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 .

Answer

4

Explanation

`p_(A)V=(W_(A))/(M_(A))RT`
`"2 bar "xxV=(1)/(M_(A))RT" …(i)"`
After addition of 2g of gas B
`"3 bar"xxV=((1g)/(M_(A))+(2g)/(M_(B)))RT" …(ii)"`
From (ii) and (i)
`(3)/(2)=(((1)/(M_(A))+(2)/(M_(B))))/(((1)/(M_(A))))`
`=((M_(B)+2M_(A))/(M_(A).M_(B)))/((1)/(M_(A)))=(M_(B)+2M_(A))/(M_(B))=1+(2M_(A))/(M_(B))`
`or (2M_(A))/(M_(B))=(3)/(2)-1=(1)/(2)`
`or (M_(A))/(M_(B))=(1)/(4)`
`or M_(B)=4M_(A)`