When $$2$$ $$g$$ of a gas $$A$$ is introduced into an evaluated flask kept at $${25^ \circ }C,$$ the pressure is found to be one atmosphere. If $$3$$ $$gm$$ of another gas $$B$$ is then added to the same flask, the total pressure becomes $$1.5$$ $$atm.$$ Assuming ideal gas behaviour, calculate the ratio of the molecular weights $${M_A}:{M_B}.$$
A.
1 : 3
B.
1 : 1
C.
2 : 1
D.
3 : 1
Answer :
1 : 3
Solution :
From ideal gas equation, $$PV = nRT$$
$$PV = \left( {\frac{m}{M}} \right)RT\,\,{\text{or}}\,\,M = m\frac{{RT}}{{PV}}$$
Let the molecular $$wt.$$ of $$A$$ and $$B$$ be $${M_A}$$ and $${M_B}$$ respectively.
$$\eqalign{
& {\text{Then}}\,{M_A} = 2 \times \frac{{RT}}{{1 \times V}};\,\,{M_B} = \frac{{3 \times RT}}{{0.5 \times V}} \cr
& \therefore \,\,\frac{{{M_A}}}{{{M_B}}} = \frac{{2RT}}{V} \times \frac{{0.5V}}{{3RT}} = \frac{{2 \times 0.5}}{3} = \frac{1}{3} \cr} $$
Therefore, the ratio $${M_A}:{M_B} = 1:3$$
Releted MCQ Question on Physical Chemistry >> States of Matter Solid, Liquid and Gas
Releted Question 1
Equal weights of methane and oxygen are mixed in an empty container at $${25^ \circ }C.$$ The fraction of the total pressure exerted by oxygen is