Two chambers containing m_1 ~g and m_2 ~g of a gas at pressures P_1...

According to Boyle'slaw, $P V=k$ (a constant) or $P \frac{m}{\rho}=k$ or $\rho=\frac{P m}{k}$ $\left(\because V=\frac{m}{\rho}\right)$\rho=\frac{P}{K} \left(\because \frac{k}{m}=K\right.$a constant$)$So,$\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} \therefore \quad$Total volume$=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$$\begin{aligned} & \rho=\frac{m_1+m_2}{V_1+V_2}=\frac{m_1+m_2}{K\left(\frac{m_1}{P_1}+\frac{m_2}{P_2}\right)} \\ \therefore \quad P & =K \rho=\frac{m_1+m_2}{\frac{m_1}{P_1}+\frac{m_2}{P_2}}=\frac{P_1 P_2\left(m_1+m_2\right)}{\left(m_1 P_2+m_2 P_1\right)}\end{aligned}$$