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Q. A modified form of Vander Waals' equation of state for $1$ mole is:
$\left(P + \frac{a}{V^{2}}\right)\left(V - b\right)=RT$
Report Boyle's Temperature $T_{b}$ .

NTA AbhyasNTA Abhyas 2020

Solution:

$\text{P} = \frac{\text{RT}}{\text{V} - b} - \frac{a}{\text{TV}^{2}}$
$\frac{\text{PV}}{\text{RT}} = \frac{\text{V}}{\text{V} - b} - \frac{a}{\text{RT}^{2} \text{V}} = \frac{1}{1 - \frac{b}{\text{V}}} - \frac{a}{\text{RT}^{2} \text{V}}$
$\text{Z} = \left(1 - \frac{b}{\text{V}}\right)^{- 1} - \frac{a}{\left(\text{RT}\right)^{2} \text{V}}$
$= 1 + \frac{b}{\text{V}} + \frac{b^{2}}{\text{V}^{2}} + \text{...} - \frac{a}{\text{RT}^{2} \text{V}}$
$= 1 + \left(b - \frac{a}{\left(\text{RT}\right)^{2}}\right) \frac{1}{\text{V}} + \frac{b^{2}}{\left(\text{V}\right)^{2}} + \text{...}$
$\therefore $ Second virial co efficient $B = b - \frac{\alpha }{\text{RT}^{2}}$
Third virial co efficient $\text{C} = b^{2}$
At Boyles' Temperature second virial co efficient = o
$\therefore $ $b = \frac{a}{\text{RT}_{\text{b}^{2}}}$
$\therefore $ $\text{T}_{\text{b}} = \sqrt{\frac{a}{\text{R} b}}$