Mean Free Path
The average of these free paths in between the successive collision is called mean free path and is denoted by lambda(\(\lambda \)\). This note provides us an information on Mean Free Path.
Summary
The average of these free paths in between the successive collision is called mean free path and is denoted by lambda(\(\lambda \)\). This note provides us an information on Mean Free Path.
Things to Remember
The average of these free paths in between the successive collision is called mean free path and is denoted by lambda(\(\lambda \)\).
The mean free path is inversely proportional to the density of the gas.
According to this relation mean the mean free path is directly proportional to the absolute temperature of the gas and inversely proportional to the pressure of the gas.
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Mean Free Path
According to kinetic theory, the gas molecule does not exert force except during the collision so, they move in straight line with the different magnitude which forms zig-zag. The average of these free paths in between the successive collision is called mean free path and is denoted by \(\lambda\)(\lambda \). If \(\lambda _1, \lambda_2, \: \text {and}\: \lambda _3\) be the path of molecules then mean free path.
$$ \lambda = \frac {\lambda _1 +\lambda_2+ \lambda _3 +\dots +\lambda_n}{n} $$
Kinetic theory of matter combines in itself the two components, heat is a form of energy and matter is molecular in nature.
The following are the fundamental assumptions of kinetic theory of gasses:
- A gas is formed of a large number of tiny invisible, perfectly elastic particles, called the molecules.
- The molecules are always in the state of continuous motion with varying velocities in all possible direction.
- The time of collisions is negligible as compared with the time taken to traverse the free path.
- The molecules traverse straight paths between any two collisions.
- The collisions between molecules and with the walls are perfectly elastic so that there is no loss of kinetic energy in the collisions.
- The size of the molecules is infinitely small compared to the average distance traversed by a molecule between any two consecutive collisions.
- The volume of the molecules is negligible as compared to the volume of the vessel containing gas.
- The molecules exert no force on each other except when they collide and the whole of the molecular energy is kinetic.
- The intermolecular distance in a gas is much larger than of a solid or liquid and the molecules of gas are free to move in the entire space available to them.
Expression for mean free path
Let us consider a gas processing n molecules per unit volume. Let us consider that only one molecule which is under consideration is in motion while others are at rest. If \(\sigma \) is the diameter of each molecule, then the moving molecule will collide with all those molecules.
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If v is the velocity of the moving molecule, then in one second it will collide with the entire molecules whose centers lie in a cylinder of radius \(\sigma\)) and length \(\mu \), i.e., within a volume (\(\pi \sigma ^2 \mu\)). The number of collisions made by moving molecule in one second is \(mu\), therefore, the mean path is given by
\begin{align*}\lambda &= \frac {\text {distance traversed by molecule}}{\text {number of collision made by that molecule in one second}} \\ &= \frac {\mu t}{\pi\sigma ^2 \mu n} \\ &= \frac {1}{\pi \sigma ^2 n}\dots (1)\end{align*}
The expression has been derived by assuming that, only one molecule is considered as moving while others are at rest. Maxwell assumed the realistic assumption that all the molecules are moving with all possible velocities in all possible direction and found the result by applying distribution law of molecular speeds. The correct mean free path obtained by him
$$ \lambda = \frac {1}{\sqrt {2}(\pi \sigma^2 n)} \dots (2)$$
If m is the mass of the molecule, then \(mn =\rho\), where \(\rho\) is the density of the gas. The expression for the mean free path in terms of density becomes.
$$ \lambda = \frac {m}{\sqrt {2}(\pi \sigma^2 \rho)} \dots (3)$$
Thus the mean free path is inversely proportional to the density of the gas.
According to kinetic theory of gases, the pressure of the gas is given by
$$ p = \frac 13 \rho \vec {v^2} $$
where \(\vec {{v}^2}\) is the mean square speed of the molecules.
\begin{align*} p &= \frac 13 \rho \frac {3kT}{m} [\because \vec {v^2} = \frac {3kT}{m}\\ \text {or,}\: \frac {m}{\rho} &= \frac {kT}{p} \\ \text {Substituting this value in}\: (3), \: \text {we have}\\\lambda &= \frac {kT}{\sqrt {2}(\pi \sigma^2 \rho)} \\ \lambda &\propto T \\ \lambda &\propto \frac 1p \end{align*}
This is the expression for mean free path in terms of absolute temperature and pressure of the gas. According to this relation mean the mean free path is directly proportional to the absolute temperature of the gas and inversely proportional to the pressure of the gas.
Bibliography
S.S. Singhal, J.P. Agarwal, Satya Prakash. heat and thermodynamics and statistical physics. Pragati Prakashan, 2010.
—. Heat and Thermodynamics and Statistical Physics. Pragati Prakashan, 2010.
Vatsyayan, Dr. Rakesh Ranjan. Refresher Course in Physics. Kathmandu: Surya Book Traders, 2015.
Lesson
Transport Phenomenon
Subject
Physics
Grade
Bachelor of Science
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