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(C) The internal energy of a monoatomic ideal gas is $U = \frac{3}{2} n R T$,so $U \propto T$.Density $\rho = \frac{m}{V}$,so $\rho \propto \frac{1}{V

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Molar heat capacity for the process $BC$ is $\frac{2 R}{3}$

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(C) The internal energy of a monoatomic ideal gas is $U = \frac{3}{2} n R T$,so $U \propto T$.Density $ho = \frac{m}{V}$,so $ho \propto \frac{1}{V}$.For process $AB$,the line passes through origin in $U$-$ho$ plot,so $U = k ho$. Substituting $U \propto T$ and $ho \propto 1/V$,we get $T \propto 1/V$,or $TV = 	ext{constant}$.Using $PV = nRT$,we have $(PV/nR)V = 	ext{constant}$,so $PV^2 = 	ext{constant}$. This is a polytropic process with index $m = 2$.The molar heat capacity is $C = C_V + \frac{R}{1-m} = \frac{3R}{2} + \frac{R}{1-2} = \frac{3R}{2} - R = \frac{R}{2}$. Thus,option $A$ is correct.In process $BC$,$ho$ is constant,so it is an isochoric process. Since $U$ decreases,temperature decreases,so heat is rejected. Thus,option $B$ is correct.For an isochoric process,the molar heat capacity is $C_V = \frac{3}{2} R$. Option $C$ states it is $\frac{2R}{3}$,which is incorrect.In process $CA$,$U$ is constant,so $T$ is constant (isothermal). $W_{CA} = nRT \ln(V_A/V_C) = nRT \ln(ho_C/ho_A) = (\frac{2}{3} U_0) \ln(4ho_0/ho_0) = \frac{2}{3} U_0 \ln 4$. Option $D$ is correct.

For a monoatomic ideal gas following the cyclic process $ABCA$ shown in the $U$ vs $\rho$ plot,identify the incorrect option:

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