Question

The relative rate of effusion of ch4 to so2 through the container containing ch4 and so2 in 3:2 mass ratio is

Answer 1

Answer: The relative rate of effusion of $CH_4$ to $SO_2$ is 3 :1

Explanation:

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

Mathematically,

$\text{Rate}\propto \sqrt{\frac{1}{M}}$

The relation between the rate of effusion and volume is given by the equation:

$\text{Rate}=\frac{V}{t}$

The relationship between the rate of effusion and number of moles is given by Avogadro's law. This law states that the volume of the gas is directly proportional to the number of moles of gas.

The equation follows:

$V\propto n$

The above equation becomes:

$\text{Rate}=\frac{n}{t}$

Also, number of moles is directly proportional to the mass.

$n=\frac{m}{M}$

Now, the equation becomes:

$\frac{R_{CH_4}}{R_{SO_2}}=\frac{n_{CH_4}}{n_{SO_2}}\times \sqrt{\frac{M_{SO_2}}{M_{CH_4}}}$

$\frac{R_{CH_4}}{R_{SO_2}}=\frac{3}{2}\times \sqrt{\frac{64}{16}}$

$\frac{R_{CH_4}}{R_{SO_2}}=\frac{3}{1}$

Hence, the relative rate of effusion of $CH_4$ to $SO_2$ is 3 :1