1. Key idea: internal energy of an ideal monoatomic gas

For an ideal gas, $U = \frac{3}{2}nRT$ so the change in internal energy depends only on temperature: $\Delta U = \frac{3}{2}nR\Delta T$ Using the ideal gas law, $PV = nRT$ we see that temperature depends only on the product $PV$.

So if the final states $B, C, D$ are such that $P_BV_B = P_CV_C = P_DV_D,$ then $T_B = T_C = T_D$ and hence $E_{AB} = E_{AC} = E_{AD}.$

This is the standard interpretation of such a diagram: the three processes start from the same initial point $A$ and end at different points lying on the same isotherm.

2. Work done in each process

Work done by the gas is $W = \int P\,dV$ So the sign of work depends on whether volume increases, stays constant, or decreases.

• Process $A \to B$: volume increases, so $W_{AB} > 0$

• Process $A \to C$: this is vertical in the $P$-$V$ diagram (isochoric), so $dV=0 \implies W_{AC}=0$

• Process $A \to D$: volume decreases, so $W_{AD} < 0$

3. Compare with options

We obtained: $E_{AB}=E_{AC}=E_{AD}, \qquad W_{AB}>0, \qquad W_{AC}=0, \qquad W_{AD}<0$

This matches Option B exactly.

4. Final answer

The correct relation is: $E_{AB}=E_{AC}=E_{AD}, \quad W_{AB}>0, \quad W_{AC}=0, \quad W_{AD}<0$ So the correct option is B.