Question

Two vessels separately contain two ideal gases $A$ and $B$ at the same temperature. The pressure of gas $A$ is three times the pressure of gas $B$. Under these conditions, the density of gas $A$ is found to be two times the density of $B$. The ratio of molecular weights of gas $A$ and $B$, i.e. $\frac{M_{A}}{M_{B}}$ is

• A. $\frac{2}{3}$
• B. $\bigcirc \frac{3}{2}$
• C. $\frac{3}{4}$
• D. $\bigcirc \frac{4}{3}$

Answer 1

Answer: (A)

We are given with following condition,

Let $n_{A}$ and $n_{B}$ are the number of moles of gases in vessel $A$ and vessel $B$ respectively,
then by gas equation, we have
$p_{A} V_{A}=n_{A} R T_{A}$ and $p_{B} V_{B}=n_{B} R T_{B}$ but given $T_{A}=T_{B}$
$\Rightarrow \frac{p_{A} V_{A}}{n_{A}}=\frac{p_{B} V_{B}}{n_{B}} $
$\Rightarrow \frac{p_{A} V_{A}}{m_{A} / M_{A}}=\frac{p_{B} V_{B}}{m_{B} / M_{B}}$
[ As, number of moles $\left.=\frac{\text { mass of gas }}{\text { molar mass }}\right]$
$\Rightarrow \frac{p_{A} M_{A}}{\rho_{A}}=\frac{p_{B} M_{B}}{\rho_{B}}\left[\because \frac{m}{V}=\rho=\right.$ density $]$
Subtituting values, we have
$\frac{3 p \cdot M_{A}}{2 p}=\frac{p \cdot M_{B}}{\rho}$
$ \Rightarrow \frac{M_{A}}{M_{B}}=\frac{2}{3}$