Question

The relative rates of diffusion of $C{O_2}$ and $C{l_2}$ gases is $1:1.267$.
A.True
B.False

Answer 1

Hint: Diffusion: It is defined as the movement of one type of gas molecules to the empty space around it.
Graham’s law of diffusion / effusion: It is defined as the rate of diffusion or effusion of a gas is inversely proportional to the square root of the molar mass of the particles in the gas.

Complete answer:
First of all we will discuss about diffusion and effusion:
Diffusion: It is defined as the movement of one type of gas molecule to the empty space around it. For example: the smell of food, tea and perfume in the atmosphere.
Effusion: It is defined as the process in which gas escapes from a container through hole of diameter which is smaller than the mean free path of the molecules. For example: the escape of gas from a hole in a balloon to the atmosphere.
Mean free path: It is defined as the average distance travelled by the moving molecule between collisions.
Molecular mass of the gas: It is defined as the mass of the gas i.e. the mass of the every atom present in the gas.
Graham’s law of diffusion or effusion: It is defined as the rate of diffusion or effusion of a gas is inversely proportional to the square root of the molar mass of the particles in the gas.
Now in the question we are given with the gases i.e. carbon dioxide and chlorine gas. We know that the molecular mass of chlorine atoms is $35.5$. So the molecular weight of chlorine gas i.e. $C{l_2}$ will be $71$. Also we know that the molecular carbon in carbon dioxide is $12$ and that of oxygen is $16$. So the molecular weight of carbon dioxide gas i.e. $C{O_2}$ will be $12 + 32 = 44$. So the relative rate of diffusion according to Graham’s law will be: $\dfrac{{{r_{C{O_2}}}}}{{{r_{C{l_2}}}}} = \sqrt {\dfrac{{71}}{{44}}} $. The ratio will be $1.267$.
So the statement that the relative rate of diffusion of $C{O_2}$ and $C{l_2}$ gases is $1:1.267$ is false.

Hence option B is correct.

Note:
Mean free path for a particle decreases on increasing the pressure at constant temperature. But if we increase the temperature then molecules will move faster and hence collision time will decrease but the average distance travelled between the moving molecules remains the same so the free path will remain the same.