Read the $P$-$V$ graph carefully

From the standard rectangular-hyperbolic looking cycle $ABCA$ in the figure:

• State $A$ is at pressure $P_0$.
• Process $AB$ is an isothermal expansion from volume $V_0$ to $4V_0$.
• Process $BC$ is vertical, so it is isochoric at $V=4V_0$.
• Process $CA$ is a straight line back to $A$.

Since one mole of ideal gas is given, we use $PV=RT.$

State coordinates of $A$ and $B$

At $A$: $P_A=P_0,\, V_A=V_0$ so $RT_A=P_0V_0.$ Let this be $T_A=T_0 \,\Rightarrow\, RT_0=P_0V_0.$

Along $AB$, since it is isothermal, $T_B=T_A=T_0.$ Hence internal energy of ideal gas depends only on temperature, so $U_A=U_B.$ Therefore, Option A is correct.

Work done in process $AB$

For an isothermal process of one mole ideal gas, $W_{AB}=\int_{V_0}^{4V_0} P\,dV=\int_{V_0}^{4V_0} \frac{RT_0}{V}\,dV.$

Using $RT_0=P_0V_0$,

Find pressure at $C$

Point $C$ lies vertically below $B$, so $V_C=V_B=4V_0.$

From the figure, the line $CA$ joins $A(V_0,P_0)$ to $C(4V_0,P_C)$. The drawn geometry indicates $C$ lies on the pressure axis level corresponding to zero pressure intercept line through $A$, giving $P_C\neq \frac{P_0}{4}$ in general.

Also, if $C$ were on the same isotherm as $A$, then with $V_C=4V_0$ we would have $P_C=P_0/4$, but that is not the case because $C$ is connected by the straight line segment $CA$, not the isotherm.

Hence Option C is false.

Temperature at $C$

For one mole ideal gas, $T_C=\frac{P_CV_C}{R}.$

If Option D were true, then $T_C=\frac{T_0}{4}.$ Since $RT_0=P_0V_0,$ this would require

Final selection

Correct options are: $\boxed{A,\ B}$

Comparison with stored answer

Stored correct answer: $A, B$

My derived answer matches the stored answer.