Question

Pressure of $1\text{g}$ of an ideal gas $A$ at ${27}^{\circ }\text{C}$ is found to be 2 bar. When $2\text{g}$ of another ideal gas $B$ is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Find a relationship between their molecular masses.

• A. $M_A=4M_B$
• B. $M_B=4M_A$
• C. $M_A=2M_B$
• D. $M_B=2M_A$

Answer 1

The correct option is B $M_B=4M_A$

Given,
Mass of gas A, $W_A=1\text{g}$
Mass of gas B, $W_B=2\text{g}$
$P_A=2\text{bar}$
$P_{total}=3\text{bar}$
$P_{total}=P_A+P_B$
$P_B=3-2=1\text{bar}$
$PV=nRT\Rightarrow PV=\frac{W}{M}RT$
$\frac{P_A{M}_A}{W_A}=\frac{P_B{M}_B}{W_B}\text{or}\frac{2\times M_A}{1}=\frac{1\times M_B}{2}\text{or}\frac{M_A}{M_B}=\frac{1}{4}\text{or}{M}_B=4M_A$