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. $◯\frac{3}{2}$
• C. $\frac{3}{4}$
• D. $◯\frac{4}{3}$

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 $=\frac{\text{ mass of gas }}{\text{ molar mass }}]$
$\Rightarrow \frac{p_A{M}_A}{{\rho }_A}=\frac{p_B{M}_B}{{\rho }_B}[\because \frac{m}{V}=\rho =$ density $]$
Subtituting values, we have
$\frac{3p\cdot M_A}{2p}=\frac{p\cdot M_B}{\rho }$
$\Rightarrow \frac{M_A}{M_B}=\frac{2}{3}$