For a monoatomic ideal gas following the cyclic process A B C A shown in the U vs ρ plot, identify the…

The internal energy of a system is a function of temperature only, i.e., $U\propto T$

Density is inversely proportional to volume, $\rho \propto \frac{1}{V}$

So, for process AB, slope is constant, therefore, $\frac{U_0}{{\rho }_0}=\frac{T}{{\displaystyle \frac{1}{V}}}\Rightarrow TV\text{= constant, }$using ideal gas equation, $TV=\left(\frac{PV}{nR}\right)V\Rightarrow P{V}^2\text{= constsnt}$

Therefore, AB is a polytropic process, $P{V}^m=\text{constant}$of polytropic index $2.$

For a monoatomic gas, $C_V=\frac{3}{2}R$

So, the molar heat capacity for a polytropic process is, $C=C_V+\frac{R}{1-m}=\frac{3R}{2}+\frac{R}{1-2}\Rightarrow C=\frac{R}{2}$

Now, In process BC the internal energy of the system is decreasing therefore, heat must be rejected by the system.

Now for molar heat capacity for process BC,

BC is an isochoric process, therefore, molar heat capacity in process BC is equal to molar heat capacity at constant volume, therefore, $C_{BC}=C_V=\frac{3}{2}R$

Now, for work done by the system in the process CA,

Since CA is an isothermal process, $W_{CA}=nRT\ln \left(\frac{V_A}{V_C}\right)$

Internal energy $U_0=n{C}_{V}T=n\frac{3}{2}RT\Rightarrow nRT=\frac{2}{3}{U}_0$

$V_A\propto \frac{1}{{\rho }_A}\propto \frac{1}{{\rho }_0}$and $V_C\propto \frac{1}{{\rho }_C}\propto \frac{1}{4{\rho }_0}$

Putting the above results in work done formula,

$W_{CA}=\left(\frac{2}{3}{U}_0\right)\ln \left(\frac{{\displaystyle \frac{1}{{\rho }_0}}}{\frac{1}{4{\rho }_0}}\right)=\frac{2}{3}{U}_0\ln 4$