Chapter 19 - Heat Engines and Refrigerators - Exercises and Problems - Page 553: 61

See the detailed answer below.

$$\bf (a)$$ We need to find the temperature at point 2 by using the ideal gas law; $$T_2=\dfrac{P_2V_2}{nR}=\dfrac{(400\times 10^3)(1000\times 10^{-6})}{(0.02)(8.31)}=\color{red}{\bf 2407}\;\rm K$$ From 1 to 2, the gas undergoes an adiabatic process. So, $$P_1V_1^\gamma=P_2V_2^\gamma$$ Hence, $$V_1=V_2\left[\dfrac{P_2}{P_1}\right]^\frac{1}{\gamma}$$ Since this is a diatomic gas, so $\gamma=7/5$ $$V_1=1000\left[\dfrac{400}{100}\right]^\frac{5}{7}=\bf 2692\;\rm cm^3$$ Thus, $$T_1=\dfrac{P_1V_1}{nR}=\dfrac{(100\times 10^3)(2692\times 10^{-6})}{(0.02)(8.31)}=\color{red}{\bf 1620}\;\rm K$$ $$T_3=\dfrac{P_3V_3}{nR}=\dfrac{(400\times 10^3)(2692\times 10^{-6})}{(0.02)(8.31)}=\color{red}{\bf 6479}\;\rm K$$ $$\bf (b)$$ $\textbf{In the first process}$, which is an adiabatic one, the heat exchange is zero. $$Q_{1\rightarrow 2}= \color{red}{\bf 0}\;\rm J $$ Thus, $$(\Delta E_{th})_{1\rightarrow 2}=-(W_s)_{1\rightarrow 2}=\frac{5}{2}nR(T_2-T_1)$$ $$(\Delta E_{th})_{1\rightarrow 2}=\frac{5}{2}(0.02)(8.31)(2407-1620)$$ $$(\Delta E_{th})_{1\rightarrow 2}= \color{red}{\bf 327}\;\rm J $$ Anf hence, $$(W_s)_{1\rightarrow 2}= \color{red}{\bf -327}\;\rm J $$ $\textbf{In the second process}$, which is an isobaric process, $$(W_s)_{2\rightarrow 3}=P_2(V_3-V_2)$$ $$(W_s)_{2\rightarrow 3}=(400\times 10^3)(2692-1000)\times 10^{-6}$$ $$(W_s)_{2\rightarrow 3}= \color{red}{\bf 677}\;\rm J $$ $$(\Delta E_{th})_{2\rightarrow 3}= \frac{5}{2}nR(T_3-T_2)$$ $$(\Delta E_{th})_{2\rightarrow 3}=\frac{5}{2}(0.02)(8.31)(6479-2407)$$ $$(\Delta E_{th})_{2\rightarrow 3}=\color{red}{\bf 1692 }\;\rm J $$ $$Q_{2\rightarrow 3}= \frac{7}{2}nR(T_3-T_2)$$ $$Q_{2\rightarrow 3}=\frac{7}{2}(0.02)(8.31)(6479-2407)$$ $$Q_{2\rightarrow 3}=\color{red}{\bf 2369}\;\rm J $$ $\textbf{In the third process}$, which is an isochoric process, $$(W_s)_{3\rightarrow 1}= \color{red}{\bf 0}\;\rm J $$ Hence, $$(\Delta E_{th})_{3\rightarrow 1}=Q_{3\rightarrow 1}= \frac{5}{2}nR(T_1-T_3)$$ $$(\Delta E_{th})_{3\rightarrow 1}=Q_{3\rightarrow 1}=\frac{5}{2}(0.02)(8.31)(1620-6479)$$ $$(\Delta E_{th})_{3\rightarrow 1}=Q_{3\rightarrow 1}= \color{red}{\bf -2019}\;\rm J $$ See the table below. \begin{array}{|c|c|c|c|} \hline & \Delta E_{th}\;{\rm (J)}& W_s\;{\rm (J)}&Q\;{\rm (J)}\\ \hline 1\rightarrow 2& 327& -327&0 \\ \hline 2\rightarrow 3& 1692 &677 &2369\\ \hline 3\rightarrow1&-2019& 0&-2019\\ \hline {\rm Sum}& 0 & 350&Q_{in}=2369\\ \hline \end{array} $$\bf (c)$$ The engine’s thermal efficiency is given by $$\eta=\dfrac{(W_s)_{total}}{Q_{in}}=\dfrac{350}{2369}$$ $$\eta=\color{red}{\bf 0.15}$$