### Mean Free Path Calculator:

This mean free path calculator will help you calculate the mean free path of a particle of specified diameter in an ideal gas.

An ideal gas consists of a large number of particles, atoms or molecules that are in constant rapid motion, and they can collide with each other. During these purely elastic collisions, the particles change their directions or energies.

The mean free path is the average distance a particle travels in a material medium between successive collisions with particles of that medium. The term “mean free path” is used in many areas of physics.

Use our calculator to estimate the average distance between collisions of particles of each ideal gas.

The ideal gas parameters you need should provide the pressure and temperature and the effective particle diameter for the mean free path equation.

Get help solving the mean free path of a particle in an ideal gas on this page. An ideal gas contains a large number of particles or atoms or molecules,

They are in constant fast motion and may collide with each other. Details about mean free path and its formula can be found here.

### Mean Free Path Calculator:

Mean free path formula is Λ _{(m)} in meter is equal to the Boltzmann constant K _{(j/k)} in joule per kelvin and multiply the diameter of the particle T_{(k)} in kelvin and divided by the root of two √2 and multiply the pi π in value 3.14 and multiply the square of distance d^{2}_{(m)} in meter and multiply the pressure of the gas p _{(N/m2)} in Newton per meter^{2}. Hence the mean free path formula can be written as

Λ _{(m)}=K _{(j/k)}*T_{(k)} /( √2 * π * d^{2}_{(m)} *p _{(N/m2)})

Where

Λ=is the mean free path in m

K=is the Boltzmann constant k=1.380649 *10^{-23} j /K

T=is the diameter of the particle in kelvin

P=is the pressure of the gas in N/m2

### Example:1

Calculate the mean free path of nitrogren molecule at 30^{}C when pressure is 2.5 atm.If the diameter of the nitrogen molecule is 2.5 A, the average speed of the nitrogen molecule is 250 m/s .Find the time taken by the molecule between two successive collision?

### Answer:

Given that

Temperature T =30^{}C =30+273=303

Pressure p =2.5 atm =2.525 *10 ^{5 } N/m^{2}

Nitrogen diameter d =2.5A=2.5 *10^{-10} m

Boltzmann constant k =1.380649 *10^{-23} J/k

Mean free path

Λ _{(m)}=K _{(j/k)}*T_{(k)} /( √2 * π * d^{2}_{(m)} *p _{(N/m2)})

=(1.380649 *10^{-23 }*303 )/ (√2 * π (2.5 *10^{-10})^{2} * (2.2525 *10^{5} )

=4.1833665 *10^{-21} / 6.254746 *10^{-14}

=1.49 *10^{-37} m

The time interval between two successive collisions t =distance /speed

= Λ /v

=1.49 *10^{-37} /250

=3.725 *10^{-35}

Therefore the mean free path of nitrogen 3.725 *10^{-35} m

### Example:2

Calculate the mean free path of nitrogren molecule at 50^{}C when pressure is 4.5 atm.If the diameter of the nitrogen molecule is 4.5 A, the average speed of the nitrogen molecule is 350 m/s .Find the time taken by the molecule between two successive collision?

### Answer:

Given that

Temperature T =30^{}C =50+273=323

Pressure p =4.5 atm =4.545 *10 ^{5 } N/m^{2}

Nitrogen diameter d =4.5A=4.5 *10^{-10} m

Boltzmann constant k =1.380649 *10^{-23} J/k

Mean free path

Λ _{(m)}=K _{(j/k)}*T_{(k)} /( √2 * π * d^{2}_{(m)} *p _{(N/m2)})

=(1.380649 *10^{-23 }*323 )/ (√2 * π (4.5*10^{-10})^{2} * (4.545*10^{5} )

=4.4594 *10^{-21} / 4.0890 *10^{-13}

=1.09058 *10^{-34}m

The time interval between two successive collisions t =distance /speed

= Λ /v

=1.09058 *10^{-34} / 350

=3.1159 *10^{-37}

Therefore the mean free path of nitrogen 3.1159 *10^{-37} m