Two samples A and B are initially kept in the same state. The...

Let the initial state of both samples be $P_0, V_0, T_0$ and the final volume be $V_f$.

Since the samples are expanded, $V_f > V_0$.

For sample $B$, the process is isothermal, so the temperature remains constant. Thus, $T_b = T_0$.

For sample $A$, the process is adiabatic. The relation between temperature and volume is given by $T V^{\gamma-1} = \text{constant}$.

$T_a V_f^{\gamma-1} = T_0 V_0^{\gamma-1}$

$T_a = T_0 \left( \dfrac{V_0}{V_f} \right)^{\gamma-1}$

Since $V_f > V_0$ and $\gamma > 1$, we have \left( \dfrac{V_0}{V_f} \right)^{\gamma-1} < 1$.<br><br>Therefore,$T_a < T_0$.<br><br>Since$T_b = T_0$, it follows that$T_a < T_b$.$