The work performed by a sample of an ideal gas during process A is...

According to the first law of thermodynamics:

$Q = \Delta U + W$

For an ideal gas, the change in internal energy depends only on the change in temperature, so $\Delta U = n C_V \Delta T$.

The molar heat capacity for any process is given by $C = \dfrac{Q}{n \Delta T}$.

Substituting the expression for $Q$, we get:

$C = \dfrac{\Delta U + W}{n \Delta T} = C_V + \dfrac{W}{n \Delta T}$

For process $A$ and process $B$, the molar heat capacities are:

$C_A = C_V + \dfrac{W_A}{n \Delta T}$

$C_B = C_V + \dfrac{W_B}{n \Delta T}$

Subtracting the two equations yields:

$C_A - C_B = \dfrac{W_A - W_B}{n \Delta T}$

It is given that the work performed by the gas in process $A$ is twice that in process $B$, so $W_A = 2W_B$. Since work is performed by the gas, it is positive ($W_B > 0$), which implies $W_A > W_B$.

It is also given that the temperature increases in both processes, meaning $\Delta T > 0$.

Therefore, $C_A - C_B = \dfrac{W_B}{n \Delta T} > 0 \Rightarrow C_A > C_B$.