Tag: law of equipartition of energy and mean free path

Quizzes (0) Questions (93)

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

Multiple choice 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

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

  1. $F _1 > F _2$

  2. $F _1 < F _2$

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

  4. $F _1=F _2=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $F _1 > F _2$

Multiple choice 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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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:
$l.p=\dfrac{kT}{\sqrt{2}\pi d _m^2}$

Multiple choice 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

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. $4m$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\lambda =\dfrac{v _{rms}}{f}=\dfrac{10}{10}=1 m$

Multiple choice 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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice 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

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

  1. zero

  2. less

  3. more

  4. $\infty$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.   

Multiple choice 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

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

  1. T

  2. T$^{-1}$

  3. T$^2$

  4. T$^4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice 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

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. $10^{-5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,
$\rho = \dfrac{1}{\sqrt 2 (3.14)(0.1)^2 (2.25)}$
$\rho = \dfrac{1}{0.0999}$
$\rho = 10 ^{-3}$
Multiple choice 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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice 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

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. $4.56\times 10^{12}\ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mean free path formula is lambda = 1 / (sqrt(2) * pi * n * d^2). Given n = 2.5 * 10^6 m^-3 and d = 0.24 * 10^-9 m, calculating this yields approximately 1.56 * 10^12 m.

Multiple choice 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

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. $22.5x{10^{ - 8}}m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the ideal gas law to find number density n = P/kT, then applying the mean free path formula lambda = 1 / (sqrt(2) * pi * n * d^2) with d = 2 * 10^-10 m, we get approximately 1.12 * 10^-7 m.