Question

When $4g$ of an ideal gas $A$ is introduced into an evacuated flask kept at ${25}^{\circ }C$, the pressure is found to be one atmosphere. If $6g$ of another ideal gas $B$ is then added to the same flask, the pressure becomes $2$ atm at the same temperature. The ratio of molecular weight $(M_A:M_B)$ of the two gases would be :

• A. $1:2$
• B. $2:1$
• C. $2:3$
• D. $3:2$
• E. $1:4$

Answer 1

The correct option is C $2:3$

Given: $w_A=4g$,$w_B=6g$,$P_A=1atm$,$P_B=2atm$,

Solution:

No. of moles of A, $n_A=\frac{w_A}{M_A}=\frac{4}{M_A}$

No. of moles of $B=\frac{w_B}{M_B}=\frac{6}{M_B}$

Total no. of moles after adding B, $n_B=\frac{4}{M_A}+\frac{6}{M_B}$

From ideal gas equation, we have

$\frac{P_A{V}_A}{n_{A}R{T}_A}=\frac{P_B{V}_B}{n_{B}R{T}_B}$

Since Volume and Temperature are constant:

$\Rightarrow \frac{P_A}{n_A}=\frac{P_B}{n_B}$

$\Rightarrow \frac{1}{\frac{4}{M_A}}=\frac{2}{\frac{4}{M_A}+\frac{6}{M_B}}$

$\Rightarrow \frac{4}{M_A}+\frac{6}{M_B}=2\times \frac{4}{M_A}$

$\Rightarrow \frac{6}{M_B}=\frac{8}{M_A}-\frac{4}{M_A}$

$\Rightarrow \frac{M_A}{M_B}=\frac{4}{6}=\frac{2}{3}$

So, correct answer is option (C) 2:3