Chapter 19 - Heat Engines and Refrigerators - Exercises and Problems - Page 554: 72

See the detailed answer below.

First of all, we need to find the number of moles of the gas, so we need to use the ideal gas law. $$PV=nRT$$ $$n=\dfrac{P_1V_1}{RT_1}=\dfrac{(1.013\times 10^5)(1050\times 10^{-6})}{(8.31)(25+273)}=\bf 0.043\;\rm mol$$ And we know that $\gamma=\dfrac{C_{\rm P}}{C_{\rm V}}=1.4$, and that $C_{\rm P}-C_{\rm V}=R$, so $\dfrac{C_{\rm V}+R}{C_{\rm V}}=1.4$. Thus, $$C_{\rm V}=\frac{5}{2}R~~~~{\rm and}~~~~C_{\rm P}=\frac{7}{2}R$$ $$\bf (a)$$ We have all the data for point 1. Let's work on point 2, from 1 to 2 it is an adiabatic process where $$P_1V_1^\gamma=P_2V_2^\gamma$$ Hence, $$P_2=P_1\left[\dfrac{V_1}{V_2}\right]^\gamma=1.0\left[\dfrac{1050}{50}\right]^{1.4}=\bf 71\;\rm atm$$ With the same approach, for the same process, $$T_2=T_1\left[\dfrac{V_1}{V_2}\right]^{\gamma-1}=(25+273)\left[\dfrac{1050}{50}\right]^{1.4-1}=\bf 1007\;\rm K=\bf 734^\circ\rm C$$ At point 3, we don't know the volume or the temperature, but we know the pressure. Noting that from 2 to 3 there is an isobaric process where $Q=1000\;\rm J$. Thus, $$Q=nC_{\rm P}(T_3-T_2)$$ Thus, $$T_3=\dfrac{Q}{nC_{\rm P}}+T_2 =\dfrac{1000}{(0.043)(\frac{7}{2})(8.31)}+(734+273)=\bf 1807\;\rm K=\bf 1534^\circ \rm C$$ Now it is easy to find $V_3$ by using the ideal gas law, $$V_3=\dfrac{nRT_3}{P_3}=\dfrac{(0.043)(8.31)(1807)}{71\times 1.013\times 10^5}=\bf 89.8\;\rm cm^3$$ Now from 3 to 4, we have an adiabatic process, so $$P_3V_3^\gamma=P_4V_4^\gamma$$ Hence, $$P_4=P_3\left[\dfrac{V_3}{V_4}\right]^{\gamma}=(71)\left[\dfrac{89.8 }{1050 }\right]^{1.4}=\bf 2.27\;\rm atm$$ And $T_4$ is given by $$T_4=\dfrac{P_4V_4}{nR}=\dfrac{ (2.27 \times 1.013\times 10^5)(1050\times 10^{-6})}{(0.043)(8.31)}=\bf 675.7\;\rm K=\bf 402.7^\circ\rm C$$ See the table below: \begin{array}{|c|c|c|c|} \hline &P\;({\rm atm})& V\;({\rm cm^3})&T\;({\rm ^\circ C})\\ \hline \rm Point 1& 1 & 1050&25 \\ \hline \rm Point 2 & 71& 50&734\\ \hline \rm Point 3 & 71& 89.8&1534 \\ \hline \rm Point 4 &2.27 & 1050&402.7\\ \hline \end{array} $$\bf (b)$$ The net work done in one full cycle is given by $$W=W_{1\rightarrow2}+W_{2\rightarrow3}+W_{3\rightarrow4}+W_{4\rightarrow1}$$ From 4 to 1, is an isochoric process, so $W=0$ $$W=W_{1\rightarrow2}+W_{2\rightarrow3}+W_{3\rightarrow4}+0$$ From 1 to 2, and 3 to 4 are adiabatic processes, so $Q=0$, and hence, $\Delta E_{th}=-W$. and the work done on the isobaric process is given by the area under the curve. $$W=-nC_{\rm V}(T_2-T_1)+P_{2}(V_3-V_2)+-nC_{\rm V}(T_4-T_3)$$ $$W=-\frac{5}{2}nR(T_2-T_1+T_4-T_3)+P_{2}(V_3-V_2) $$ $$W=-\frac{5}{2}(0.043)(8.31)(1007-298+675.7-1808)\\ +(71\times 1.013\times 10^5)(89.8-50)\times 10^{-6} $$ $$W=\color{red}{\bf 664}\;\rm J$$ $$\bf (c)$$ The thermal efficiency is given by $$\eta=\dfrac{W}{Q_H}=\dfrac{664}{1000}=\color{red}{\bf0.664}$$ $$\bf (d)$$ The power output of an engine of 8 cylinders is given by $${\rm Power}=8\cdot \dfrac{W}{t}=8\cdot\dfrac{664}{60}(2400)$$ $${\rm Power}=\color{red}{\bf 212 }\;\rm kW=\color{red}{\bf 284}\;\rm hp$$