Tag: behavior of perfect gas and kinetic theory

Questions Related to behavior of perfect gas and kinetic theory

Mean free path depends on which of the following?

  1. size of the molecule

  2. density of the molecule

  3. diameter of the molecule

  4. All of the above


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?

  1. $1m$

  2. $2 m$

  3. $3 m$`

  4. $4m$


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 :

  1. is decreased

  2. is increased

  3. remains unchanged

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


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 :

  1. zero

  2. less

  3. more

  4. $\infty$


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 :

  1. T

  2. T$^{-1}$

  3. T$^2$

  4. T$^4$


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?

  1. $10^{-3}$

  2. $10^{-2}$

  3. $10^{-4}$

  4. $10^{-5}$


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 ?

  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


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 $d=0.24\ nm$.

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

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

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

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


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

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

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

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

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


Correct Option: B

The mean free path of a molecule of He gas is $\alpha $. Its mean free path along any arbitrary coordinate axis will be

  1. $\alpha $

  2. $\dfrac { \alpha }{ 3 } $

  3. $\dfrac { \alpha }{ \sqrt { 3 } } $

  4. $3\alpha $


Correct Option: C