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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 the relationship between molecular masses. |
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Answer» Solution :Suppose MOLECULAR masses of A and B are `M_(A) and M_(B)` RESPECTIVELY. Then their number of moles will be `n_(A) =(1)/(M_(A)), n_(B)=(2)/(M_(B))` `P_(A) =2" bar ," P_(A) +P_(B) =3` bar i.e, `P_(B)=1` bar Applying the relation PV =nRT `THEREFORE (P_(A))/(P_(B))=(n_(A))/(n_(B))=(1//M_(A))/(2//M_(B))=(M_(B))/(2M_(A))` or `(M_(B))/(M_(A))=2xx(P_(A))/(P_(B))=2xx(2)/(1)=4` or `M_(B) =4 M_(A)` |
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