Pressure of `1g` of an ideal gas `A` at `27^(@)C` is found to be 2 bar when `2g` of another ideal gas `B` is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship thieir molecular masses .

Explanation

Given of gas `A: w = 1g, T = 300 K,P =2`bar
for gas `B: w = 2g.T = 300 K: P =3 -2 = 1`bar
`(gas B +gas A)` mixture shows a pressure =3 bar
For `A : p_(A) V =(w_(A))/(m_(A))RT`
`P_(B) V =(w_(B))/(m_(B))RT`
`(P_(A))/(P_(B)) = (w_(A))/(w_(B)) xx (m_(B))/(m_(A))`
`(2)/(1) = (1)/(2) xx (m_(B))/(m_(A))`
`(m_(A))/(m_(B)) = (1)/(4) :. m_(B) = 4m_(A)` .