Chapter 19 - Heat Engines and Refrigerators - Exercises and Problems - Page 552: 52

a) \begin{array}{|c|c|c|c|} \hline & \Delta E_{th}\;{\rm (J)}& W_s\;{\rm (J)}&Q\;{\rm (J)}\\ \hline 1\rightarrow 2& 3750 & 0&3750 \\ \hline 2\rightarrow 3 & 7489& 4993&12482 \\ \hline 3\rightarrow4 & -7499 & 0&-7499 \\ \hline 4\rightarrow1& -2493& -2493&-6232 \\ \hline {\rm Sum}& &2500 &\\ \hline \end{array} b) $0.15$

a) First we need to find $V_{\rm min}$ at point 1, $$P_1V_1=nRT_1$$ Hence, $$V_1=V_{\rm min}=\dfrac{nRT_1}{P_1}=\dfrac{(1)(8.31)(300)}{(300\times 10^3)}$$ $$V_1=V_2=\color{blue}{\bf 8.31\times 10^{-3}}\;\rm m^3$$ Hence, $$V_3=V_4=2V_2=\color{blue}{\bf 16.62\times 10^{-3}}\;\rm m^3$$ $\bf\text{The first process 1$\rightarrow$2: [Isochoric]}$ We know that the volume of the gas is constant, so the work done is zero. and hence $$(\Delta E_{th})_{1\rightarrow2}=Q_{1\rightarrow2}=nC_{\rm V}(T_2-T_1)\tag 1$$ And we are given the heat entered already, so $$(\Delta E_{th})_{1\rightarrow2}=Q_{1\rightarrow2}=\color{red}{\bf 3750}\;\rm J$$ $$(W_s)_{1\rightarrow2} =\color{red}{\bf 0}\;\rm J$$ We can use (1) to find $T_2$; $$3750=nC_{\rm V}(T_2-T_1)=\frac{3}{3}n R(T_2-T_1)$$ $$T_2=\dfrac{3750}{nC_{\rm V}}+T_1$$ Plugging the known; $$T_2=\dfrac{3750}{(\frac{3}{2})(1)(8.31)}+300= \color{blue}{\bf 600.8}\;\rm K$$ Now we can find $P_{max}=P_2$, $$P_2=P_3=\dfrac{nRT_2}{V_2}=\dfrac{(1)(8.31)(600.8)}{(8.31\times 10^{-3})}$$ $$P_2=P_3=\color{blue}{\bf 600.8}\;\rm kPa$$ Let's find $T_3$ and $T_4$, $$T_3=\dfrac{P_3V_3}{nR}=\dfrac{(600.8\times 10^3)(16.62\times 10^{-3})}{(1)(8.31)}=\color{blue}{\bf 1201.6}\;\rm K$$ $$T_4=\dfrac{P_4V_4}{nR}=\dfrac{(300\times 10^3)(16.62\times 10^{-3})}{(1)(8.31)}=\color{blue}{\bf 600}\;\rm K$$ $\bf\text{The second process 2$\rightarrow$3: [Isobaric]}$ $$(W_s)_{2\rightarrow3}=P_2dV=P_2(V_3-V_2)=P_2(2V_2-V_2)$$ $$(W_s)_{2\rightarrow3}=P_2V_2=(600.8\times 10^3)(8.31\times 10^{-3})$$ $$(W_s)_{2\rightarrow3} =\color{red}{\bf 4993}\;\rm J$$ $$(\Delta E_{th})_{2\rightarrow3}=\frac{3}{2}nR(T_3-T_2)=\frac{3}{2}(1)(8.31)(1201.6-600.8)$$ $$(\Delta E_{th})_{2\rightarrow3} =\color{red}{\bf 7489}\;\rm J$$ $$Q_{2\rightarrow 3}=nC_{\rm P}\Delta T=\frac{5}{2}nR(T_3-T_2)$$ $$Q_{2\rightarrow 3}=\frac{5}{2}(1)(8.31)(1201.6-600.8)$$ $$Q_{2\rightarrow 3} =\color{red}{\bf 12482}\;\rm J$$ $\bf\text{The third process 3$\rightarrow$4: [Isochoric]}$ $$(W_s)_{3\rightarrow4} =\color{red}{\bf 0}\;\rm J$$ $$(\Delta E_{th})_{3\rightarrow4}=\frac{3}{2}nR(T_4-T_3)=\frac{3}{2}(1)(8.31)(600-1201.6)$$ $$(\Delta E_{th})_{3\rightarrow4} =\color{red}{\bf -7499}\;\rm J$$ Hence, $$Q_{3\rightarrow 4} =\color{red}{\bf -7499}\;\rm J$$ $\bf\text{The fourth process 4$\rightarrow$1: [Isobaric]}$ $$(W_s)_{4\rightarrow1}=P_1dV=P_1(V_1-V_4)=P_1(V_2-2V_2)$$ $$(W_s)_{4\rightarrow1}=-P_1V_2=-(300\times 10^3)(8.31\times 10^{-3})$$ $$(W_s)_{4\rightarrow1}=\color{red}{\bf -2493}\;\rm J$$ $$(\Delta E_{th})_{4\rightarrow1}=\frac{3}{2}nR(T_1-T_4)=\frac{3}{2}(1)(8.31)(300-600)$$ $$(\Delta E_{th})_{4\rightarrow1} =\color{red}{\bf -3740}\;\rm J$$ $$Q_{4\rightarrow1}=nC_{\rm P}\Delta T=\frac{5}{2}nR(T_1-T_4)$$ $$Q_{4\rightarrow1}=\frac{5}{2}(1)(8.31)(300-600)$$ $$Q_{4\rightarrow1} =\color{red}{\bf -6232}\;\rm J$$ \begin{array}{|c|c|c|c|} \hline & \Delta E_{th}\;{\rm (J)}& W_s\;{\rm (J)}&Q\;{\rm (J)}\\ \hline 1\rightarrow 2& 3750 & 0&3750 \\ \hline 2\rightarrow 3 & 7489& 4993&12482 \\ \hline 3\rightarrow4 & -7499 & 0&-7499 \\ \hline 4\rightarrow1& -2493& -2493&-6232 \\ \hline {\rm Sum}& &2500 &\\ \hline \end{array} --- b) The thermal efficiency is given by $$\eta=\dfrac{W_{out}}{Q_H}$$ where $Q_H=Q_{1\rightarrow2}+Q_{2\rightarrow 3}$ Plugging from the table; $$\eta=\dfrac{2500}{3750+12482}$$ $$\eta=\color{red}{\bf 0.15}$$