$$ \begin{aligned} & \mathrm{n}=5, \mathrm{~T}=293 \mathrm{~K}=\text { const, } \Delta \mathrm{U}=0, \\ & \mathrm{P}_1=2.1 \mathrm{MPa}, \mathrm{P}_2=1.3 \mathrm{MPa} \\ & \mathrm{P}_{\text {ext }}=4.3 \mathrm{MPa}=\text { const. } \\ & \mathrm{W}=-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_2-\mathrm{V}_1\right)=-\mathrm{P}_{\text {ext }}\left(\frac{\mathrm{nRT}}{\mathrm{P}_2}-\frac{\mathrm{nRT}}{\mathrm{P}_1}\right) \\ & \mathrm{W}=-\mathrm{P}_{\text {ext }} \mathrm{nRT}\left(\frac{1}{\mathrm{P}_2}-\frac{1}{\mathrm{P}_1}\right) \\ & \text { or, } \\ & =-4.3 \times 5 \times 8.314 \times 293\left(\frac{1}{1.3}-\frac{1}{2.1}\right) \\ & =-4.3 \times 5 \times 8.314 \times 293\left(\frac{2.1-1.3}{1.3 \times 2.1}\right) \\ & =-15347.7 \mathrm{~J} \\ & \text { or, } \quad \mathrm{W}=-15.35 \mathrm{~kJ} \\ & \Delta \mathrm{U}^0=\mathrm{q}+\mathrm{W} \\ & \therefore \quad \mathrm{q}=-\mathrm{W} \\ & \text { or, } \quad q=15.35 \mathrm{~kJ} \text { (for } 5 \text { moles) } \\ & \therefore \quad q / \text { mole }=\frac{15.35}{5}=3 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$