Question

500 mL of a monoatomic gas X diffuses through a hole at a pressure of 4 atm and 127.1500 mL of another elemental diatomic gas Y diffuses through the same hole at pressure of 2 atm and 27. If the time taken for both the diffusion process is same, then compare the atomic weights of the two gases.

Answer 1

Graham's law of diffusion: It states that the rate of diffusion if different gases are inversely proportional to the square root of their densities at constant temperature and pressure.

Step-1: Calculate the ratio of rate of diffusion with respect to volume

Mathematically, $r=\sqrt{\frac{1}{d}}$

r=rate of diffusion of the gas

d= density of the gas

The rate of diffusion for two gases is given as:

$r_X=\frac{V_X}{t_X};r_Y=\frac{V_Y}{t_Y}$

$\frac{r_X}{r_Y}=\frac{{\displaystyle \frac{V_X}{t_X}}}{{\displaystyle \frac{V_Y}{t_Y}}}$

where r_X = rate of diffusion of gas X

r_Y= rate of diffusion of gas Y

V_X= Volume of gas X

V_Y=Volume of gas Y

t_X =Time taken for diffusion by gas X

t_Y =Time taken for diffusion by gas Y

Given, $t_X=t_Y$

$\frac{r_X}{r_Y}=\frac{V_X}{V_Y}$

$\frac{r_X}{r_Y}=\frac{500}{1500}=\frac{1}{3}$

Step-1: Calculate the ratio of rate of diffusion with respect to molecular mass

$\frac{r_x}{r_y}=\sqrt{\frac{d_y}{d_x}}=\sqrt{\frac{M_y}{M_x}}$

Given,

$\sqrt{\frac{M_Y}{M_X}}=\frac{1}{3}$

On squaring both sides, $\frac{M_y}{M_x}=\frac{1}{9}$

Step-3: Calculate the ratio of atomic masses

As gas X is monoatomic, therefore Molecular mass is the atomic mass.

$Atomicmass=\frac{Molecularmass}{2}=\frac{1}{2}=0.5$

Substituting,

$\frac{A_Y}{A_X}=\frac{0.5}{9}=\frac{1}{18}$

Therefore, the ratio of atomic masses of gas Y: Gas X=1:18