The internal energy of a gas is given by U = 2PV. It expands from ${\mathrm{V}}_0$ to $2{\mathrm{V}}_0$. Against a constant pressure ${\mathrm{P}}_0$, the heat absorbed by the gas in the process is:

As it is a constant pressure process volume becomes double temperature also becomes double so increase in temperature is $T_0$.

given U = 2PV; $∆\mathrm{U}$ =  2${\mathrm{P}}_0{\mathrm{V}}_0$

Work done at constant pressure  ${\mathrm{P}}_0\left(2{\mathrm{V}}_0-{\mathrm{V}}_0\right) ={\mathrm{P}}_0{\mathrm{V}}_0$

dQ = dU + dW = 3${\mathrm{P}}_0{\mathrm{V}}_0$