Question

Pressure of 1g of a solid of a ideal gas $A$ at $27$ is found to be $2$bar. When 2 g another ideal gas $B$ is introduces in the same flask at same temperature, the pressure becomes 3 bar. Find a relationship between their molecular masses.

Answer 1

Solution:-

Mass of gas $A\left(W_A\right)=1g$

Mass of gas $B\left(W_B\right)=2g$

Pressure exerted by the gas $A=2\text{bar}$

Total pressure due to both the gases $=3\text{bar}$

Here, volume and temperature remains constant.

Let $M_A$ and $M_B$ be the molecular masses of the gases $A$ and $B$ respectively.

Therefore,

$p_{A}V=\left(\frac{W_A}{MA}\right)RT$

$2\times V=1\times \frac{RT}{M_A}.....\left(1\right)$

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

$3\times V=(\frac{1}{M_A}+\frac{2}{M_B})RT.....\left(2\right)$

From equation $\left(1\right)\&\left(2\right)$, we have

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

$\frac{2M_A}{M_B}=\frac{3}{2}-1=\frac{1}{2}$

$M_B=4M_A$

Hence the relation between molecular masses of gas $A$ and $B$ is $4M_A=M_B$.