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(A) Using the ideal gas equation $PV = nRT$,where $n = \frac{w}{M}$.For gas $A$: $P_A V = \frac{w_A}{M_A} RT \implies 1 	imes V = \frac{2}{M_A} RT \i

Vedclass · Accepted Answer

$1: 3$

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(A) Using the ideal gas equation $PV = nRT$,where $n = \frac{w}{M}$.For gas $A$: $P_A V = \frac{w_A}{M_A} RT \implies 1 	imes V = \frac{2}{M_A} RT \implies V = \frac{2RT}{M_A} \quad (1)$.When gas $B$ is added,the total pressure $P_{total} = P_A + P_B = 1.5 \ atm$. Since $P_A = 1 \ atm$,then $P_B = 0.5 \ atm$.For gas $B$: $P_B V = \frac{w_B}{M_B} RT \implies 0.5 	imes V = \frac{3}{M_B} RT \implies V = \frac{6RT}{M_B} \quad (2)$.Equating $(1)$ and $(2)$: $\frac{2RT}{M_A} = \frac{6RT}{M_B}$.$\frac{2}{M_A} = \frac{6}{M_B} \implies \frac{M_A}{M_B} = \frac{2}{6} = \frac{1}{3}$.Thus,the ratio of molar masses $M_A : M_B$ is $1: 3$.

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