
• For this particular gas, are you aware of whether $C_v$ is a function of specific volume or not? Do you know how to determine whether it is? – Chet Miller Jun 26 '17 at 20:49