Two gases occupy two containers $$A$$ and $$B$$ the gas in $$A,$$ of volume $$0.10\,{m^3},$$  exerts a pressure of $$1.40\,MPa$$   and that in $$B$$ of volume $$0.15\,{m^3}$$  exerts a pressure $$0.7\,MPa.$$  The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in $$MPa$$ )

Solution :
We know that
$$\eqalign{ & {P_A}{V_A} = nRT,{P_B}{V_B} = {n_B}RT\,\,{\text{and}} \cr & {P_f}\left( {{V_A} + {V_B}} \right) = \left( {{n_A} + {n_B}} \right)RT \cr} $$
$$\eqalign{ & {P_f}\left( {{V_A} + {V_B}} \right) = {P_A}{V_A} + {P_B}{V_B} \cr & \therefore {P_f} = \left( {\frac{{{P_A}{V_A} + {P_B}{V_B}}}{{{V_A} + {V_B}}}} \right) \cr & = \frac{{1.4 \times 0.1 + 0.7 \times 0.15}}{{0.1 + 0.15}}MPa \cr & = 0.98MPa \cr} $$