In this problem, we will use the Ideal Gas Constant \left(0.08314 \frac{\mathrm{L} \,\mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right)\left(\frac{1 \mathrm{~m}^{3}}{1000 \mathrm{~L}}\right)=8.314 \times10^{-5} \frac{\mathrm{m}^{3} \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}
A) 25^{\circ} \mathrm{C}=298 \mathrm{~K} \quad 15^{\circ} \mathrm{C}=288 \mathrm{~K}
\mathrm{PV}=\mathrm{NRT}
\frac{\mathrm{T}}{\mathrm{P}}=\frac{\mathrm{V}}{\mathrm{NR}}
With N, R and V all constant:
\begin{aligned}& \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}_{\mathrm{initial}}}{\mathrm{T}_{\text {initial }}}=\frac{(298 \mathrm{~K})(5 \mathrm{bar})}{(288 \mathrm{~K})}=\mathbf{5 . 1 7 \mathbf { b a r }}\end{aligned}
B) Find the molar volume of gas at the initial temperature and pressure, using the VDW equation:
\begin{aligned}& \mathrm{P}=\frac{\mathrm{RT}}{\underline{\mathrm{V}}-\mathrm{b}}-\frac{\mathrm{a}}{\underline{\mathrm{V}}^{2}}\\& 5 \mathrm{bar}=\frac{\left(8.314 \times 10^{-5} \frac{\mathrm{m}^{3} \mathrm{bar}}{\mathrm{mol} \mathrm{K}}\right) 288 \mathrm{~K}}{(\underline{\mathrm{V}})-\left(\frac{100 \mathrm{~cm}^{3}}{\mathrm{~mol}}\right)\left(\frac{\mathrm{m}}{100 \mathrm{~cm}}\right)^{3}}-\frac{\left(8.0 \times 10^{6} \frac{\mathrm{cm}^{6} \mathrm{bar}}{\mathrm{mol}^{2}}\right)\left(\frac{\mathrm{m}}{100 \mathrm{~cm}}\right)^{6}}{(\underline{\mathrm{V}})^{2}}\end{aligned}
Solving this cubic equation for ” V ” yields three answers, only one of which is real. Therefore,
\underline{V}=0.004544 \frac{\mathrm{m}^{3}}{\mathrm{~mol}}
Solving for N using known total volume:
\begin{aligned}& V=N \underline{V}\\& N=\frac{5 \mathrm{~m}^{3}}{0.004544 \frac{\mathrm{m}^{3}}{\mathrm{~mol}}}=1100 . \mathrm{mol}\end{aligned}
Since system is closed and isochoric, N, V and V are all constant. At final conditions:
\begin{aligned}& P=\frac{R T}{\underline{V}-b}-\frac{a}{\underline{V}^{2}}\\& P=\frac{\left(8.314 \times 10^{-5} \frac{\mathrm{m}^{3} \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}\right) 298 \mathrm{~K}}{\left(\frac{5 \mathrm{~m}^{3}}{1100 \mathrm{mols}}\right)-\left(\frac{100 \mathrm{~cm}^{3}}{\mathrm{~mol}}\right)\left(\frac{\mathrm{m}}{100 \mathrm{~cm}}\right)^{3}}-\frac{\left(8.0 \times 10^{6} \frac{\mathrm{cm}^{6} \mathrm{bar}}{\mathrm{mol}^{2}}\right)\left(\frac{\mathrm{m}}{100 \mathrm{~cm}}\right)^{6}}{\left(\frac{5 \mathrm{~m}^{3}}{1100 \mathrm{mols}}\right)^{2}}\\& \qquad =\bf 5.19 \,\mathbf{bar}\end{aligned}
C) No work was done on the gas by the environment. Heat was added, but there was no expansion or contraction, and therefore no \mathrm{W}_{\mathrm{EC}}.