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(C) For an ideal gas,the equation of state is $PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$.Rearranging this,we get $P = \frac{m}{V

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$3: 4$

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(C) For an ideal gas,the equation of state is $PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$.Rearranging this,we get $P = \frac{m}{V} \cdot \frac{RT}{M} = ho \frac{RT}{M}$,where $ho$ is the density and $M$ is the molecular weight.Thus,$M = \frac{ho RT}{P}$.Since the temperature $T$ is the same for both gases,we have:$\frac{M_A}{M_B} = \frac{ho_A}{ho_B} 	imes \frac{P_B}{P_A}$Given that $P_A = 2P_B$ and $ho_A = 1.5ho_B = \frac{3}{2}ho_B$,we substitute these values:$\frac{M_A}{M_B} = \frac{1.5ho_B}{ho_B} 	imes \frac{P_B}{2P_B} = 1.5 	imes \frac{1}{2} = \frac{3}{2} 	imes \frac{1}{2} = \frac{3}{4}$Therefore,the ratio of the molecular weights of $A$ and $B$ is $3: 4$.

Two vessels separately contain two ideal gases $A$ and $B$ at the same temperature. The pressure of $A$ is twice that of $B$. Under these conditions,the density of $A$ is found to be $1.5$ times the density of $B$. The ratio of the molecular weights of $A$ and $B$ is:

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