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laws

Question

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

Moderate

A

2
B

$\frac{1}{2}$
C

$\frac{2}{3}$
D

$\frac{3}{4}$

Solution

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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