Question
The compression factor (compressibility factor) for one mole of a van der Waals gas at 0°C and 100 atm pressure is found to be 0.5. Assuming that the volume of a gas molecule is negligible, calculate the van der Waals' constant a.
✓
Answer
$\text{Z}=\frac{\text{pV}}{\text{nRT}},$ i.e.
$0.5=\frac{100\times\text{V}}{1\times0.082\times273}$
$\text{V}=0.1119\text{L}$
$\Big(\text{p}+\frac{\text{a}}{\text{V}^2}\Big)(\text{V}-\text{b})=\text{RT for }1\text{mol}$
Neglecting b, $\Big(\text{p}+\frac{\text{a}}{\text{V}^2}\Big)\text{V}=\text{RT}$
$\text{pV}+\frac{\text{a}}{\text{V}}=\text{RT}$
$\frac{\text{pV}}{\text{RT}}+\frac{\text{a}}{\text{VRT}}=1$
$\text{a}=\Big(1-\frac{\text{pV}}{\text{RT}}\Big)\text{VRT}$
$=(1-0.5)0.1119\times0.082\times273$
$=1.252\text{atm L}^2\text{mol}^{-2}$
$0.5=\frac{100\times\text{V}}{1\times0.082\times273}$
$\text{V}=0.1119\text{L}$
$\Big(\text{p}+\frac{\text{a}}{\text{V}^2}\Big)(\text{V}-\text{b})=\text{RT for }1\text{mol}$
Neglecting b, $\Big(\text{p}+\frac{\text{a}}{\text{V}^2}\Big)\text{V}=\text{RT}$
$\text{pV}+\frac{\text{a}}{\text{V}}=\text{RT}$
$\frac{\text{pV}}{\text{RT}}+\frac{\text{a}}{\text{VRT}}=1$
$\text{a}=\Big(1-\frac{\text{pV}}{\text{RT}}\Big)\text{VRT}$
$=(1-0.5)0.1119\times0.082\times273$
$=1.252\text{atm L}^2\text{mol}^{-2}$
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