A)
\begin{aligned}& \mathrm{V}_{\text {initial }}=\frac{\mathrm{nRT}_{\text {initial }}}{\mathrm{P}_{\text {initial }}}=\frac{(1 \mathrm{~kg})\left(\frac{\mathrm{mol}}{28.02 \mathrm{~g}}\right)\left(\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}}\right)\left(83.14 \frac{\mathrm{bar} \mathrm{~cm}^{3}}{\mathrm{~mol} \mathrm{~K}}\right) 250 \mathrm{~K}}{1 \mathrm{bar}} \\& =\bf 741,800 \mathrm{~cm}^{3} \\& \mathrm{~V}_{\text {final }}=\frac{\mathrm{nRT}_{\text {final }}}{\mathrm{P}_{\text {final }}}=\frac{(1 \mathrm{~kg})\left(\frac{\mathrm{mol}}{28.02 \mathrm{~g}}\right)\left(\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}}\right)\left(83.14 \frac{\mathrm{bar} \mathrm{~cm}^{3}}{\mathrm{~mol} \mathrm{~K}}\right) 250 \mathrm{~K}}{10 \mathrm{bar}}=\mathbf{7 4}, \mathbf{1 8 0} \mathbf{~c m}^{\mathbf{3}}\end{aligned}
Set up an energy balance around the gas
\begin{aligned}\Delta\left\{M\left(\widehat {\rm U}+\frac{\rm v^2}{2}+\rm gh\right) \right\} \\& =\mathrm{m}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\mathrm{m}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+\mathrm{W}_{\mathrm{EC}}+\mathrm{W}_{\mathrm{S}}+\mathrm{Q}\end{aligned}
Cancelling Terms for this closed system:
\Delta \mathrm{M}(\widehat{\mathrm{U}})=\mathrm{W}_{\mathrm{EC}}+\mathrm{Q}
M\left(\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}\right)=W_{E C}+Q
For an ideal gas:
\mathrm{d} \widehat{\mathrm{U}}=\mathrm{C}_{\mathrm{v}} \mathrm{dT}=0
Thus:
\begin{aligned}& \mathrm{W}_{\mathrm{EC}}=-\mathrm{Q}\\&\rm W_{EC}=-\int_{V=741,800 \mathrm{~cm}^{3}}^{V=74,180 \mathrm{~cm}^{3}} \mathrm{PdV}\\& \qquad\qquad\qquad=-\int_{V=741,800 \mathrm{~cm}^{3}}^{V=74,180 \mathrm{~cm}^{3}} \frac{\mathrm{NRT}}{\mathrm{V}} \mathrm{dV}\\& \qquad\qquad\qquad=(1 \mathrm{~kg})\left(\frac{\mathrm{mol}}{28.02 \mathrm{~g}}\right)\left(\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}}\right)\left(8.314 \frac{\mathrm{J}}{\mathrm{mol} \mathrm{K}}\right)(250 \mathrm{~K}) \ln \frac{74180}{741800}\\& \qquad\qquad\qquad\bf=169.7 \mathrm{~kJ}\\& \bf Q=-\mathbf{1 6 9 . 7 ~ k J}\end{aligned}
B) Solve for V_{\text {initial }}
\begin{aligned}& \mathrm{P}=\frac{\mathrm{RT}}{\underline{\mathrm{V}}-\mathrm{b}}-\frac{\mathrm{a}}{\underline{\mathrm{V}}^{2}}\\& 1 \mathrm{bar}=\frac{\left(83.14 \frac{\mathrm{cm}^{3} \mathrm{bar}}{\mathrm{mol} \mathrm{K}}\right)(250 \mathrm{~K})}{(\underline{\mathrm{V}})-\left(\frac{38.6 \mathrm{~cm}^{3}}{\mathrm{~mol}}\right)}-\frac{\left(1.37 \times 10^{6} \frac{\mathrm{cm}^{6} \mathrm{bar}}{\mathrm{mol}^{2}}\right)}{(\underline{\mathrm{V}})^{2}}\\& \underline{V}=20758 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}\\& 1 \mathrm{~kg}\left(20758 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}\right)\left(\frac{1 \mathrm{~mol}}{28.02 \mathrm{~g}}\right)\left(\frac{1000 \mathrm{~g}}{\mathrm{~kg}}\right)=\mathbf{7 4 0 ,828 }\mathbf{c m}^{3}\end{aligned}
Solve for V_{\text {final }}
\begin{aligned}& \mathrm{P}=\frac{\mathrm{RT}}{\underline{\mathrm{V}}-\mathrm{b}}-\frac{\mathrm{a}}{\underline{\mathrm{V}}^{2}}\\& 10 \mathrm{bar}=\frac{\left(83.14 \frac{\mathrm{cm}^{3} \mathrm{bar}}{\mathrm{mol} \mathrm{K}}\right)(250 \mathrm{~K})}{(\underline{\mathrm{V}})-\left(\frac{38.6 \mathrm{~cm}^{3}}{\mathrm{~mol}}\right)}-\frac{\left(1.37 \times 10^{6} \frac{\mathrm{cm}^{6} \mathrm{bar}}{\mathrm{mol}^{2}}\right)}{(\underline{\mathrm{V}})^{2}}\\\\& \underline{\mathrm{V}}=2052 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}\\& 1 \mathrm{~kg}\left(2052 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}\right)\left(\frac{1 \mathrm{~mol}}{28.02 \mathrm{~g}}\right)\left(\frac{1000 \mathrm{~g}}{\mathrm{~kg}}\right)=\bf 73,233 \mathbf{c m}^{3}\end{aligned}
Departure from ideal gas volume is more significant at the higher pressure.
Set up an energy balance around the gas
\begin{aligned}\Delta\left\{M\left(\widehat {\rm U}+\frac{\rm v^2}{2}+\rm gh\right) \right\} \\&=\mathrm{m}_{\mathrm{in}}\left(\widehat{\mathrm{H}}_{\mathrm{in}}+\frac{\mathrm{v}_{\text {in }}^{2}}{2}+\mathrm{gh}_{\mathrm{in}}\right)-\mathrm{m}_{\mathrm{out}}\left(\widehat{\mathrm{H}}_{\mathrm{out}}+\frac{\mathrm{v}_{\mathrm{out}}^{2}}{2}+\mathrm{gh}_{\mathrm{out}}\right)+\mathrm{W}_{\mathrm{EC}}+\mathrm{W}_{\mathrm{S}}+\mathrm{Q}\end{aligned}
Cancelling Terms
\Delta \mathrm{M}(\widehat{\mathrm{U}})=\mathrm{W}_{\mathrm{EC}}+\mathrm{Q}
M\left(\widehat{U}_{\text {final }}-\widehat{U}_{\text {initial }}\right)=W_{E C}+Q
Since Figure 2-3 does not have Specific Internal Energies, we will find each Specific Enthalpy and convert based on calculated Volumes and known Pressures using the equation:
\widehat{H}-P \widehat{V}=\widehat{U}
Nitrogen at 1 bar and 250 \mathrm{~K} \rightarrow \widehat{\mathrm{H}}_{\text {initial }}=408 \frac{\mathrm{kJ}}{\mathrm{kg}}
Nitrogen at 10 bar and 250 \mathrm{~K} \rightarrow \widehat{\mathrm{H}}_{\text {final }}=402 \mathrm{~K} \frac{\mathrm{kJ}}{\mathrm{kg}}
\begin{aligned}& \widehat{U}_{\text {initial }}=\widehat{H}_{\text {initial }}-P_{\text {initial }} \times \widehat{V}_{\text {initial }} \\& \widehat{U}_{\text {initial }}=408 \frac{\mathrm{kJ}}{\mathrm{kg}}-1 \mathrm{bar} \times\left(\frac{740828 \mathrm{~cm}^{3}}{1 \mathrm{~kg}}\right)\left(\frac{1 \mathrm{~kJ}}{10000~ \mathrm{bar}~ \mathrm{cm}^{3}}\right)=333 \frac{\mathrm{kJ}}{\mathrm{kg}} \\& \widehat{U}_{\text {final }}=\widehat{H}_{\text {final }}-P_{\text {final }} \times \widehat{V}_{\text {final }} \\& \widehat{U}_{\text {final }}=402 \frac{\mathrm{kJ}}{\mathrm{kg}}-10~ \mathrm{bar} \times\left(\frac{73233 \mathrm{~cm}^{3}}{1 \mathrm{~kg}}\right)\left(\frac{1 \mathrm{~kJ}}{10000~ \mathrm{bar}~\textrm {cm} ^3}\right)=329 \frac{\mathrm{kJ}}{\mathrm{kg}}\end{aligned}
Solve for \mathrm{W}_{\mathrm{EC}}
\begin{aligned}& \frac{\mathrm{W}_{\mathrm{EC}}}{\mathrm{N}}=-\int_{\text {initial }}^{\text {final }} \mathrm{Pd} \underline{\mathrm{V}}\\& \frac{W_{\mathrm{EC}}}{\mathrm{N}}=-\int_{\text {initial }}^{\text {final }}\left(\frac{\mathrm{RT}}{\underline{\mathrm{V}}-\mathrm{b}}-\frac{\mathrm{a}}{\underline{\mathrm{V}}^{2}}\right) \mathrm{dV}\\& \frac{\mathrm{W}_{\mathrm{EC}}}{\mathrm{N}}=-\left(83.14 \frac{\mathrm{cm}^{3} \mathrm{bar}}{\mathrm{mol} \mathrm{K}}\right)(250 \mathrm{~K})\left(\int_{\text {initial }}^{\text {final }}\left(\frac{1}{\underline{\mathrm{V}}-\mathrm{b}}\right) \mathrm{dV}\right)-\left(\mathrm{a} \int_{\text {initial }}^{\text {final }}\left(\frac{1}{\underline{V}^{2}}\right) \mathrm{dV}\right)\\& \frac{\mathrm{W}_{\mathrm{EC}}}{\mathrm{N}}=-\left(83.14 \frac{\mathrm{cm}^{3} \mathrm{bar}}{\mathrm{mol} \mathrm{K}}\right)(250 \mathrm{~K})\left\{\ln \left(\frac{\mathrm{V}_{\text {final }}-\mathrm{b}}{\underline{V}_{\text {initial }}-\mathrm{b}}\right)-\mathrm{a}\left(\frac{1}{\underline{V}_{\text {initial }}}-\frac{1}{\underline{V}_{\text {final }}}\right)\right\}\\& \frac{\mathrm{W}_{\mathrm{EC}}}{\mathrm{N}}=-\left(83.14 \frac{\mathrm{cm}^{3} \text { bar }}{\mathrm{mol} \mathrm{K}}\right)(250 \mathrm{~K})\left\{\ln \left(\frac{2052 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}-\frac{38.6 \mathrm{~cm}^{3}}{\mathrm{~mol}}}{20758 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}-\frac{38.6 \mathrm{~cm}^{3}}{\mathrm{~mol}}}\right)-1.37\right.\\& \qquad\qquad \left.\times 10^{6} \frac{\mathrm{cm}^{6} \mathrm{bar}}{\mathrm{mol}^{2}}\left(\frac{1}{20758 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}}-\frac{1}{2052 \frac{\mathrm{cm}^{3}}{\mathrm{~mol}}}\right)\right\}\\& \frac{\mathrm{W}_{\mathrm{EC}}}{\mathrm{N}}=49056 \frac{\mathrm{cm}^{3} \mathrm{bar}}{\mathrm{mol}}= \mathbf{4905. 6} \frac{\mathbf{J}}{\mathbf{mol}}\\& \mathrm{W}_{\mathrm{EC}}=\left(4905.6 \frac{\mathrm{J}}{\mathrm{mol}}\right)\left(\frac{1 \mathrm{~mol}}{28.02 \mathrm{~g}}\right)(1000 \mathrm{~g})=175,100 \mathrm{~J}=\mathbf{175.1} \bf{k J}\end{aligned}
Plug everything back into the energy balance
\begin{aligned}& 1 \mathrm{~kg}\left(329 \frac{\mathrm{kJ}}{\mathrm{kg}}-333 \frac{\mathrm{kJ}}{\mathrm{kg}}\right)=175.1 \mathrm{~kJ}+\mathrm{Q}\\& Q=-179.1 \,\mathbf{k J}\end{aligned}