Interpret the $PV$ diagram

• The cycle is traversed as $A \to B \to C \to D \to A$.
• The upper part $ABC$ is a semicircle above the horizontal axis.
• The lower part $CDA$ is half of an ellipse below it.
• Hence the loop is clockwise.

On a $PV$ diagram:

• Work done by gas along a path is $W=\int P\,dV$
• Net work done in a cycle equals the area enclosed by the loop.
• For a clockwise cycle, net work done by gas is positive.

Check option D

Since the cycle $ABCDA$ is clockwise, the enclosed area is positive. Therefore, $W_{\text{cycle}} > 0$ So positive work is done by the gas in the cycle.

Hence, D is correct.

Check option C: work done during $A \to B \to C$

Along $A \to B \to C$, volume increases continuously from the left end to the right end. Thus $dV>0$ throughout, and pressure is positive, so $W_{ABC}=\int_A^C P\,dV >0$ It is definitely not zero.

Hence, C is false.

Check option A: is $A \to B$ isothermal?

For an ideal gas, an isothermal process satisfies $PV=\text{constant}$ and on a $PV$ diagram it is a rectangular hyperbola, not a circular arc in general.

Here $AB$ is part of a semicircle. A circular arc does not represent an isotherm for an ideal gas.

Hence, A is false.

Check option B: heat flow during $B \to C \to D$

Use the first law: $Q=\Delta U + W$ For an ideal gas, $\Delta U \propto \Delta T \propto \Delta(PV)$

Compare states $B$ and $D$ from the figure:

• They lie roughly one above the other, so their volumes are the same.
• Pressure at $B$ is greater than pressure at $D$. Therefore, $P_BV_B > P_DV_D \implies T_B > T_D$ so from $B$ to $D$, $\Delta U_{B\to C\to D} <0$

Now examine work along $B \to C \to D$:

• From $B \to C$, $dV>0$, so work is positive.
• From $C \to D$, $dV<0$, so work is negative.
• The path lies on the right side and then lower arc; overall, the negative contribution dominates sufficiently so that the gas ends at lower $P$ and same $V$, and heat must be rejected on this part of the cycle.

Equivalently, since the full cycle is clockwise, heat is absorbed mainly on the upper path and rejected on the return path. Thus on $B \to C \to D$, heat flows out of the gas.

Hence, B is correct.

Final selection

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