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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:
A
$\frac{1}{2}$
B
$\frac{2}{3}$
C
$\frac{3}{4}$
D
2
Detailed Solution
$P = \frac{\rho RT}{M} \Rightarrow M = \frac{\rho RT}{P}$
$\frac{M_A}{M_B} = \frac{\rho_A}{\rho_B}\cdot\frac{T_A}{T_B}\cdot\frac{P_B}{P_A} = 1.5\times 1\times\frac{1}{2} = \frac{3}{4}$
$\frac{M_A}{M_B} = \frac{\rho_A}{\rho_B}\cdot\frac{T_A}{T_B}\cdot\frac{P_B}{P_A} = 1.5\times 1\times\frac{1}{2} = \frac{3}{4}$