Question 2.21: Biosphere II is an experimental structure that was designed ...

А) 70^{\circ} \mathrm{F}=(70-32) \frac{5}{9}=21.11+273.15=294.26 \mathrm{~K}

0^{\circ} \mathrm{F}=(0-32) \frac{5}{9}=-17.78+273.15=255.37 \mathrm{~K}

PV=NRT

Consider N is constant (closed system), V is constant (given), and R is constant (always true). So:

\frac{\mathrm{T}_{\text {final }}}{\mathrm{P}_{\text {final }}}=\frac{\mathrm{T}_{\text {initial }}}{\mathrm{P}_{\text {initial }}}

\mathrm{P}_{\text {final }}=\frac{\mathrm{T}_{\text {final }} \mathrm{P}_{\text {initial }}}{\mathrm{T}_{\text {initial }}}=\frac{(255.37 \mathrm{~K})(1 \mathrm{~atm})}{(294.26 \mathrm{~K})}=\mathbf{0 . 8 6 7 8} \mathbf{~ a t m}

B) 105^{\circ} \mathrm{F}=(105-32) \frac{5}{9}=40.56+273.15=313.71 \mathrm{~K}

 P_{\text {final }}=\frac{T_{\text {final }} P_{\text {initial }}}{T_{\text {initial }}}=\frac{(313.71 \mathrm{~K})(1 \mathrm{~atm})}{(294.26 \mathrm{~K})}=\mathbf{1.0661} \mathbf{~ a t m}

C) Volume of Air in Lungs = Max Volume – Minimum Volume

Minimum Volume =125000 m³

Max Volume  =\frac{\mathrm{nRT}_{\max }}{\mathrm{P}}

n is unknown so find it from the known air volume at the minimum T :

\begin{aligned}& \mathrm{n}=\frac{\mathrm{PV}_{\text {minimum }}}{\mathrm{RT}_{\text {minimum }}}=\frac{(1 \mathrm{~atm})\left(125000 \mathrm{~m}^{3}\right)\left(\frac{1000 \mathrm{~L}}{1 \mathrm{~m}^{3}}\right)}{\left(0.08206 \frac{\mathrm{L} \,\mathrm{atm}}{\mathrm{mol}~ \mathrm{K}}\right)(255.37 \mathrm{~K})}=5968481 \mathrm{~mol} \\& \text { Max Volume }=\frac{(5968481 \mathrm{~mol})\left(0.08206 \frac{\mathrm{L} \,\mathrm{atm}}{\mathrm{mol}~ \mathrm{K}}\right)(313.71 \mathrm{~K})}{(1 \mathrm{~atm})\left(\frac{1000 \mathrm{~L}}{1 \mathrm{~m}^{3}}\right)}=153573 \mathrm{~m}^{3}\\& \text{Volume of Air in Lungs} =153573 \mathrm{~m}^{3}-125000 \mathrm{~m}^{3}=\mathbf{2 8 5 7 3 ~\mathbf { m } ^ { 3 }}\end{aligned}

D) Since this is being modeled an ideal gas, Internal Energy is dependent upon only temperature.

\begin{gathered}\mathrm{d} \widehat{U}=\mathrm{C}_{\mathrm{V}} \mathrm{dT}\\\int_{\min }^{\max } d \widehat{U}=\int_{\min }^{\max } C_{V} d T\\\mathrm{C}_{\mathrm{v}}=\mathrm{C}_{\mathrm{p}}-\mathrm{R}\\C V_{\text {air }}=\left(0.79 \mathrm{C}_{\mathrm{pN}_{2}}+0.21 \mathrm{C}_{\mathrm{pO}_{2}}\right)-\mathrm{R}\\\int_{\min }^{\max } \mathrm{d} \widehat{U}=\int_{\min }^{\max }\left(\left(0.79 \mathrm{C}_{\mathrm{PN}_{2}}+0.21 \mathrm{C}_{\mathrm{pO}_{2}}\right)-\mathrm{R}\right) \mathrm{dT}\\\int_{\min }^{\max } \mathrm{d} \widehat{\mathrm{U}}=\left(0.79 \int_{\min }^{\max } \mathrm{C}_{\mathrm{pN}_{2}} \mathrm{dT}\right)+\left(0.21 \int_{\min }^{\max } \mathrm{C}_{\mathrm{pO}_{2}} \mathrm{dT}\right)-\left(\mathrm{R} \int_{\min }^{\max } \mathrm{dT}\right) \\ \mathrm{T}_{\max }=313.71 \mathrm{~K} \\ \mathrm{~T}_{\min }=255.37 \mathrm{~K}\\\frac{C_{P}^{*}}{R}=A+B T+C T^{2}+D T^{3}+E T^{4}\end{gathered}

\rightarrow Nitrogen Integral

\begin{gathered} \left(0.79 \int_{\min }^{\max } \mathrm{C}_{\mathrm{pN}_{2}} \mathrm{dT}\right)=0.79 \mathrm{R} \int_{255.37 \mathrm{~K}}^{313.71 \mathrm{~K}}\left(A+B T+C T^{2}+D T^{3}+E T^{4}\right) \mathrm{dT} \\ =0.79\left(8.314 \frac{\mathrm{kJ}}{\mathrm{kmol} \,\mathrm{K}}\right)\{(3.539)(313.71-255.37) \\ +\frac{1}{2}\left(-2.61 \times 10^{-4}\right)\left(313.71^{2}-255.37^{2}\right) \\ +\frac{1}{3}\left(7.00 \times 10^{-8}\right)\left(313.71^{3}-255.37^{3}\right) \\ +\frac{1}{4}\left(1.57 \times 10^{-9}\right)\left(313.71^{4}-255.37^{4}\right) \\ \left.+\frac{1}{5}\left(-9.9 \times 10^{-13}\right)\left(313.71^{5}-255.37^{5}\right)\right\} \\ \left(0.79 \int_{\min }^{\max } \mathrm{C}_{\mathrm{pN}_{2}} \mathrm{dT}\right)=1341.26 \frac{\mathrm{kJ}}{\mathrm{kmol}} \\\end{gathered}

\begin{gathered}& \left(0.21 \int_{\min }^{\max } \mathrm{C}_{\mathrm{pO}_{2}} \mathrm{dT}\right)=0.21 \mathrm{R} \int_{255.22 \mathrm{~K}}^{313.56 \mathrm{~K}}\left(A+B T+C T^{2}+D T^{3}+E T^{4}\right) \mathrm{dT} \\& =0.21\left(8.314 \frac{\mathrm{kJ}}{\mathrm{kmol} \,\mathrm{K}}\right)\{(3.630)(313.71-255.37) \\& +\frac{1}{2}\left(-1.794 \times 10^{-3}\right)\left(313.71^{2}-255.37^{2}\right) \\& +\frac{1}{3}\left(6.58 \times 10^{-6}\right)\left(313.71^{3}-255.37^{3}\right) \\& +\frac{1}{4}\left(-6.01 \times 10^{-9}\right)\left(313.71^{4}-255.37^{4}\right) \\& \left.+\frac{1}{5}\left(-1.79 \times 10^{-12}\right)\left(313.71^{5}-255.37^{5}\right)\right\} \\& \left(0.21 \int_{\min }^{\max } \mathrm{C}_{\mathrm{pO}_{2}} \mathrm{dT}\right)=359.28 \frac{\mathrm{kJ}}{\mathrm{kmol}} \\& \left(\mathrm{R} \int_{\min }^{\max } \mathrm{dT}\right)=\left(8.314 \frac{\mathrm{kJ}}{\mathrm{kmol}}\right)(313.71-255.37)=485.04 \frac{\mathrm{kJ}}{\mathrm{kmol}} \\& \int_{\min }^{\max } \mathrm{d} \widehat{\mathrm{U}}=\left(0.79 \int_{\min }^{\max } \mathrm{C}_{\mathrm{pN}_{2}} \mathrm{dT}\right)+\left(0.21 \int_{\min }^{\max } \mathrm{C}_{\mathrm{pO}_{2}} \mathrm{dT}\right)-\left(\mathrm{R} \int_{\min }^{\max } \mathrm{dT}\right) \\& \int_{\min }^{\max } \mathrm{d} \widehat{\mathrm{U}}=1341.26 \frac{\mathrm{kJ}}{\,\mathrm{kmol}}+359.28 \frac{\mathrm{kJ}}{\mathrm{kmol}}-485.04 \frac{\mathrm{kJ}}{\mathrm{kmol}~ \mathrm{K}}=\mathbf{1 2 1 5 . 5} \frac{\mathbf{k J}}{\mathbf{k m o l}}\\& \Delta U=(5968481 \mathrm{~mol})\left(\frac{1 \,\mathrm{kmol}}{1000 \mathrm{~mol}}\right)\left(1215.5 \frac{\mathrm{kJ}}{\mathrm{kmol}}\right)=\mathbf{7 . 2 5} \times \mathbf{1 0}^6 \,\mathbf{k J}\end{gathered}

E)

\begin{gathered}& \mathrm{W}_{\mathrm{EC}}=-\int \mathrm{PdV} \\& \mathrm{W}_{\mathrm{EC}}=-\mathrm{P} \int_{0}^{28573 \mathrm{~m}^{3}} \mathrm{dV}=-(1 \mathrm{~atm})\left(28573 \mathrm{~m}^{3}\right)\left(\frac{101325 \frac{\mathrm{N}}{\mathrm{m}^{2}}}{1 \mathrm{~atm}}\right)\left(\frac{1 \mathrm{~J}}{1 \,\mathrm{Nm}}\right) \\& =\bf -2.895 \times 10^{9} ~J\end{gathered}