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

Explanation

Here we are given that,
`P_(A) `= 2 bar `W_(B) = 2g`
`T = 27^(@)C = 300 K P_(B) = ?`
`w_(A) = 1g " "(M_(A))/(M_(B))= ?`
From Dalton’s law of partial pressure we now that,
`P_("Total") = P_(A) + P_(B)` ..... (i)
Now, when gas B is introduced in the flas , pressure becomes 3 bar means that, `P_(T) = 3 "bar"` , also `P_(A) = 2 "bar"` (given), therefore from equation (i) we get,
`P_(B) = P_("Total")- P_(A) = 3 - 2 = 1 "bar"`
Since both the gases are present in same flask at same temperature. Their V and T are same and we can write,  `P_(A) = n_(A)(RT)/(V)` and `P_(B) = n_(B) (RT)/(V)`
`P_(A) = (w_(A))/(M_(A)) (RT)/(V)` and `P_(B) = (w_(B))/(M_(B)) (RT)/(V)`
`2 =(1)/(M_(A)) xx (RT)/(V)` ..... (ii) 
`1 =(2)/(M_(B)) xx (RT)/(V)` ……(iii) Dividing equation (iii) by (ii), we get, 
`(M_(A))/(M_(B))= (1)/(4)` or `M_(B)=4M_(A)`