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(C) According to the ideal gas equation,the molecular weight $M$ of an ideal gas is given by:$M = \frac{\rho R T}{P}$where $P$ is the pressure,$T$ is

Vedclass · Accepted Answer

$0.75$

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(C) According to the ideal gas equation,the molecular weight $M$ of an ideal gas is given by:$M = \frac{ho R T}{P}$where $P$ is the pressure,$T$ is the temperature,$ho$ is the density,and $R$ is the universal gas constant.For gases $A$ and $B$:$M_A = \frac{ho_A R T_A}{P_A}$ and $M_B = \frac{ho_B R T_B}{P_B}$The ratio of their molecular weights is:$\frac{M_A}{M_B} = \left( \frac{ho_A}{ho_B} ight) \left( \frac{T_A}{T_B} ight) \left( \frac{P_B}{P_A} ight)$Given:$\frac{ho_A}{ho_B} = 1.5 = \frac{3}{2}$$T_A = T_B \implies \frac{T_A}{T_B} = 1$$P_A = 2 P_B \implies \frac{P_B}{P_A} = \frac{1}{2}$Substituting these values:$\frac{M_A}{M_B} = \left( \frac{3}{2} ight) 	imes (1) 	imes \left( \frac{1}{2} ight) = \frac{3}{4} = 0.75$

Two vessels separately contain two ideal gases $A$ and $B$ at the same temperature. The pressure of $A$ is 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 the molecular weight of $A$ to that of $B$ is

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