Answer
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Hint: From the kinetic theory of gasses it is assumed that gas contains several molecules that are in random motion and these molecules collide with each other. The average distance between two successive collisions is called the mean free path.
Formula used:
Mean free path$ = \dfrac{{kT}}{{\sqrt 2 \pi {d_m}^2\dfrac{N}{V}}}$
\[\begin{array}{*{20}{l}}
{k - Boltzmann{\text{ }}constant} \\
{T - Temperature} \\
{d - Diameter} \\
{N/V-no{\text{ }}of{\text{ }}molecules{\text{ }}per{\text{ }}unit{\text{ }}volume{\text{ }}\left( {density} \right)}
\end{array}\]
Complete step by step answer:
Mean free path$ = \dfrac{{kT}}{{\sqrt 2 \pi {d_m}^2\dfrac{N}{V}}}$
From the equation it is observed that the mean free path is directly proportional to the temperature but inversely proportional to the diameter and density of the molecule. From this relation it is understood that the mean free path depends on the diameter, size of the molecule and density of the molecule.
(A) With the increase in size of the molecule, the diameter of the molecule increases, therefore the space between the molecules decreases. This results in increased collisions and thereby decreases the mean free path.
(B) When density increases, the molecules are closer to each other. Thereby, increasing the number of collisions and resulting in decreasing mean free path
(C) With the increase in diameter of the molecule, the space between the molecules decreases. This results in increased collisions and thereby decreases the mean free path.
(D) Mean free path is affected by size, density and diameter of the molecule.
So, the correct answer is “Option D”.
Note: Mean free path is the average distance travelled by a particle between two successive collisions in a system. This collision causes a change in the direction of motion, energy and physical property of the particle. Mean free path is influenced by the density, radius of the molecule and also pressure and temperature. As the pressure increases the mean free path decreases.
Formula used:
Mean free path$ = \dfrac{{kT}}{{\sqrt 2 \pi {d_m}^2\dfrac{N}{V}}}$
\[\begin{array}{*{20}{l}}
{k - Boltzmann{\text{ }}constant} \\
{T - Temperature} \\
{d - Diameter} \\
{N/V-no{\text{ }}of{\text{ }}molecules{\text{ }}per{\text{ }}unit{\text{ }}volume{\text{ }}\left( {density} \right)}
\end{array}\]
Complete step by step answer:
Mean free path$ = \dfrac{{kT}}{{\sqrt 2 \pi {d_m}^2\dfrac{N}{V}}}$
From the equation it is observed that the mean free path is directly proportional to the temperature but inversely proportional to the diameter and density of the molecule. From this relation it is understood that the mean free path depends on the diameter, size of the molecule and density of the molecule.
(A) With the increase in size of the molecule, the diameter of the molecule increases, therefore the space between the molecules decreases. This results in increased collisions and thereby decreases the mean free path.
(B) When density increases, the molecules are closer to each other. Thereby, increasing the number of collisions and resulting in decreasing mean free path
(C) With the increase in diameter of the molecule, the space between the molecules decreases. This results in increased collisions and thereby decreases the mean free path.
(D) Mean free path is affected by size, density and diameter of the molecule.
So, the correct answer is “Option D”.
Note: Mean free path is the average distance travelled by a particle between two successive collisions in a system. This collision causes a change in the direction of motion, energy and physical property of the particle. Mean free path is influenced by the density, radius of the molecule and also pressure and temperature. As the pressure increases the mean free path decreases.
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