Three different processes that can occur in an ideal monoatomic gas are shown in the $$P$$ vs $$V$$ diagram. The paths are labelled as $$A \to B$$, $$A \to C$$ and $$A \to D$$. The change in internal energies during these processes are taken as $$E_{AB}$$, $$E_{AC}$$ and $$E_{AD}$$ and the work done as $$W_{AB}$$, $$W_{AC}$$ and $$W_{AD}$$. The correct relation between these parameters are:

We are dealing with an ideal mono-atomic gas. For such a gas the internal energy is a pure function of temperature alone and is given by the well-known expression

$$U=\tfrac32\,nRT.$$

Consequently, the change in internal energy between any two states depends only on the initial and final temperatures:

$$\Delta U=\tfrac32\,nR\,(T_{\text{final}}-T_{\text{initial}}).$$

In the $$P\!-\!V$$ diagram all three processes start from the common state $$A(P_A,V_A,T_A)$$ and terminate at the three distinct points $$B,\;C,\;D$$. A careful inspection of the diagram shows that the points $$B,\;C,\;D$$ lie on the same isothermal curve that passes through none of the other points. Algebraically that statement reads

$$P_BV_B=P_CV_C=P_DV_D=k\quad(\text{constant}).$$

Using the ideal-gas equation $$PV=nRT$$ this instantly gives

$$T_B=T_C=T_D=T',$$

where $$T'$$ is some temperature (which may or may not equal $$T_A$$). Because all three final temperatures are equal, the change in internal energy from the common initial state $$A$$ to any of the three final states is identical:

$$E_{AB}=\Delta U_{AB}=\tfrac32\,nR\,(T'-T_A),$$

$$E_{AC}=\Delta U_{AC}=\tfrac32\,nR\,(T'-T_A),$$

$$E_{AD}=\Delta U_{AD}=\tfrac32\,nR\,(T'-T_A).$$

Hence

$$E_{AB}=E_{AC}=E_{AD}.$$

Next we examine the work done, remembering the definition

$$W=\int_{V_{\text{initial}}}^{V_{\text{final}}}P\,dV.$$

Sign conventions to keep in mind:

• When the gas expands, $$dV>0$$ so $$W>0$$(positive work done by the gas).
• When the gas undergoes an isochoric (constant-volume) change, $$dV=0$$ so $$W=0$$(no work).
• When the gas is compressed, $$dV<0$$ so $$W<0$$(negative work, work done on the gas).

Looking again at the geometry of the three paths:

• Along $$A\to B$$ the curve clearly moves to a larger volume: $$V_B>V_A$$. Therefore $$W_{AB}>0.$

• Along $$A\to C$$ the path is vertical, i.e. the volume is unchanged: $$V_C=V_A$$. Hence $$W_{AC}=0.$

• Along $$A\to D$$ the path proceeds to a smaller volume: $$V_D<V_A$$. Consequently $$W_{AD}<0.$

Collecting all the deductions we have

$$E_{AB}=E_{AC}=E_{AD},\qquad W_{AB}>0,\qquad W_{AC}=0,\qquad W_{AD}<0.$$

This set of equalities and inequalities is exactly the one listed in Option B of the question.

Hence, the correct answer is Option B.