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:
$\lambda ={\displaystyle \frac{1}{\sqrt{2}\pi d^{2}P}}$m
$$CollisionFrequency={\displaystyle \frac{v_{rms}}{\lambda }}$$
$$CollisionTime={\displaystyle \frac{d}{v_{rms}}}$$
$$Timebetweensuccessivecollision={\displaystyle \frac{l}{v_{rms}}}$$

Complete step by step solution:
The mean free path of a molecule is calculated as follows:
$\lambda ={\displaystyle \frac{1}{\sqrt{2}\pi d^{2}P}}$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.
$$r=1A$$
But $$d=2r$$
Hence $$d=2A$$
But we need the diameter in meters.
So $$1A={10}^{-10}m$$
Hence $d=2\times {10}^{-10}m$
We need to convert P and T in its SI units.
$$P=2atm$$
But $$1atm={10}^{5}Pa$$
So, P = $2\times {10}^{5}Pa$
Similarly,
$$T={17}^{o}C$$
But $$1^{o}C=273K$$
So, $$T=273+17K$$
$$T=290K$$
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,
$\lambda ={\displaystyle \frac{1}{\sqrt{2}\pi \times {\left(2\times {10}^{-10}\right)}^{2}2\times {10}^5}}m$
$\lambda =1.11\times {10}^{-7}m$
Collision frequency is given by the formula,
$$CollisionFrequency={\displaystyle \frac{v_{rms}}{\lambda }}$$
First, we need to calculate vrms
$v_{rms}=\sqrt{{\displaystyle \frac{3RT}{M}}}$ (Equation 2)
$v_{rms}=\sqrt{{\displaystyle \frac{3\times 8.314\times 290}{28\times {10}^3}}}$
$v_{rms}=508m{s}^{-1}$
Inserting all the values in Equation 2,
We get,
$$CollisionFrequency={\displaystyle \frac{508}{1.11\times {10}^{-7}}}$$
$$CollisionFrequency=4.58\times {10}^9{\sec }^{-1}$$
Now, collision time is given by the formula,
$$CollisionTime={\displaystyle \frac{d}{v_{rms}}}$$
Inserting the values in the above equation,
We get,
$$CollisionTime={\displaystyle \frac{2\times {10}^{-7}}{508}}$$
$$CollisionTime=3.93\times {10}^{13}$$ seconds
Time taken in successive collision is given by,
$$Timebetweensuccessivecollision={\displaystyle \frac{l}{v_{rms}}}$$
$$Timebetweensuccessivecollision={\displaystyle \frac{1.11\times {10}^{-7}}{508}}$$
$$Timebetweensuccessivecollision=2.18\times {10}^{-10}$$
Finally, we will compare Collision Time and Time taken between successive collision as follows,
${\displaystyle \frac{CollisionFrequency}{Timebetweensuccessivecollision}}$
= ${\displaystyle \frac{2.18\times {10}^{-10}}{3.93\times {10}^{-13}}}$
= $500$ (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).