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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.

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 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 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The mean free path in 3D is lambda = 1 / (sqrt(2) * pi * n * d^2). When considering motion along a single coordinate axis, the effective mean free path is reduced by a factor of sqrt(3) due to the projection of the velocity components.

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 and rms velocity of a nitrogen molecule at a temperature 17C are $1.2 \times 10^{-7}$ m and $5 \times 10^2$ m/s respectively.The time between two successive collisions

  1. $2.4 \times 10^{-10}$ S

  2. $1.2 \times 10^{-10}$ S

  3. $3.4 \times 10^{-13}$ S

  4. $3.4 \times 10^{-10}$ S
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mean free path $\lambda=1.2 \times 10^{-7}$

rms velocity  $V _{rms}=5\times 10^{2} m/s$
Time betweem succesive collisions:
$T=\dfrac{\lambda}{V _{rms}}$
$=\dfrac{1.2 \times 10^{-7}}{5\times 10^2}$
$=0.24 \times 10^{-9}$
$=2.4 \times 10^{-10} s$

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

Modern vacuum pumps can evacuate a vessel down to a pressure of $4.0\times { 10 }^{ -15 }atm$. At room temperature $(300K)$, taking $R=8.3J{ K }^{ -1 }\quad { mole }^{ -1 },1\quad atm={ 10 }^{ 5 }Pa\quad \quad $ and ${ N } _{ Avagadro }=6\times { 10 }^{ 23 }{ mole }^{ -1 }$, the mean distance between the molecules of gas in an evacuated vessel will be of the order of :

  1. $0.2\mu$ $m$

  2. $0.3\mu$ $m$

  3. $0.2$mm

  4. $0.2nm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
As we know formula for mean free path
$Y=\dfrac{KT}{\sqrt{2}\pi{\sigma}^{2}p}$
where $\sigma=$diameter of the molecule
$p=$pressure of the gas
$T=$Temperature
$K=$Boltzmann's constant.
Let intermolecular distance be $D$ then in a volume $\dfrac{4\pi}{3}{D}^{3}$ there is only one
$\dfrac{4\pi}{3}{D}^{3}p=\dfrac{1}{{N} _{A}}={R} _{T}$
or $D={\left(\dfrac{3RT}{4\pi{N} _{A}p}\right)}^{\frac{1}{3}}$
Put $p=4\times{10}^{-10}$Pa
$R=83$,${N} _{A}=6\times{10}^{23}$ and $T=300$K
$D={\left(\dfrac{3\times 83 \times 300}{4\times\dfrac{22}{7}\times 6\times{10}^{23}\times 4\times{10}^{-10}}\right)}^{\frac{1}{3}}$
$=0.2$mm
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 the molecules of a gas depends on 

  1. the diamter of molecules

  2. molecular density of gas

  3. both'a' and 'b'

  4. neither 'a' nor 'b'
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The mean free path formula lambda = 1 / (sqrt(2) * pi * n * d^2) shows that it depends on both the molecular diameter (d) and the molecular number density (n).