Question

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

Answer 1

Mass of gas $A$,$W_A$ $=1g$

Mass of gas $B$, $W_B$ $=2g$

Pressure exerted by the gas $A=2$bar

Total pressure due to both the gases $=3$ bar

In this case, the temperature and volume becomes constant

Now, if $M_A$ and $M_B$ are molar masses of gases $A$ and $B$, then

$P_{A}V=\frac{W_{A}RT}{M_A}$ and $P_{total}V=(\frac{W_A}{M_A}+\frac{W_B}{M_B})RT$

$=2V=1\times \frac{RT}{M_A}$ and $3V=(\frac{1^{}}{M_A}+\frac{2}{M_B})RT$

From these equations, we get

$\frac{3}{2}=\frac{(\frac{1^{}}{M_A}+\frac{2}{M_B})}{\frac{1}{M_A}}$

This result in $\frac{2M_A}{M_B}=\left(\frac{3}{2}\right)-1=\frac{1}{2}$

Or $M_B=4M_A$

Thus, the relationship between the molecular masses of A &B is given by

$4M_A=M_B$