Question

If the pressure in a closed vessel is reduced by drawing out some gas, the mean free path of molecules:
(A) is decreased
(B) is increased
(C) remains unchanged
(D) increases or decreases according to the nature of the gas

Answer 1

Hint: When a gas is kept in a closed container, the molecules inside the container collide with each other. These molecules also collide with the surface of the container. Now, the average distance travelled by a molecule between two successive collisions is known as the mean free path of the molecule. We shall use its formula and see its dependence on pressure to answer our question.

Complete answer:
Let the mean free path of molecules kept inside a closed container be denoted by $M$.
Now, the formula for mean free path is given by:
$\Rightarrow M=\dfrac{1}{\sqrt{2}\pi {{n}_{0}}{{d}^{2}}}$
Or, $M\propto \dfrac{1}{{{n}_{0}}}$ [Let this expression be equation number (1)]
Where,
${{n}_{0}}$ is the number of molecules per unit volume.
$d$ is the diameter of the molecules kept inside the container.
Now, from ideal gas equation, we have:
$\Rightarrow PV=nRT$
$\Rightarrow PV=\dfrac{{{n}_{0}}RT}{{{N}_{A}}}$ [Where, ${{N}_{A}}$ is the Avogadro’s Number]
$\Rightarrow PV={{n}_{0}}kT$ [Where, $k$ is the Boltzmann’s constant]
Here, we can clearly see that the Pressure is directly proportional to number of molecules per unit volume, that is:
$\Rightarrow P\propto {{n}_{0}}$ [Let this expression be equation number (2)]
From equation number (1) and (2), we can say that the mean free path of the molecules is inversely proportional to the pressure inside the container. That means:
$\Rightarrow M\propto \dfrac{1}{P}$
Therefore, on decreasing the pressure in a closed vessel by drawing out some gas, we are increasing the average distance travelled by the molecules.
Hence, the mean free path of molecules is increased.

Hence, option (B) is the correct option.

Note:
This condition is true for both ideal as well as non- ideal gases. As, on drawing out gas we are not only reducing the pressure inside the container but we are also decreasing the probability of collision between two molecules which in turn increases the average distance travelled by the particles.