Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is

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Solution

Q.
Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is [2015]

1 2

2 1/2

3 2/3

4 3/4

Ans.
(4)

According to an ideal gas equation, the molecular weight of an ideal gas is

   $M = \frac{ρ R T}{P}$    $(as P = \frac{ρ R T}{M})$

where $P, T$ and $ρ$ are the pressure, temperature and density of the gas respectively, and $R$ is the universal gas constant.

$∴$ The molecular weight of A is

   $M_{A} = \frac{ρ_{A} R T_{A}}{P_{A}} and that of B is M_{B} = \frac{ρ_{B} R T_{B}}{P_{B}}$

Hence, their corresponding ratio is

   $\frac{M_{A}}{M_{B}} = (\frac{ρ_{A}}{ρ_{B}}) (\frac{T_{A}}{T_{B}}) (\frac{P_{B}}{P_{A}})$

Here, $\frac{ρ_{A}}{ρ_{B}} = 1.5 = \frac{3}{2}, \frac{T_{A}}{T_{B}} = 1 and \frac{P_{A}}{P_{B}} = 2$

$∴$    $\frac{M_{A}}{M_{B}} = (\frac{3}{2}) (1) (\frac{1}{2}) = \frac{3}{4}$

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