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:
Answer: (B) $E_{AB} = E_{AC} = E_{AD},\ W_{AB} > 0,\ W_{AC} = 0,\ W_{AD} < 0$
$B$, $C$ and $D$ all lie on the same isotherm $T_1$ while $A$ is on $T_2$. Internal energy of an ideal gas depends only on $T$, so $E_{AB} = E_{AC} = E_{AD}$.
$A\to B$: volume increases, $W_{AB} > 0$. $A\to C$: constant volume, $W_{AC} = 0$. $A\to D$: volume decreases, $W_{AD} < 0$.
Solution by Sreeraj P, M.Sc Physics