States of Matter - Result Question 70
70. The compression factor (compressibility factor) for one mole of a van der Waals’ gas at $0^{\circ} C$ and $100 \hspace {1mm} atm$ 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. (1.25)
Solution:
In case of negligible molecular volume, $b=0$. For 1 mole of gas
$ \left(p+\dfrac{a}{V^{2}}\right) V=R T $
$\Rightarrow \quad p V+\dfrac{a}{V}=R T $
$\Rightarrow \quad \dfrac{p V}{R T}+\dfrac{a}{V R T}=1 \quad \quad\left[\because \dfrac{p V}{R T}=Z\right] $
$\Rightarrow \quad \quad Z+\dfrac{a}{\left(\dfrac{Z R T}{p}\right) R T}=1 \Rightarrow Z+\dfrac{a p}{Z R^{2} T^{2}}=1 $
$\Rightarrow \quad a=\dfrac{Z R^{2} T^{2}(1-Z)}{p}=\dfrac{0.5(0.082 \times 273)^{2}(1-0.5)}{100} $
$ a=1.25 \hspace {1mm} atm \hspace {1mm} L^{2} \hspace {1mm} mol^{-2}$
BETA
