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Q13.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm  and temperature 17^{0}C. Take the radius of a nitrogen molecule to be roughly 1.0A . Compare the collision time with the time the
molecule moves freely between two successive collisions (Molecular mass of N_{2}=28.0\mu).

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Pressure, P = 2atm

Temperature, T = 17 oC

The radius of the Nitrogen molecule , r=1\ \AA

The molecular mass of  N= 28 u

The molar mass of N= 28 g

From ideal gas equation

\\PV=nRT\\ \frac{n}{V}=\frac{P}{RT}\\

The above tells us about the number of moles per unit volume, the number of molecules per unit volume would be given as

\\n'=\frac{N_{A}n}{V}=\frac{6.022\times 10^{23}\times 2\times 1.013\times 10^{5}}{8.314\times (17+273)}\\ n'=5.06\times 10^{25}

The mean free path \lambda is given as 

\\\lambda =\frac{1}{\sqrt{2}\pi n'd^{2}}\\ \lambda =\frac{1}{\sqrt{2}\times \pi \times 5.06\times 10^{25}\times (2\times 1\times 10^{-10})^{2}}\\ \lambda =1.11\times 10^{-7}\ m

The root mean square velocity vrms is given as 

\\v_{rms}=\sqrt{\frac{3RT}{M}}\\ v_{rms}=\sqrt{\frac{3\times 8.314\times 290}{28\times 10^{-3}}}\\ v_{rms}=508.26\ m\ s^{-1}

The time between collisions T is given as

\\T=\frac{1}{Collision\ Frequency}\\ T=\frac{1}{\nu }\\ T=\frac{\lambda }{v_{rms}}\\ T=\frac{1.11\times 10^{-7}}{508.26}\\ T=2.18\times 10^{-10}s

Collision time T' is equal average time taken by a molecule to travel a distance equal to its diameter

\\T'=\frac{d}{v_{rms}}\\ T'=\frac{2\times 1\times 10^{-10}}{508.26}\\ T'=3.935\times 10^{-13}s

The ratio of the average time between collisions to the collision time is

\\\frac{T}{T'}=\frac{2.18\times 10^{-10}}{3.935\times 10^{-13}}\\ \frac{T}{T'}=554

Thus we can see time between collisions is much larger than the collision time.

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