Explanation

(a) According to Boyle’s law, $pV = k\ $(a constant)
Or $p\frac{m}{p} = k\ \ \ \ \ \ \ \ \ or\ \ \ \ \ p = \frac{pm}{k}$
Or $p\ = \frac{p}{k}\ (where,\frac{k}{m} = k$ a constant)
So, $\rho_{1} = \frac{p_{1}}{k}\ \ and\ V_{1}\frac{p_{1}}{k} = \frac{m_{1}}{p_{1}} = \frac{m_{1}}{p_{1}/k} = \frac{km_{1}}{\rho_{1}}$
Similarly , $V_{2} = \frac{km_{2}}{p_{2}}\ $
Total volume = $V_{1} + V_{2} = k\left( \frac{m_{1}}{p_{1}} + \frac{m_{2}}{p_{2}} \right)$
Let $p$ be the common pressure and $\rho$ be the common density of mixture. Then
$\rho = \ \frac{m_{1} + m_{2}}{V_{1} + V_{2}} = \ \frac{m_{1} + m_{2}}{k\left( \frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}} \right)}$
$\therefore\ p = k\rho = \ \frac{m_{1} + m_{2}}{\frac{m_{1}}{P_{1}} + \frac{m_{2}}{P_{2}}} = \ \frac{P_{1}P_{2}(m_{1} + m_{2})}{(m_{1}P_{2} + m_{2}P_{1})}$