Tag: physics

Questions Related to physics

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. $\infty$

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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$^4$

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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

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

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

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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

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. $\alpha $

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

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

  4. $3\alpha $

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Correct Option: C

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

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

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

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

  4. $3.4 \times 10^{-10}$ S

Show answer
Correct Option: A
Explanation:

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

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

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 :

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. $0.2\mu$ $m$

  2. $0.3\mu$ $m$

  3. $0.2$mm

  4. $0.2nm$

Show answer
Correct Option: C
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

The mean free path of the molecules of a gas depends on 

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. the diamter of molecules

  2. molecular density of gas

  3. both'a' and 'b'

  4. neither 'a' nor 'b'

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Correct Option: C