Explanation

(b) Vander Waal’s gas equation for $\mu$ mole of real gas
$\left( P + \frac{\mu^{2}a}{V^{2}} \right)(V - \mu b) = \mu RT$
$P = \left( \frac{\mu RT}{V - \mu b} - \frac{\mu^{2}a}{V^{2}} \right)$
Given equation,
$P = \left( \frac{RT}{2V - b} = \frac{a}{4b^{2}} \right)$
On comparing the given equation with this standard equation, we get
$\mu = \frac{1}{2}$
Hence $,\ \mu = \frac{m}{M}$
$\Rightarrow mass\ of\ gas,\ m = \mu M = \frac{1}{2} \times 44 = 22g$