During a particular process, both the pressure p and the volume V...

For an ideal gas, the equation of state is $pV = nRT$.

Since both pressure $p$ and volume $V$ increase, their product $pV$ increases. Therefore, the temperature $T$ of the gas must increase.

The work done by the gas is given by $W = \int p dV$. Since the volume of the gas increases, $dV > 0$, which implies that the work done by the system is positive.

According to the first law of thermodynamics, $Q = \Delta U + W$. Since $W > 0$, the heat supplied $Q$ is greater than the change in internal energy $\Delta U$.

Such a process is physically possible, for example, $p \propto V$.