Question
Two vessels separately contain two ideal gases $$A$$ and $$B$$ at the same temperature, the pressure of $$A$$ being twice that of $$B.$$ Under such conditions, the density of $$A$$ is found to be 1.5 times the density of $$B.$$ The ratio of molecular weight of $$A$$ and $$B$$ is
A.
$$\frac{2}{3}$$
B.
$$\frac{3}{4}$$
C.
$$2$$
D.
$$\frac{1}{2}$$
Answer :
$$\frac{3}{4}$$
Solution :
$$\eqalign{ & {\rho _A} = 1.5\,{\rho _{B}}\,\,\,\,{\rho _B} \cr & {\rho _A} = 2{\rho _B}\,\,\,\,\,\,\,{p_B} \cr} $$
According to ideal gas equation, we have
Pressure, $$p = \frac{{\rho RT}}{M},$$ where $$M$$ is molecular weight of ideal gas.
Such that, $$\frac{p}{\rho } = \frac{{RT}}{M}$$
$$ \Rightarrow M = \frac{{\rho RT}}{P}$$
where, $$R$$ and $$T$$ are constants.
So, $$M \propto \frac{\rho }{p} \Rightarrow \frac{{{M_A}}}{{{M_B}}} = \frac{{{\rho _A}}}{{{\rho _B}}} \times \frac{{{p_B}}}{{{p_A}}}$$
$$ = 1.5 \times \frac{1}{2} = 0.75 = \frac{3}{4}$$
$$\eqalign{ & {\rho _A} = 1.5\,{\rho _{B}}\,\,\,\,{\rho _B} \cr & {\rho _A} = 2{\rho _B}\,\,\,\,\,\,\,{p_B} \cr} $$
According to ideal gas equation, we have
Pressure, $$p = \frac{{\rho RT}}{M},$$ where $$M$$ is molecular weight of ideal gas.
Such that, $$\frac{p}{\rho } = \frac{{RT}}{M}$$
$$ \Rightarrow M = \frac{{\rho RT}}{P}$$
where, $$R$$ and $$T$$ are constants.
So, $$M \propto \frac{\rho }{p} \Rightarrow \frac{{{M_A}}}{{{M_B}}} = \frac{{{\rho _A}}}{{{\rho _B}}} \times \frac{{{p_B}}}{{{p_A}}}$$
$$ = 1.5 \times \frac{1}{2} = 0.75 = \frac{3}{4}$$