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.

Pressure of 1 g of an ideal gas A at 27^(@)C is found to be 2 bar. Whe

1.	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.
Answer» Solution :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)`