States of Matter 1 Question 70
70. The compression factor (compressibility factor) for one mole of a van der Waals’ gas at $0^{\circ} \mathrm{C}$ and $100 \mathrm{~atm}$ pressure is found to be 0.5 . Assuming that the volume of a gas molecule is negligible, calculate the van der Waals’ constant ’ $a$ ‘.
(2001, 5M)
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Answer:
Correct Answer: 70. $(0.99 \mathrm{~atm})$
Solution:
- In case of negligible molecular volume, $b=0$. For 1 mole of gas
$$ \begin{aligned} & \left(p+\frac{a}{V^{2}}\right) V=R T \ \Rightarrow \quad & p V+\frac{a}{V}=R T \ \Rightarrow \quad & \frac{p V}{R T}+\frac{a}{V R T}=1 \quad \quad\left[\because \frac{p V}{R T}=Z\right] \ \Rightarrow \quad & \quad Z+\frac{a}{\left(\frac{Z R T}{p}\right) R T}=1 \Rightarrow Z+\frac{a p}{Z R^{2} T^{2}}=1 \ \Rightarrow \quad & a=\frac{Z R^{2} T^{2}(1-Z)}{p}=\frac{0.5(0.082 \times 273)^{2}(1-0.5)}{100} \ & a=1.25 \mathrm{~atm} \mathrm{~L}^{2} \mathrm{~mol}^{-2} \end{aligned} $$
