Two chamber containing $m_1$and $m_2$ grams of a gas at pressures $P_1$and $P_2$ respectively are put in communication with each other, temperature remaining constant. The common pressure reached will be 

According to Boyle's law, $PV = kT$(a constant)

or   $p\frac{m}{\rho }=kT$ or $\rho =\frac{Pm}{kT}$

or $\rho =\frac{P}{K}$                  (where $\frac{kT}{m}=K$ a constant)

So, ${\rho }_1=\frac{P_1}{K}$ and $V_1=\frac{m_1}{{\rho }_1}=\frac{m_1}{P_1/K}=\frac{K{m}_1}{P_1}$

Similarly, $V_2=\frac{K{m}_2}{P_2}$

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,

$\rho =\frac{m_1+m_2}{V_1+V_2}=\frac{m_1+m_2}{K\left({\displaystyle \frac{m_1}{P_1}}+{\displaystyle \frac{m_2}{P_2}}\right)}$

Therefore , $P=K\rho =\frac{m_1+m_2}{{\displaystyle \frac{m_1}{P_1}}+{\displaystyle \frac{m_2}{P_2}}}=\frac{P_1{P}_2\left(m_1+m_2\right)}{\left(m_1{P}_2+m_2{P}_1\right)}$