Question 2.24: When one lb-mol is heated from 25°F to 200°F at a constant p...

A) This process can be broken into five steps

\begin{aligned}& \Delta \text { Enthalpy }=\left(25^{\circ} \mathrm{F} \text { to } 50^{\circ} \mathrm{F}\right)+\text { Melting }+\left(50^{\circ} \mathrm{F} \text { to } 150^{\circ} \mathrm{F}\right)+\text { Vaporizing } \\&\qquad\qquad\qquad +\left(150^{\circ} \mathrm{F} \text { to } 200^{\circ} \mathrm{F}\right) \\& \Delta \widehat{\mathrm{H}}=\int_{25^{\circ} \mathrm{F}}^{50^{\circ} \mathrm{F}} \mathrm{C}_{\mathrm{p}, \text { solid }} \mathrm{dT}+\Delta \widehat{\mathrm{H}}_{\text {fus }}+\int_{50^{\circ} \mathrm{F}}^{150^{\circ} \mathrm{F}} \mathrm{C}_{\mathrm{p}, \text { liquid }} \mathrm{dT}+\Delta \widehat{\mathrm{H}}_{\mathrm{vap}}+\int_{150^{\circ} \mathrm{F}}^{200^{\circ} \mathrm{F}} \mathrm{C}_{\mathrm{p}, \text { gas }}^{*} \mathrm{dT}\\&\Delta \widehat{\mathrm{H}}=\int_{25+459.67^{\circ} \mathrm{R}}^{50+459.67^{\circ} \mathrm{R}}\left(0.7 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}{ }^{\circ} \mathrm{R}}\right) \mathrm{dT}+\left(75 \frac{\mathrm{BTU}}{\mathrm{lb} \mathrm{m}}\right)+\int_{50+459.67^{\circ} \mathrm{R}}^{150+459.67^{\circ} \mathrm{R}}\left(0.85 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}{ }^{\circ} \mathrm{R}}\right) \mathrm{dT} \\&\qquad +\left(125 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right)+\int_{150+459.67^{\circ} \mathrm{R}}^{200+459.67^{\circ} \mathrm{R}}\left(1-\frac{\mathrm{T}}{1200}+\mathrm{T}^{2} \times 10^{-6}\right) \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}{ }^{\circ} \mathrm{R}} \mathrm{dT}\\&\Delta \widehat{\mathrm{H}}=17.5 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+75 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+85 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+125 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}+43.7 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}=246.2 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\\& \Delta \mathrm{H}=\Delta \widehat{\mathrm{H}} \times \mathrm{m}\\& \Delta \mathrm{H}=246.2 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}} \times\left(1 \,\mathrm{lbmol} \times \frac{50 \mathrm{lb}_{\mathrm{m}}}{\mathrm{lbmol}}\right)= \mathbf{12,3 0 0} \,\bf BTU\end{aligned}

B) Give your best estimate of the change in internal energy

\Delta \widehat{\mathrm{U}}=\Delta \widehat{\mathrm{H}}-\mathrm{P}\left(\widehat{\mathrm{V}}_{2, \text { vapour }}-\widehat{\mathrm{V}}_{1, \text { solid }}\right)

From Problem 2-22, \widehat{V}_{\text {solid/liquid }}=\frac{1}{\rho}

From Problem 2-23,\widehat{\mathrm{V}}_{g a s}=\frac{\underline{\mathrm{V}}}{\mathrm{mw}}=\left(\frac{\mathrm{RT}}{\mathrm{P\,mw}}\right)

\begin{aligned}& \Delta \widehat{\mathrm{U}}=\Delta \widehat{\mathrm{H}}-\mathrm{P}\left(\frac{\mathrm{RT}}{\mathrm{Pmw}}-\frac{1}{\rho}\right) \\& \Delta \widehat{\mathrm{U}}=\left(246.2 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right) \\& \qquad -(1 \mathrm{~atm})\left(\frac{\left(\frac{1 \mathrm{~kmol}}{2.20 \mathrm{~lbmol}}\right)\left(0.08206 \frac{\mathrm{m}^{3} \mathrm{~atm}}{\mathrm{~K} \mathrm{~kmol}}\right)(366 \mathrm{~K})}{(1 \mathrm{~atm})\left(50 \frac{\mathrm{lb}_{\mathrm{m}}}{\mathrm{lbmol}}\right)}\right. \\& \qquad \left.-\frac{1}{\left(40 \frac{\mathrm{lb}}{\mathrm{ft}^{3}}\right)\left(\frac{3.28 \mathrm{ft}}{1 \mathrm{~m}}\right)^{3}}\right) \\& \Delta \widehat{\mathrm{U}}=\left(246.2 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\right) \\& \qquad-\left(\frac{0.273 ~m^{3} a t m}{\rm l b_{m}}\right. \\& \qquad\left.-\frac{0.00071 \mathrm{~m}^{3} \mathrm{~atm}}{\mathrm{lb}_{\mathrm{m}}}\right)\left(\frac{\frac{8.314 \mathrm{~kJ}}{\mathrm{~K} \mathrm{~kmol}}}{\frac{0.08206 \mathrm{~m}^{3} \mathrm{~atm}}{\mathrm{kmol} \mathrm{~K}}}\right)\left(\frac{1000 \mathrm{~J}}{\mathrm{~kJ}}\right)\left(\frac{9.48 \times 10^{-4} \mathrm{BTU}}{\mathrm{J}}\right) \\& \Delta \widehat{\mathrm{U}}=246.2-26.2 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}}\\& \Delta \mathrm{U}=\Delta \widehat{\mathrm{U}} \times \mathrm{m} \\& \Delta \mathrm{U}=220.0 \frac{\mathrm{BTU}}{\mathrm{lb}_{\mathrm{m}}} \times\left(1 \,\mathrm{lbmol} \times \frac{50~ \mathrm{lb}_{\mathrm{m}}}{\mathrm{lbmol}}\right)=\mathbf{11, 000} \,\bf { BTU }\end{aligned}