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

Ideal gas A :
Mass `(m_(1))=1 g`
Pressure `p_(A)=2` bar
Molecular mass `= M_(A)`
Ideal gas B :
Mass `(m_(2))=2g`
Pressure `p_(B)=1` bar
Molecular mass `= M_(B)`
Temperature T and volume V of both gases are same
Total pressure `p_(A)+p_(B)=3` bar
`therefore p_(B)=3` bar `- p_(A)`
= 3 bar - 2 bar = 1 bar
Mole `(n_(A))=(1g)/(M_(A))` and Mole `(n_(B))=(2g)/(M_(B))`
Total mole `= (n_(A)+n_(B))=(1)/(M_(A))+(2)/(M_(B))`
`pV=nRT`
`pV=(m)/(M)RT`
`therefore p=(m)/(M)((RT)/(V))`
`therefore p_(A)=(m_(A))/(M_(A))((RT)/(V))" "` .....(i)
`therefore p_(B)=(m_(B))/(M_(B))((RT)/(V)) " "` .....(ii)
where, `(RT)/(V)=` constasnt (`because` R constant and T, V are not changed)
Now ratio of (i) and (ii) as under so,
`(p_(A))/(p_(B))=(m_(A))/(M_(A))xx(M_(B))/(m_(B))`
`therefore ("2 bar")/("1 bar")=(1g xx M_(B))/(M_(A)xx 2g)`
`therefore (M_(B))/(M_(A))=(2xx2)/(1xx1)=4 " so " M_(B)=4M_(A)`