Physics - Heat & Thermo Question with Solution | TestHub

$$\begin{array}{r}{\mathrm{V}}_1=20\times 10^{-3}{\mathrm{m}}^3,\\ {\mathrm{T}}_3={\mathrm{T}}_1=300\mathrm{K}\end{array}$$

$$\begin{array}{r}\text{ For process AB }(P=\text{ constant })T_2=2T_1=600\\ \text{ For process BC }T{V}^{\gamma -1}=\text{ constant }\text{ Therefore }\left(2T_1\right)\\ \Rightarrow V_3=\left(V_2\right)(B{)}^{3/2}=2\sqrt{2}{V}_2=80\sqrt{2}\times 10^{-3}{\mathrm{m}}^3=80\sqrt{2}\text{ litres }\\ P_3=\frac{\mu R{T}_3}{V_3}=\frac{(2)(8.3)(300)}{80\sqrt{2}\times 10^{-3}}=0.44\times 10^5\mathrm{N}/{\mathrm{m}}^2\end{array}$$

Work done under isobaric process $=\mathrm{P}\Delta \mathrm{V}={\mathrm{P}}_1\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)$

$$=\frac{\mu {\mathrm{RT}}_1}{{\mathrm{V}}_1}\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)=\frac{2\times 8.3\times 300}{20\times 10^{-3}}\left(40\times 10^{-3}-20\times 10^{-3}\right)=4980\mathrm{J}$$

Work done under adiabatic process $=\frac{\mu \mathrm{R}\left({\mathrm{T}}_2-{\mathrm{T}}_3\right)}{(\gamma -1)}=\frac{2\times 8.3\times (300)}{(5/3-1)}=7470\mathrm{J}$