Chapter 19 - Heat Engines and Refrigerators - Exercises and Problems - Page 553: 60

See the detailed answer below.

a) We can use the ideal gas law, $PV=nRT$, to find the temperatures. Thus, $$T_1=\dfrac{P_1V_1}{nR}=\dfrac{(400\times 10^3)(0.025)}{(2)(8.31)}=\color{red}{\bf 602}\;\rm K$$ $$T_2=\dfrac{P_2V_2}{nR}=\dfrac{(600\times 10^3)(0.050)}{(2)(8.31)}=\color{red}{\bf 1805}\;\rm K$$ $$T_3=\dfrac{P_3V_3}{nR}=\dfrac{(400\times 10^3)(0.050)}{(2)(8.31)}=\color{red}{\bf 1203}\;\rm K$$ __________________________________________________________ b) $\textbf{In the first process,}$ We know that the work done by the system in one process is equal to the area under the curve. Hence, $$(W_s)_{1\rightarrow2}=\text{Area under the $(P-V)$ curve}$$ $$(W_s)_{1\rightarrow2}=\frac{1}{2}(0.05-0.025)(600-400)\times 10^3+(400\times 10^3)(0.05-0.025)$$ $$(W_s)_{1\rightarrow2}=\color{blue}{\bf 12,500}\;\rm J$$ $$(\Delta E_{th})_{1\rightarrow2}=\frac{3}{2}nR(T_2-T_1)$$ $$(\Delta E_{th})_{1\rightarrow2}=\frac{3}{2}(2)(8.31)(1805-602)$$ $$(\Delta E_{th})_{1\rightarrow2}=\color{blue}{\bf 29,990}\;\rm J$$ $$Q_{1\rightarrow2}=(\Delta E_{th})_{1\rightarrow2}+(W_s)_{1\rightarrow2}=29,990+12,500 $$ $$Q_{1\rightarrow2}=\color{blue}{\bf 42,490}\;\rm J$$ $\textbf{In the second process,}$ It is an isochoric process and the volume of the gas is constant. $$(W_s)_{2\rightarrow3}=\color{blue}{\bf 0}\;\rm J$$ $$(\Delta E_{th})_{2\rightarrow3}=\frac{3}{2}nR(T_3-T_2)$$ $$(\Delta E_{th})_{2\rightarrow3}=\frac{3}{2}(2)(8.31)(1203-1805)$$ $$(\Delta E_{th})_{2\rightarrow3}=\color{blue}{\bf -15,008}\;\rm J$$ $$Q_{2\rightarrow3}=(\Delta E_{th})_{2\rightarrow3}+(W_s)_{2\rightarrow3}=-15,008+0$$ $$Q_{2\rightarrow3}=\color{blue}{\bf -15,008}\;\rm J$$ $\textbf{In the third process,}$ It is an isobaric process: $$(W_s)_{3\rightarrow1}=P_1\Delta V=P_1(V_1-V_3)$$ $$(W_s)_{3\rightarrow1}=(400\times 10^3) (0.025-0.050)$$ $$(W_s)_{3\rightarrow1}=\color{blue}{\bf -10,000}\;\rm J$$ $$(\Delta E_{th})_{3\rightarrow1}=\frac{3}{2}nR(T_1-T_3)$$ $$(\Delta E_{th})_{3\rightarrow1}=\frac{3}{2}(2)(8.31)(602-1203)$$ $$(\Delta E_{th})_{3\rightarrow1}=\color{blue}{\bf -14,983}\;\rm J$$ $$Q_{3\rightarrow1}=(\Delta E_{th})_{3\rightarrow1}+(W_s)_{3\rightarrow1}= -14,983-10,000$$ $$Q_{2\rightarrow3}=\color{blue}{\bf -24,983}\;\rm J$$ 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& 29,990&12,500 &42,490\\ \hline 2\rightarrow 3& -15,008& 0 &-15,008\\ \hline 3\rightarrow1&-14,983& -10,000 &-24,983\\ \hline {\rm Sum}& 0 & 2,500&Q_{in}=42,490\\ \hline \end{array} __________________________________________________________ c) The engine’s thermal efficiency is given by $$\eta=\dfrac{W_{net}}{Q_{in}}=\dfrac{2,500}{42,490}$$ $$\eta=\color{red}{\bf 0.06}$$