Two samples A and B of the same gas have equal volumes and...

Let the initial pressure and volume of both samples be $P_0$ and $V_0$.

For sample A, the gas undergoes isothermal expansion to double its volume. The work done is given by:

$W_A = nRT_0 \ln\left(\dfrac{V_f}{V_i}\right) = P_0 V_0 \ln\left(\dfrac{2V_0}{V_0}\right) = P_0 V_0 \ln 2$

For sample B, the gas undergoes adiabatic expansion to double its volume. The work done is given by:

$W_B = \dfrac{P_i V_i - P_f V_f}{\gamma - 1}$

Using the adiabatic equation $P_i V_i^{\gamma} = P_f V_f^{\gamma}$, we find the final pressure $P_f$:

$P_f = P_0 \left(\dfrac{V_0}{2V_0}\right)^{\gamma} = P_0 2^{-\gamma}$

Substituting $P_f$ and $V_f = 2V_0$ into the work equation:

$W_B = \dfrac{P_0 V_0 - (P_0 2^{-\gamma})(2V_0)}{\gamma - 1} = \dfrac{P_0 V_0 (1 - 2^{1-\gamma})}{\gamma - 1}$

Given that the work done in both cases is identical, $W_A = W_B$:

$P_0 V_0 \ln 2 = \dfrac{P_0 V_0 (1 - 2^{1-\gamma})}{\gamma - 1}$

$1 - 2^{1-\gamma} = (\gamma - 1) \ln 2$