Question

At \[20\;{\rm{^\circ C}}\] temperature, an argon gas at atmospheric pressure is confined in a vessel with a volume of \[1\;{{\rm{m}}^{\rm{3}}}\], the effective hard-sphere diameter of argon atom is \[3.10 \times {10^{ - 10}}\;{\rm{m}}\]. Its means free path is:
A. \[100\;{\rm{nm}}\]
B. \[90\;{\rm{nm}}\]
C. \[93.6\;{\rm{nm}}\]
D. \[95\;{\rm{nm}}\]

Answer 1

Hint: The above problem can be resolved by using the concept and application to express the mean free path. The mean free path of any particle is the average length of distance required to be covered by the particles when undergoing the collision. The standard mathematical formula for the mean free path is used in this condition, and the variables in the expression have their usual meaning.

Complete step by step solution
Given:
The temperature of argon gas is, \[T = 20\;{\rm{^\circ C}} = \left( {20 + 273} \right)\;{\rm{K}} = 293\;{\rm{K}}\].
The volume of the vessel is, \[V = 1\;{{\rm{m}}^{\rm{3}}}\].
The diameter of the argon atom is, \[d = 3.10 \times {10^{ - 10}}\;{\rm{m}}\].
The expression for the mean free path is,
\[\lambda = \dfrac{{RT}}{{\sqrt 2 \pi {d^2}{N_A}P}}\]
Here, R is the universal gas constant and its value is \[8.314\;{\rm{J/K}}\], \[{N_A}\] is the Avogadro’s number and its value is \[6.022 \times {10^{23}}\] and P is the pressure in standard condition and its value is \[1.0135 \times {10^5}\;{\rm{dynes}}\].
Solve by substituting the values in above expression as,
 \[
\lambda = \dfrac{{RT}}{{\sqrt 2 \pi {d^2}{N_A}P}}\\
\lambda = \dfrac{{8.314\;{\rm{J/K}} \times 298\;{\rm{K}}}}{{\sqrt 2 \pi {{\left( {3.10 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2} \times 6.022 \times {{10}^{23}} \times 1.01325 \times {{10}^5}\;{\rm{dynes}}}}\\
\lambda = 93.6 \times {10^{ - 9}}\;{\rm{m}} \times \dfrac{{1\;{\rm{nm}}}}{{{{10}^{ - 9}}\;{\rm{m}}}} = 93.6\;{\rm{nm}}\\
\lambda = 93.6\;{\rm{nm}}\\
\]
Therefore, the mean free path is of \[93.6\;{\rm{nm}}\]and option (C) is correct.

Note:
To solve the given problem, it is essential to remember the mean free path's mathematical formula. The mean free path concept can be understood by taking the example of the particle moving randomly. When the particle is subjected to some external effort, such that all possible ways through which motion of the particle is possible is known as the particle's degree of freedom. The degrees of freedom have various practical applications including the analysis of particle motion in 3-D.