
The relative rate of diffusion of a gas (Molecular weight\[=\text{ }128\]) as compared to oxygen is:
A.$2\times \sqrt{2}~~times$
B.$2~~times$
C.$\dfrac{1}{2}$
D.$\dfrac{1}{2\times \sqrt{2}}$
Answer
461.7k+ views
Hint: We know that Diffusion is defined as the movement of one type of gas molecules to the empty space around it. Graham’s law 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 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 and effusion 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 is defined as the average distance travelled by the moving molecule between collisions. Molecular mass of the gas is defined as the mass of the gas i.e. the mass of the every atom present in the gas.
The relative rate of diffusion of gas (Molecular weight \[=\text{ }128\]) as compared to oxygen is \[1/2.~\] The rate of diffusion is inversely proportional to the square root of molar mass.
$\dfrac{{{r}_{gas}}}{{{r}_{{{O}_{2}}}}}=\sqrt{\dfrac{{{M}_{{{O}_{2}}}}}{{{M}_{gas}}}}=\sqrt{\dfrac{32}{128}}=\dfrac{1}{2}$
Therefore, the correct answer is option C.
Additional Information:
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
Note:
Remember that 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.
Complete answer:
First of all we will discuss about diffusion and effusion:
Diffusion 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 and effusion 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 is defined as the average distance travelled by the moving molecule between collisions. Molecular mass of the gas is defined as the mass of the gas i.e. the mass of the every atom present in the gas.
The relative rate of diffusion of gas (Molecular weight \[=\text{ }128\]) as compared to oxygen is \[1/2.~\] The rate of diffusion is inversely proportional to the square root of molar mass.
$\dfrac{{{r}_{gas}}}{{{r}_{{{O}_{2}}}}}=\sqrt{\dfrac{{{M}_{{{O}_{2}}}}}{{{M}_{gas}}}}=\sqrt{\dfrac{32}{128}}=\dfrac{1}{2}$
Therefore, the correct answer is option C.
Additional Information:
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
Note:
Remember that 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.
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