Tag: law of equipartition of energy and mean free path

Questions Related to law of equipartition of energy and mean free path

A molecule of gas in a container hits one wall (1) normally and rebounds back. It suffers no collision and hits the opposite wall (2) which is at an angle of $30^o$ with wall 1.
Assuming the collisions to be elastic and the small collision time to be the same for both the walls, the magnitude of average force by wall 2. $(F _2)$ provided the molecule during collision satisfy

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $F _1 > F _2$

  2. $F _1 < F _2$

  3. $F _1=F _2$, both non-zero

  4. $F _1=F _2=0$

Show answer
Correct Option: A
Explanation:

Initial momentum, $P _1=mvcos 30$
and final momentum, $P _2 = mvcos30$
change in momentum
$\Delta P = -2mv cos30$
$\Delta P =-\sqrt 3 mv$
Force on wall-1
$F _1=\frac {2mv}{\Delta t}$
Force on wall-2
$F _2=\frac {\sqrt 3mv}{\Delta t}$, so $F _1 > F _2$

Mean free path depends on which of the following?

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. size of the molecule

  2. density of the molecule

  3. diameter of the molecule

  4. All of the above

Show answer
Correct Option: D
Explanation:

The mean free path or average distance between collisions for a gas molecule may be estimated from kinetic theory.

Mean free path displays linear proportionality to the temperature and inverse proportionality to the pressure and molecular diameter.
Mathematically it is expressed as:
$l.p=\dfrac{kT}{\sqrt{2}\pi d _m^2}$

A gas has an average speed of 10 m/s and a collision frequency of 10 $s^{-1}$. What is its mean free path?

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $1m$

  2. $2 m$

  3. $3 m$`

  4. $4m$

Show answer
Correct Option: A
Explanation:

The collision frequency $(f)$ is the ratio of rms velocity $(v _{rms})$ to mean free path $(\lambda)$.

Here, collision frequency $f=10 s^{-1}$ and $v _{avg}=10 m/s$
So, $v _{rms}=\sqrt{v _{avg}^2}=\sqrt{10^2}=10 m/s$
Thus, mean free path $\lambda =\dfrac{v _{rms}}{f}=\dfrac{10}{10}=1 m$

If the pressure in a closed vessel is reduced by drawing out some gas, the mean-free path of molecules :

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. is decreased

  2. is increased

  3. remains unchanged

  4. increases or decreases according to the nature of the gas

Show answer
Correct Option: B
Explanation:

The mean-free path of molecule is the distance traveled by a molecule in two consecutive collision. If pressure is reduced and there are less particle then a molecule will travel longer distance before collision, so mean free path is increased.

If the pressure of a gas is increased then its mean free path becomes :

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. zero

  2. less

  3. more

  4. $\infty$

Show answer
Correct Option: B
Explanation:

As gas pressure increases mean free path of the gas decreases. 
Mean free path is the distance traveled by a gas molecule between two successive collisions.
So, as pressure increases number of collisions increase. Hence, mean free path decreases.   

The mean free path of a gas varies with absolute temperature as :

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. T

  2. T$^{-1}$

  3. T$^2$

  4. T$^4$

Show answer
Correct Option: A
Explanation:

The expression for mean free path $\lambda =\dfrac { RT }{ \sqrt { 2 } \pi { d }^{ 2 }NP } $ mean free path is directly proportional to Temperature

Hence, option A is correct

A gas has a molecular diameter of 0.1 m. It also has a mean free path of 2.25 m. What is its density?

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $10^{-3}$

  2. $10^{-2}$

  3. $10^{-4}$

  4. $10^{-5}$

Show answer
Correct Option: A
Explanation:
Given :
Molecular diameter of a gas, $d = 0.1 m$.
Mean free path, $l = 2.25 m$.
The mean free path traversed by the molecules is given by
$l = \dfrac{1}{\sqrt 2 \pi d^2 \rho}$
Therefor,
$\rho = \dfrac{1}{\sqrt 2 \pi d^2 l}$
Using the given values we get,
$\rho = \dfrac{1}{\sqrt 2 (3.14)(0.1)^2 (2.25)}$
$\rho = \dfrac{1}{0.0999}$
$\rho = 10 ^{-3}$

In physics, the mean free path is the average distance traveled by a moving particle (such as an atom , a molecule, a photon) between successive impacts (collisions), which modify its direction or energy or other particle properties. In which of the following mean free path is used ?

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. to estimate the resistivity of a material

  2. to design a chemical apparatus

  3. It can be used in optics and in acoustics

  4. All of the above

Show answer
Correct Option: D
Explanation:

The concept of mean free path is popular in all branches of physics and it has a number of applications including estimation of resistivity of material, designing a chemical apparatus, and optics and acoustics, etc.

A satellite sent into space samples the density of matter within the solar system and gets a value $2.5$ hydrogen atoms per cubic centimeter. What is the mean free path of the hydrogen atoms? Take the diameter of a hydrogen atoms as $d=0.24\ nm$.

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $1.56\times 10^{12}\ m$

  2. $2.56\times 10^{12}\ m$

  3. $3.56\times 10^{12}\ m$

  4. $4.56\times 10^{12}\ m$

Show answer
Correct Option: A

Estimate the mean free path of nitrogen molecule in a cylinder containing nitrogen at 2.0atm pressure and temperature ${17^o}C$.(take the radius of nitrogen molecule to be 1.0A, Molecular mass=28gm

physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

  1. $2.25x{10^{ - 8}}m$

  2. $1.12x{10^{ - 7}}m$

  3. $11.2x{10^{ - 7}}m$

  4. $22.5x{10^{ - 8}}m$

Show answer
Correct Option: B