Let T_a and T_b be the final temperatures of the samples A and B...

For two identical samples of an ideal gas expanding from the same initial state to the same final volume, the work done is given by the area under the $P-V$ curve, $\Delta W = \int P dV$.

The equation for an isothermal process (system A) is $PV = \text{constant}$, and for an adiabatic process (system B) is $PV^{\gamma} = \text{constant}$.

Since $\gamma > 1$, the adiabatic curve is steeper than the isothermal curve. During expansion, the isothermal curve lies above the adiabatic curve, meaning $P_{\text{isothermal}} > P_{\text{adiabatic}}$ at any given volume.

As a result, the area under the isothermal curve is greater than the area under the adiabatic curve.

Therefore, the work done by system A is greater than the work done by system B, giving $\Delta W_a > \Delta W_b$.