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Question: How does the rate of effusion of sulfur dioxide, \(S{{O}_{2}}\), compare to that of helium?...

How does the rate of effusion of sulfur dioxide, SO2S{{O}_{2}}, compare to that of helium?

Explanation

Solution

First, we need to understand what is effusion. When gas escapes from a container through a small hole (a hole that has a diameter smaller than the average distance traveled by a gas molecule between successive collisions or the mean free path of the molecule), the movement of gas is known as effusion.

Complete answer:
The rate at which a gas molecule effuses is known as the rate of effusion of that gas. A Scottish scientist Thomas Graham experimentally determined that the rate of effusion of a gas is indirectly proportional to the square root of the total mass of the gas particles.
Rate of effusion1Mass of gas particlesRate\text{ of effusion}\propto \dfrac{1}{\sqrt{Mass\text{ of gas particles}}}
This determination was used to give Graham's law of effusion, which stated that at the same temperature, the rate of effusion of two gases can be related by
Rate of effusion of gas1Rate of effusion of gas2=M2M1\dfrac{Rate\text{ of effusion of ga}{{\text{s}}_{1}}}{Rate\text{ of effusion of ga}{{\text{s}}_{2}}}=\dfrac{\sqrt{{{M}_{2}}}}{\sqrt{{{M}_{1}}}}
Where molar masses of the gases are given by M1 and M2.{{M}_{1\text{ }}}and\text{ }{{\text{M}}_{2}}.
Now to compare the rate of effusion of sulfur dioxide gas and helium,
Rate of effusion of SO2Rate of effusion of He=MHeMSO2\dfrac{Rate\text{ of effusion of S}{{\text{O}}_{2}}}{Rate\text{ of effusion of He}}=\dfrac{\sqrt{{{M}_{He}}}}{\sqrt{{{M}_{S{{O}_{2}}}}}}
The molar mass of sulfur dioxide gas is 64g/mol and the molar mass of helium gas is 4g/mol.
By using these values, we can compare the rate of effusion of these gases.

& \dfrac{Rate\text{ of effusion of S}{{\text{O}}_{2}}}{Rate\text{ of effusion of He}}=\dfrac{\sqrt{4}}{\sqrt{64}} \\\ & \dfrac{Rate\text{ of effusion of S}{{\text{O}}_{2}}}{Rate\text{ of effusion of He}}=\dfrac{2}{8} \\\ & \dfrac{Rate\text{ of effusion of S}{{\text{O}}_{2}}}{Rate\text{ of effusion of He}}=\dfrac{1}{4} \\\ \end{aligned}$$ So, we can see that the rate of effusion of He is four more than the rate of effusion of $S{{O}_{2}}$. **Note:** From Graham's law of effusion, we can infer that when two glasses are present in the same conditions of temperature and pressure, the gases which have lower molecular weight will effuse faster than the molecules that have higher molecular weight. An application of this law can be seen when the balloon which is filled with helium gas deflates faster than the balloon filled with air which contains mostly nitrogen, oxygen, and carbon dioxide.