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 between their molecular masses .

Answer

`M_(B)=4M_(A)`

Explanation

For ideal gas A, the ideal gas equation is given by,
`p_(A)=n_(A)RT……(i)`
Where, `p_(A)` and `n_(A)` represent the pressure and number of moles of gas A. For ideal gas B, the ideal gas equation is given by,
`p_(B)V=n_(8)RT.......(ii)`
Where, `p_(B) and n_(B)` represent the pressure and number of moles of gas B.
[V and T are constants for gases A and B]
From equatio (i), we have
`p_(A)V=(m_(A))/(M_(A))RT rArr(p_(A)M_(A))/(n_(A))=(RT)/V.........(ii)`
From equation (ii), we have
`p_(B)V=(m_(B))/(M_(B))RT rArr(p_(B)M_(B))/(n_(B))=(RT)/V.........(iv)`
Where, `M_(A)and M_(B)` are the molecular masses of gases A and B respectively.
Now, from equations (iii) and (iv), we have
`(p_(A)M_(A))/(m_(A))=(p_(B)M_(B))/(m_(B)).........(v)`
Given,
`m_(A)` = 2 bar
`p_(A)` = 2 bar
`m_(B)` = 2 bar
`p_(B)` = (3-2) = 1 bar
(Since total pressure is 3 bar)
Subbstituting these values in equation (v), we have
`(2xxM_(A))/(1)=(1xxM_(B))/(2)`
`rArr 4M_(A) = M_(B).`