When 2 g of a gas 'M' is introduced into an empty flask at `30^(@)C`, its pressure becomes 1 atm. Now, if 3g of another gas 'N' is introduced in the same flask at the same temperature, the total pressure becomes 1.5 atm. Find the ratio of the molar masses of the two gases.

Explanation

Let the molecular masses of gases, M and N are a and b g `mol^(-1)` respectively and volume of the flask=V L.
in case of gas M: P=1 atm, T=(273+30)K=303K,
number of moles`(n)=(2)/(a)` mol
Therefore, according to the equation PV=nRT
`1atmxxVL=(2)/(a)molxx0.0821L*atm*mol^(-1)*K^(-1)xx303K`
`therefore V=(49.75)/(a)L`
The number of moles of N gas `=(3)/(b)` mol
In case of the mixture of gases M and N: Total number of moles in the mixture of gases M and N`=((2)/(a)+(3)/(b))` mol
Total pressure of the mixture=1.5 atm
By applying equation, PV=nRT in case of the mixture of gases, we obtain
`1.5xx(49.75)/(a)=((2)/(a)+(3)/(b))xx0.0821xx303`
or, `(74.62)/(a)=((2)/(a)+(3)/(b))xx24.87=(49.75)/(a)+(74.62)/(b)`
or, `(24.87)/(a)=(74.62)/(b)" "therefore (a)/(b)=(1)/(3)`