Question

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17$^o$C. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with time the molecule moves freely between two successive collisions (Molecular mass of N$_2$ = 28.0 u)

Answer 1

Hint: Mean free path is defined as the distance moved by the molecules between two successive collisions.
The mean free path of a molecule is calculated as follows:
$\lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}P}}$m
\[Collision{\text{ }}Frequency{\text{ }} = \dfrac{{{v_{rms}}}}{\lambda }\]
\[Collision{\text{ }}Time{\text{ }} = \dfrac{d}{{{v_{rms}}}}\]
\[Time{\text{ }}between{\text{ }}successive{\text{ }}collision{\text{ }} = \dfrac{l}{{{v_{rms}}}}\]

Complete step by step solution:
The mean free path of a molecule is calculated as follows:
$\lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}P}}$m (Equation 1)
Where
$\lambda $is the mean free path in m
d is the diameter of the molecule in m
P is pressure exerted by the gas
So now it is given in the question that the radius of the molecule is 1.0 A.
\[r{\text{ }} = {\text{ }}1{\text{ }}A\]
But \[d{\text{ }} = {\text{ }}2r\]
Hence \[d{\text{ }} = {\text{ }}2{\text{ }}A\]
But we need the diameter in meters.
So \[1{\text{ }}A{\text{ }} = {\text{ }}{10^{ - 10}}m\]
Hence $d = 2 \times {10^{ - 10}}m$
We need to convert P and T in its SI units.
\[P{\text{ }} = {\text{ }}2{\text{ }}atm\]
But \[1{\text{ }}atm{\text{ }} = {\text{ }}{10^5}Pa\]
So, P = $2 \times {10^5}Pa$
Similarly,
\[T\; = {\text{ }}{17^o}C\]
But \[{1^o}C{\text{ }} = {\text{ }}273{\text{ }}K\]
So, \[T{\text{ }} = {\text{ }}273{\text{ }} + 17{\text{ }}K\]
\[T{\text{ }} = {\text{ }}290{\text{ }}K\]
We know the value of universal gas constant(R) = 8.314 J/K
Finally inserting all the values in equation 1,
We get the mean free path as follows,
$\lambda = \dfrac{1}{{\sqrt 2 \pi \times {{\left( {2 \times {{10}^{ - 10}}} \right)}^2}2 \times {{10}^5}}}m$
$\lambda = 1.11 \times {10^{ - 7}}m$
Collision frequency is given by the formula,
\[Collision{\text{ }}Frequency{\text{ }} = \dfrac{{{v_{rms}}}}{\lambda }\]
First, we need to calculate vrms
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ (Equation 2)
${v_{rms}} = \sqrt {\dfrac{{3 \times 8.314 \times 290}}{{28 \times {{10}^3}}}} $
${v_{rms}} = 508m{s^{ - 1}}$
Inserting all the values in Equation 2,
We get,
\[Collision{\text{ }}Frequency{\text{ }} = \dfrac{{508}}{{1.11 \times {{10}^{ - 7}}}}\]
\[Collision{\text{ }}Frequency{\text{ }} = 4.58 \times {10^9}{\sec ^{ - 1}}\]
Now, collision time is given by the formula,
\[Collision{\text{ }}Time{\text{ }} = \dfrac{d}{{{v_{rms}}}}\]
Inserting the values in the above equation,
We get,
\[Collision{\text{ }}Time{\text{ }} = \dfrac{{2 \times {{10}^{ - 7}}}}{{508}}\]
\[Collision{\text{ }}Time{\text{ }} = 3.93 \times {10^{13}}\] seconds
Time taken in successive collision is given by,
\[Time{\text{ }}between{\text{ }}successive{\text{ }}collision{\text{ }} = \dfrac{l}{{{v_{rms}}}}\]
\[Time{\text{ }}between{\text{ }}successive{\text{ }}collision{\text{ }} = \dfrac{{1.11 \times {{10}^{ - 7}}}}{{508}}\]
\[Time{\text{ }}between{\text{ }}successive{\text{ }}collision{\text{ }} = 2.18 \times {10^{ - 10}}\]
Finally, we will compare Collision Time and Time taken between successive collision as follows,
$\dfrac{{Collision{\text{ }}Frequency}}{{Time{\text{ }}between{\text{ }}successive{\text{ }}collision}}$
= $\dfrac{{2.18 \times {{10}^{ - 10}}}}{{3.93 \times {{10}^{ - 13}}}}$
= $500$ (Approximate)

Note: In such questions, care must be taken that we perform all calculations in SI units. Also, there is a high chance of committing silly mistakes in the calculations (since this question was calculation-intensive).