Question

One mole of a monoatomic gas of molar mass $M$ undergoes a cyclic process as shown in the figure. Here, $\rho$ is the density and $P$ is the pressure of the gas. Find the efficiency of the cycle.

• A. $\frac{3}{5}(1-\ln 2)$
• B. $\frac{2}{5}(1-\ln 2)$
• C. $\frac{3}{5}\ln 2$
• D. $\frac{2\ln 2}{5}$

Answer 1

The correct option is B $\frac{2}{5}(1-\ln 2)$

We can see that the line $1-2$ will pass through the origin if extended. Since $\rho$ is doubled on doubling $P$, it means $\rho \propto P$ i.e the process must be isothermal.
[since $\rho =\frac{PM}{RT}$]
Thus, Work done in process 1-2
$\Delta W_{12}=nRT\ln \frac{P_1}{P_2}=P_1{V}_1\ln \frac{1}{2}$
$=P_0\frac{M}{{\rho }_0}\ln \left(\frac{1}{2}\right)=-\frac{P_{0}M}{{\rho }_0}\ln 2$
Also, $\Delta U_{12}=0$ [since $T$ is constant]
$\Rightarrow \Delta Q_{12}=\Delta W_{12}=-\frac{P_{0}M}{{\rho }_0}\ln 2$ [heat rejected]

Process 2-3 is isobaric
Thus, $\Delta W_{23}=2P_0(V_3-V_2)=2P_0(\frac{M}{{\rho }_3}-\frac{M}{{\rho }_2})$
$=2P_{0}M(\frac{1}{{\rho }_0}-\frac{1}{2{\rho }_0})=\frac{P_{0}M}{{\rho }_0}$
$\Delta U_{23}=n{C}_v\Delta T=1\times \frac{3}{2}R[T_3-T_2]$
$=\frac{3}{2}[P_3{V}_3-P_2{V}_2]=\frac{3}{2}.2P_0(\frac{M}{{\rho }_0}-\frac{M}{2{\rho }_0})=\frac{3}{2}\frac{P_{0}M}{{\rho }_0}$
$\Rightarrow \Delta Q_{23}=\Delta W_{23}+\Delta U_{23}=\frac{5}{2}\frac{P_{0}M}{{\rho }_0}$ [heat supplied]

Process 3-1 is isochoric
So, $\Delta W_{31}=0$
$\Delta U_{31}=n{C}_v\Delta T=n.\frac{3}{2}R[T_1-T_3]$
$=\frac{3}{2}[P_1{V}_1-P_3{V}_3]=\frac{3}{2}[\frac{P_{0}M}{{\rho }_0}-\frac{2P_{0}M}{{\rho }_0}]$
$=-\frac{3}{2}\frac{P_{0}M}{{\rho }_0}$
$\Rightarrow \Delta Q_{31}=\Delta W_{31}=-\frac{3}{2}\frac{P_{0}M}{{\rho }_0}$ [heat rejected]

Efficiency of the cycle
$\eta =\frac{\text{Work done}}{\text{Heat supplied}}=\frac{\frac{P_{0}M}{{\rho }_0}-\frac{P_{0}M}{{\rho }_0}\ln 2}{\frac{5}{2}\frac{P_{0}M}{{\rho }_0}}$
$=\frac{1-\ln 2}{\frac{5}{2}}=\frac{2}{5}(1-\ln 2)$