Two vessels separately contain two ideal gase | vedclass

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(A) Given: The temperature $T$ is the same for both gases.Pressure of gas $A$ is $P_A = 3P_B$.Density of gas $A$ is $\rho_A = 2\rho_B$.From the ideal

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$\frac{2}{3}$

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(A) Given: The temperature $T$ is the same for both gases.Pressure of gas $A$ is $P_A = 3P_B$.Density of gas $A$ is $ho_A = 2ho_B$.From the ideal gas equation, $PV = nRT = \frac{m}{M}RT$, where $m$ is mass and $M$ is molar mass.Since density $ho = \frac{m}{V}$, we can write $P = \frac{ho RT}{M}$, which implies $M = \frac{ho RT}{P}$.For gas $A$: $M_A = \frac{ho_A RT}{P_A}$.For gas $B$: $M_B = \frac{ho_B RT}{P_B}$.Taking the ratio: $\frac{M_A}{M_B} = \frac{ho_A}{ho_B} 	imes \frac{P_B}{P_A}$.Substituting the given values: $\frac{M_A}{M_B} = \frac{2ho_B}{ho_B} 	imes \frac{P_B}{3P_B} = 2 	imes \frac{1}{3} = \frac{2}{3}$.

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