The expression for mean free path is:
A. $\lambda = \dfrac{{KT}}{{\sqrt 2 \pi {d^2}P}}$
B. $\lambda = \dfrac{{\pi dP}}{{kT}}$
C. $\lambda = \dfrac{{\pi {d^2}P}}{{kT}}$
D. $\lambda = \dfrac{{kT}}{{\pi dP}}$
Answer
588.9k+ views
Hint: The actual distance a molecule travels before a collision is called free path. The average of such distances is known as mean free path. As the size of molecules gets larger the mean free path gets shorter.
The mean free path is given as $\lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}\dfrac{N}{V}}}$
Complete step by step answer:
The motion of a gas molecule inside an ideal gas changes its direction and speed as it collides elastically with other molecules of the same gas. Between the collisions the molecules move in a straight line at some constant speed. Measuring the mean free path is difficult as it depends on a number of factors.
Let us derive the equation for the mean free path. Consider $\lambda $ to be the average distance travelled by a molecule between collisions. As the number of molecules per unit volume increases, the number of collisions increases hence the distance travelled by molecule decreases.
Thus, we have:
$\lambda \alpha \dfrac{1}{{\left( {\dfrac{N}{V}} \right)}}$ -- equation $1$
Here, $\dfrac{N}{V}$ is the number of molecules per unit volume.
Also, if we consider molecules to be point masses then there is no collision as their point mass would not have any surface. Thus, if a molecule has a larger surface that is larger in diameter than the probability of collision will increase and thus reduce the mean free path. Thus, the mean free path must be proportional to $\pi $ times the square of the diameter $d$ as the collision takes place on the surface and not only at the diameter line
$\lambda \alpha \dfrac{1}{{\pi {d^2}}}$
Thus, the number of collisions in time $t$ having when the particles is having velocity $v$ will be given as
$\pi {d^2}vt\dfrac{N}{V}$ -- equation $2$
The particle covers a path of length $vt$ in time $t$ .
Now, mean free length will be the ratio of the length of path covered in time $t$ divided by the number of collisions.
$\lambda = \dfrac{{{v_{avg}}t}}{{\pi {d^2}vt\dfrac{N}{V}}}$
Here ${v_{avg}}$ is the average velocity.
$v$ is the relative velocity.
The ratio $\dfrac{v}{{{v_{avg}}}} = \sqrt 2 $ thus
$ \Rightarrow \lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}\dfrac{N}{V}}}$ --equation $3$
Using gas equation, we have $\dfrac{N}{V} = \dfrac{P}{{KT}}$
Here, $P$ is the pressure inside the cylinder.
$K$ is the Boltzmann constant
$T$ is the temperature.
Substituting this value in equation $3$ we get
$\lambda = \dfrac{{KT}}{{\sqrt 2 \pi {d^2}P}}$
This is the expression for mean free path.
So, the correct answer is “Option A”.
Note:
Two velocities are considered in the derivation; average velocity and relative velocity. It is difficult to measure the path between collisions as the collisions take place in a random manner. As the size of the atom increases, the number of collisions increases and thus the free path decreases. Mean free path is the average of the sum of free paths.
The mean free path is given as $\lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}\dfrac{N}{V}}}$
Complete step by step answer:
The motion of a gas molecule inside an ideal gas changes its direction and speed as it collides elastically with other molecules of the same gas. Between the collisions the molecules move in a straight line at some constant speed. Measuring the mean free path is difficult as it depends on a number of factors.
Let us derive the equation for the mean free path. Consider $\lambda $ to be the average distance travelled by a molecule between collisions. As the number of molecules per unit volume increases, the number of collisions increases hence the distance travelled by molecule decreases.
Thus, we have:
$\lambda \alpha \dfrac{1}{{\left( {\dfrac{N}{V}} \right)}}$ -- equation $1$
Here, $\dfrac{N}{V}$ is the number of molecules per unit volume.
Also, if we consider molecules to be point masses then there is no collision as their point mass would not have any surface. Thus, if a molecule has a larger surface that is larger in diameter than the probability of collision will increase and thus reduce the mean free path. Thus, the mean free path must be proportional to $\pi $ times the square of the diameter $d$ as the collision takes place on the surface and not only at the diameter line
$\lambda \alpha \dfrac{1}{{\pi {d^2}}}$
Thus, the number of collisions in time $t$ having when the particles is having velocity $v$ will be given as
$\pi {d^2}vt\dfrac{N}{V}$ -- equation $2$
The particle covers a path of length $vt$ in time $t$ .
Now, mean free length will be the ratio of the length of path covered in time $t$ divided by the number of collisions.
$\lambda = \dfrac{{{v_{avg}}t}}{{\pi {d^2}vt\dfrac{N}{V}}}$
Here ${v_{avg}}$ is the average velocity.
$v$ is the relative velocity.
The ratio $\dfrac{v}{{{v_{avg}}}} = \sqrt 2 $ thus
$ \Rightarrow \lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}\dfrac{N}{V}}}$ --equation $3$
Using gas equation, we have $\dfrac{N}{V} = \dfrac{P}{{KT}}$
Here, $P$ is the pressure inside the cylinder.
$K$ is the Boltzmann constant
$T$ is the temperature.
Substituting this value in equation $3$ we get
$\lambda = \dfrac{{KT}}{{\sqrt 2 \pi {d^2}P}}$
This is the expression for mean free path.
So, the correct answer is “Option A”.
Note:
Two velocities are considered in the derivation; average velocity and relative velocity. It is difficult to measure the path between collisions as the collisions take place in a random manner. As the size of the atom increases, the number of collisions increases and thus the free path decreases. Mean free path is the average of the sum of free paths.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

