Chapter 17 - Work, Heat, and the First Law of Thermodynamics - Exercises and Problems - Page 501: 77

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First of all, we need to find the number of moles of the given gas sample. $$n=\dfrac{m}{M_{He}}=\dfrac{0.120\;\rm g}{4\;\rm g/mol}=\bf 0.03\;\rm mol$$ Applying the ideal gas law to find $T_1$, $T_2$, and $T_3$: $$PV=nRT$$ $$T_1=\dfrac{P_1V_1}{nR}=\dfrac{(3\times 1.013\times 10^5)(1000\times 10^{-6})}{(0.03)(8.31)}$$ $$T_1=T_2=\bf 1219\;\rm K=\bf 946^\circ C$$ Noting that from 1 to 2 the gas undergoes an isothermal process, so $T_2=T_2$. From (3) to (1) the gas undergoes an adiabatic process, so $$T_1V^{\gamma-1}_1=T_3V^{\gamma-1}_3$$ Hence, $$\dfrac{T_3}{T_1}=\left[\dfrac{V_1}{V_3}\right]^{\gamma-1}$$ $$ T_3 =T_1\left[\dfrac{V_1}{V_3}\right]^{\frac{C_{\rm p}}{C_{\rm v}}-1}$$ Plugging the known; $$ T_3 =(1219)\left[\dfrac{1000}{3000}\right]^{\frac{20.8}{ 12.5}-1}=\bf 588\;\rm K=\bf315^\circ C$$ From (1) to (2) is an isothermal process, so $$P_1V_1=P_2V_2$$ Hence, $$P_2=\dfrac{P_1V_1}{V_2}=\dfrac{(3)(1000)}{(3000)}=\bf 1\;\rm atm$$ From (2) to (3) is an isochoric process, so $$\dfrac{P_2}{T_2}=\dfrac{P_3}{T_3}$$ Hence, $$P_3=\dfrac{P_2T_3}{T_2}=\dfrac{(1)(588)}{(1219)}=\bf 0.48\;\rm atm$$ \begin{array}{|c|c|c|} \hline {\rm Point}& P\;{\rm (atm)}& V\;{\rm (cm^3)} &T\;{\rm (^\circ C)}\\ \hline 1 & 3 & 1000&946 \\ \hline 2 & 1 & 3000& 946\\ \hline 3 & 0.48 & 3000 &315\\ \hline \end{array} ____________________________________________________________________________ b) and C) $\bullet\bullet$ From (1) to (2): it is an isothermal process: $$W_{1\rightarrow 2}=-nRT_1\int \dfrac{dV}{V}=-nRT_1\ln\left(\dfrac{V_2}{V_1}\right)$$ $$W_{1\rightarrow 2}=-(0.03)(8.31)(1219)\ln\left(\dfrac{3000}{1000}\right)$$ $$W_{1\rightarrow 2}=\color{red}{\bf -334}\;\rm J$$ In this process $\Delta E_{th}=0$ Hence, $$Q_{1\rightarrow 2}=-W_{1\rightarrow 2}=\color{red}{\bf 334}\;\rm J$$ $\bullet\bullet$ From (2) to (3): it is an isochoric process: $$W_{2\rightarrow 3}=\color{red}{\bf 0}\;\rm J$$ Hence, $$ Q_{2\rightarrow 3}=\Delta E_{th} =nC_{\rm v}(T_3-T_2)$$ $$ Q_{2\rightarrow 3}=(0.03)(12.5) (588-1219)$$ $$ Q_{2\rightarrow 3}=\color{red}{\bf -237}\;\rm J$$ From (3) to (1): it is an adiabatic process: $$\Delta E_{th}=Q_{3\rightarrow 1}+W_{3\rightarrow 1}$$ where $Q$ in such processes is zero, so $$ Q_{3\rightarrow 1}=\color{red}{\bf 0}\;\rm J$$ $$ W_{3\rightarrow 1}=\Delta E_{th} =nC_{\rm v}(T_1-T_3)$$ $$ W_{3\rightarrow 1} =(0.03)(12.5) (1219-588)$$ $$W_{3\rightarrow 1}=\color{red}{\bf 237}\;\rm J$$ See the table below --- \begin{array}{|c|c|c|} \hline {\rm Process}& Q\;{\rm (J)}& W\;{\rm (J)} \\ \hline 1\rightarrow 2 & 334 &-334\\ \hline 2 \rightarrow 3& -237 & 0\\ \hline 3 \rightarrow 1 & 0& 237\\ \hline \end{array}