Chapter 19 - Heat Engines and Refrigerators - Exercises and Problems - Page 552: 55

See the detailed answer below.

First of all, we need to find the number of moles of the gas $n$, temperatures $T_1$, $T_3$, and $P_{max}=P_1$. The first process, from 1 to 2, it is an isobaric process. So $$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$$ Hence, $$T_1=\dfrac{V_1T_2}{V_2}=\dfrac{(100)(600)}{(600)}=\bf 100\;\rm K$$ The second process, from 2 to 3, it is an isochoric process. So So $$\dfrac{P_2}{T_2}=\dfrac{P_3}{T_3}$$ Hence, $$T_3=\dfrac{T_2P_3}{P_2}\tag 1$$ So we need to find $P_2=P_1$; The third process, from 3 to 1, it is an adiabatic process. So, $$P_3V_3^\gamma=P_1V_1^\gamma$$ where, for monatomic gas, $\gamma=C_P/C_V=(5R/2)/(3R/2)=5/3$ $$P_1=P_3 \left[\dfrac{V_3}{V_1}\right]^\gamma=(100)\left[\dfrac{600}{100}\right]^\frac{5}{3}$$ $$P_1=P_2=\bf 1981\;\rm kPa$$ Plugging into (1); $$T_3=\dfrac{(600)(100)}{(1981)}=\bf 30.3\;\rm K$$ Now we can find the number of moles of this gas, $$n=\dfrac{P_1V_1}{RT_1}=\dfrac{(1981\times 10^3)(100\times 10^{-6})}{(8.31)(100)}$$ $$n=\bf 0.2384\;\rm mol$$ --- a) $\textbf{The first process 1$\rightarrow$2: [Isobaric] }$ $$(W_s)_{1\rightarrow2}=P\Delta V=P_1(V_2-V_2)$$ $$(W_s)_{1\rightarrow2}=(1981\times 10^3)(600-100)\times 10^{-6}$$ $$(W_s)_{1\rightarrow2}=\color{red}{\bf 990.5}\;\rm J$$ $$(\Delta E_{th})_{1\rightarrow 2}=nC_V\Delta T=\frac{3}{2}nR(T_2-T_1) $$ $$(\Delta E_{th})_{1\rightarrow 2}=\frac{3}{2}(0.2384)(8.31)(600-100) $$ $$(\Delta E_{th})_{1\rightarrow2}=\color{red}{\bf 1486}\;\rm J$$ $$Q_{1\rightarrow 2}=nC_P\Delta T=\frac{5}{2}nR(T_2-T_1) $$ $$Q_{1\rightarrow 2}=\frac{3}{2}(0.2384)(8.31)(600-100) $$ $$Q_{1\rightarrow2}=\color{red}{\bf 2476}\;\rm J$$ $\textbf{The second process 2$\rightarrow$3: [Isochoric] }$ $$(W_s)_{2\rightarrow3}=P\Delta V=P_1(0)$$ $$(W_s)_{2\rightarrow3}=\color{red}{\bf 0}\;\rm J$$ $$(\Delta E_{th})_{2\rightarrow3}=nC_V\Delta T=\frac{3}{2}nR(T_3-T_2) $$ $$(\Delta E_{th})_{2\rightarrow 3}=\frac{3}{2}(0.2384)(8.31)(30.3-600) $$ $$(\Delta E_{th})_{2\rightarrow3}=\color{red}{\bf -1693}\;\rm J$$ $$Q_{2\rightarrow 3}=(\Delta E_{th})_{2\rightarrow3}=\color{red}{\bf -1693}\;\rm J$$ $\textbf{The third process 3$\rightarrow$1: [Adiabtic] }$ $$Q_{3\rightarrow 1} =\color{red}{\bf 0}\;\rm J$$ $$(\Delta E_{th})_{3\rightarrow1}=nC_V\Delta T=\frac{3}{2}nR(T_1-T_3) $$ $$(\Delta E_{th})_{3\rightarrow1}=\frac{3}{2}(0.2384)(8.31)(100-30.3) $$ $$(\Delta E_{th})_{3\rightarrow1}=\color{red}{\bf 207}\;\rm J$$ And hence, $$(W_s)_{3\rightarrow1} =-(\Delta E_{th})_{3\rightarrow1} $$ $$(W_s)_{3\rightarrow1} =\color{red}{\bf- 207}\;\rm J$$ \begin{array}{|c|c|c|c|} \hline & \Delta E_{th}\;{\rm (J)}& W_s\;{\rm (J)}&Q\;{\rm (J)}\\ \hline 1\rightarrow 2& 1486& 990.5&2476\\ \hline 2\rightarrow 3 & -1693 & 0&-1693 \\ \hline 3\rightarrow1& 207&-207 &0\\ \hline {\rm Sum}& 0&783.5&Q_{in}=2476\\ \hline \end{array} _________________________________________________________ b) The thermal efficiency of this heat engine is given by $$\eta=\dfrac{W}{Q_{in}}=\dfrac{783.5}{2476}$$ $$\eta=\color{red}{\bf 0.32}$$