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(C) The ideal gas equation is $PV = nRT$,where $n = \frac{m}{M}$.Substituting $n$,we get $PV = \frac{m}{M} RT$.Since density $d = \frac{m}{V}$,we can

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$\frac{P_A}{P_B} = 4$

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(C) The ideal gas equation is $PV = nRT$,where $n = \frac{m}{M}$.Substituting $n$,we get $PV = \frac{m}{M} RT$.Since density $d = \frac{m}{V}$,we can write $P = \frac{dRT}{M}$.For gas $A$: $P_A = \frac{d_A RT}{M_A}$.For gas $B$: $P_B = \frac{d_B RT}{M_B}$.The ratio is $\frac{P_A}{P_B} = \frac{d_A}{d_B} 	imes \frac{M_B}{M_A}$.Given $d_A = 3.0 \, g \, dm^{-3}$,$d_B = 1.5 \, g \, dm^{-3}$,and $M_A = \frac{1}{2} M_B$ (or $\frac{M_B}{M_A} = 2$).Substituting these values: $\frac{P_A}{P_B} = \frac{3.0}{1.5} 	imes 2 = 2 	imes 2 = 4$.

At the same temperature,gas $A$ is taken in a $0.5 \, dm^3$ flask and gas $B$ is taken in a $1 \, dm^3$ flask. If the density of $A$ is $3.0 \, g \, dm^{-3}$ and that of $B$ is $1.5 \, g \, dm^{-3}$,and the molar mass of $A$ is $1/2$ of $B$,what is the ratio of the pressures exerted by the gases?

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