As shown in the figure, a gas undergoes a process during which both...

From the given $p-V$ graph, as the process proceeds, both pressure ($p$) and volume ($V$) of the gas increase. Therefore, $dp > 0$ and $dV > 0$.

According to the ideal gas equation, $pV = nRT$. Differentiating both sides, we get:
$p dV + V dp = nR dT$

Since $p, V, dp$, and $dV$ are all positive quantities, the change in temperature $dT$ must also be positive ($dT > 0$).

From the first law of thermodynamics:
$dQ = dU + dW$

For $n$ moles of gas, this can be written in terms of molar heat capacities as:
$n C dT = n C_V dT + p dV$

Dividing the entire equation by $n dT$, we get the molar heat capacity $C$ for the process:
$C = C_V + \dfrac{p}{n} \dfrac{dV}{dT}$

Since $p > 0$, $n > 0$, $dV > 0$, and $dT > 0$, the term $\dfrac{p}{n} \dfrac{dV}{dT}$ is strictly positive.

Therefore, $C > C_V$.