Question

[Key Concept: In an isobaric process Q = nC_PΔT, W = nRΔT, ΔU = nC_VΔT and V is proportional to T.]

$$n = 1,{\text{ }T}_{1} = 27^{\circ}C = 300\text{ }K $$Heat added, $Q = 8.31$ watt-hr

$$\begin{matrix} & \ = 8.31 \times 3600\text{ }J \\ C_{P} & \ = \frac{5R}{2};\ C_{V} = \frac{3R}{2} \end{matrix}$$

For an isobaric process

$$\begin{matrix} & Q = {nC}_{P}\Delta\text{ }T \\ & 8.31 \times 3600 = (1)\left( \frac{5R}{2} \right)\Delta T \\ & R = 8.31 \end{matrix}$$

$\begin{matrix} & \ \Rightarrow \ \Delta T = 1440\text{ }K \\ & \ \Rightarrow \ \text{~Final temperature~} \\ & \ T_{2} = T_{1} + \Delta T = 1740\text{ }K\text{~not~}1740^{\circ}C \\ & \text{~Work done~} = P\Delta\text{ }V = nR\Delta\text{ }T \\ & \ = 1440R \end{matrix}$

Change in internal energy

$${\begin{matrix} \Delta U & \ = {nC}_{V}\Delta\text{ }T \\ & \ = (1)\left( \frac{3R}{2} \right)(1440) \\ & \ = 2160R \end{matrix} At\ constant\ pressure }\begin{matrix} & V \propto \text{ }T \\ & \frac{{\text{ }V}_{2}}{{\text{ }V}_{1}} = \frac{T_{2}}{{\text{ }T}_{1}} = \frac{1740}{300} = 5.8 \end{matrix}$$