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(C) The ideal gas equation is $PV = nRT$,where $n = \frac{w}{M_w}$.Substituting $n$,we get $PV = \frac{w}{M_w} RT$,which rearranges to $P = \frac{w}{V

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$4 : 1$

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(C) The ideal gas equation is $PV = nRT$,where $n = \frac{w}{M_w}$.Substituting $n$,we get $PV = \frac{w}{M_w} RT$,which rearranges to $P = \frac{w}{V} \frac{RT}{M_w} = \frac{dRT}{M_w}$,where $d$ is the density.Therefore,$d = \frac{PM_w}{RT}$.Given $\frac{d_A}{d_B} = 2$ and $\frac{M_{wA}}{M_{wB}} = \frac{1}{2}$.From the formula,$\frac{P_A}{P_B} = \frac{d_A}{d_B} 	imes \frac{M_{wB}}{M_{wA}}$.Substituting the values: $\frac{P_A}{P_B} = 2 	imes \frac{1}{1/2} = 2 	imes 2 = 4$.Thus,the ratio of partial pressures $\frac{P_A}{P_B}$ is $4 : 1$.

Column-$I$	Column-$II$
$A$. Hydrogen gas $(P=200 \ atm, T=273 \ K)$	$p$. Compressibility factor $Z \neq 1$
$B$. Hydrogen gas $(P \approx 0, T=273 \ K)$	$q$. Attractive forces are dominant
$C$. $CO_2$ $(P=1 \ atm, T=273 \ K)$	$r$. $PV=nRT$
$D$. Real gas with very large molar volume	$s$. $P(V_m-b)=RT$

The density of gas $A$ is twice that of $B$. The molecular mass of $A$ is half of $B$. The partial pressure ratio of $A$ and $B$ is

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