Answer
$0.303\ \mathrm{mol}$
Work Step by Step
Use $PV=nRT \quad (17- 5) \quad$
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$\displaystyle \frac{\mathrm{P}_{\mathrm{i}}\mathrm{V}_{\mathrm{i}}}{\mathrm{n}_{\mathrm{i}}\mathrm{T}_{\mathrm{i}}}=\mathrm{R}\qquad $solve for $\mathrm{n}_{\mathrm{i}}.$
$\displaystyle \mathrm{n}_{\mathrm{i}}=\frac{\mathrm{P}_{\mathrm{i}}\mathrm{V}_{\mathrm{i}}}{\mathrm{R}\mathrm{T}_{\mathrm{i}}}=\frac{(212\times1 0^{3}\ Pa)(0.0185\ \mathrm{m}^{3})}{[8.31J/ (\mathrm{mol}\cdot K)](294\ K)}=1.605\ \mathrm{mol}$
$\displaystyle \mathrm{n}_{\mathrm{f}}=\frac{\mathrm{P}_{\mathrm{f}}\mathrm{V}_{\mathrm{f}}}{\mathrm{R}\mathrm{T}_{\mathrm{f}}} = =\displaystyle \frac{(252\times1 0^{3}\ Pa)(0.0185\ \mathrm{m}^{3})}{[8.31J/(\ \mathrm{mol}\cdot K)](294\ K)}=1.908\ \mathrm{mol}$
$\mathrm{n}_{\mathrm{f}}-\mathrm{n}_{\mathrm{i}}$ = (amount of moles to be pumped in)
$=1.908\ \mathrm{mol} - 1.605\ \mathrm{mol} =0.303\ \mathrm{mol}$
