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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 in a closed vessle is reduced by drawing out some gas the mean-free path of molecules

  1. losing their kinetic energy

  2. sticking to the walls

  3. changing their momenta due to collision with the walls

  4. getting accelerated towards the wall
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since reduced pressure will result in the reduction of collision with the walls, average momentum will change and hence the 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

State whether true or false:

Mean free path order for some gases at 273 K and 1 atm P is
$He > H _2 > O _2 > N _2 > CO _2$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As atomicity of gas increases, its mean path decreases, also as attraction force between gas molecules increases, mean free path decreases so order is
$He > H _2 > O _2 > N _2 > CO _2$

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

The mean free path of the molecule of a certain gas at 300 K is $2.6\times10^{-5}:m$. The collision diameter of the molecule is 0.26 nm. Calculate
(a) pressure of the gas, and
(b) number of molecules per unit volume of the gas.

  1. (a) $1.281\times 10^{23}:m^{-3}$ (b) $5.306\times 10^{2}:Pa$

  2. (a) $1.281\times 10^{22}:m^{-3}$ (b) $5.306\times 10^{3}:Pa$

  3. (a) $12.81\times 10^{23}:m^{-3}$ (b) $53.06\times 10^{2}:Pa$

  4. (a) $2.56\times 10^{23}:m^{-3}$ (b) $10.612\times 10^{2}:Pa$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle \lambda =2.6\times 10^{-5}:m, :\sigma =0.26:nm=2.6\times 10^{-10}m$
$\displaystyle T=300:K$
$\displaystyle \lambda =\frac{1}{\sqrt{2}\pi \sigma ^{2}N^{\ast }}$
$\displaystyle 2.6\times 10^{-5}=\frac{1}{\sqrt{2}\times 3.14\times (2.6\times 10^{-10})^{2}\times N^{\ast }}$
$\displaystyle N^{\ast }=1.281\times 10^{23}m^{-3}$
$\displaystyle N^{\ast }=\frac{P}{KT}$
$\displaystyle P=1.281\times 10^{23}\times 1.38\times 10^{-23}\times 300$
$\displaystyle P=530.3:Pa$

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. $2m$

  3. $3m$

  4. $0.1m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Collision Frequency is the number of times a molecule of a gas collides with other molecules. 

Reciprocal of that frequency is the time taken by the molecule to cover the free path.
We know that distance = $ {speed} \times {time} $
Hence mean free path = $\dfrac {speed} {frequency}$

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 an average time of $0.1 s$ between collisions. What is its mean free path?

  1. $1m$

  2. $0.1m$

  3. $2m$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Mean free path of a body is defined as the distance covered by that body between two successive collisions. 
Average speed of the girl   $v _{avg} = 10$ m/s
Time between two collisions   $t = 0.1$ s
Mean free path  $\lambda = v _{avg} t = 10\times 0.1 =1$ 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

A gas has a density of $10$ particles$/m^3$ and a molecular diameter of $0.1 $m. What is its mean free path?

  1. $2.25m$

  2. $1m$

  3. $3m$

  4. $0.25m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mean free path estimated by the kinetic theory of gases is given by

$\lambda = \displaystyle \frac{1}{\sqrt{2}\pi nd^2}$
Given that number density  $n = 10\textrm{ m}^{-3}$ and diameter $d = 0.1\textrm { m}$
Thus, $\lambda = \displaystyle \frac{1}{\sqrt{2}\pi \times 10\times 0.1^2} \approx 2.25 \textrm{ m}$

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

Consider the following statements for air molecules in an air tight container.
(I) The average speed of molecules is larger than root mean square speed.
(II) Mean free path of molecules is larger than the mean distance between molecules.
(III) Mean free path of molecules increases with temperature.
(IV) The rms speed of nitrogen molecule is smaller than oxygen molecule. The true statements are.

  1. Only II

  2. II & III

  3. II & IV

  4. I, II & IV
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
(I)
$v _{avg} = \sqrt{\dfrac{8kT}{m\pi}}$
$v _{rms} = \sqrt{\dfrac{3kT}{m}}$
$ v _{avg}  \gt  v _{rms}$
so (I) is  correct.
(II)
The mean free path of a molecule is smaller  than the  distance between molecules.
( II ) is wrong.
(III) 
Mean free path is  directly prop to temp T.
$\lambda = \dfrac{RT}{\sqrt{2}\pi d^2 N _{A}P}$
so (III) is correct.
(IV)
$v _{rms} = \sqrt{\dfrac{3kT}{m}}$
Since  $ m _{N _2} < m _{O _2}$, rms speed of nitrogen is more than rms speed of oxygen molecule.
so (IV) is wrong.


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 in kinetic theory can be written as: ( every symbol has standard meaning)

  1. $l= \dfrac{\mu}{2p} \sqrt{\dfrac{\pi K _BT}{m}}$

  2. $l= \dfrac{\mu}{2p} \sqrt{\dfrac{3\pi K _BT}{2m}}$

  3. $l= \dfrac{\mu}{p} \sqrt{\dfrac{\pi K _BT}{2m}}$

  4. $l= \dfrac{3\mu}{2p} \sqrt{\dfrac{\pi K _BT}{2m}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

Mathematically it is expressed as:
$l=\dfrac{\mu}{p}\sqrt{\dfrac{\pi k _BT}{2m}}$
where, $\mu$ is the viscosity
            $m$ is the molecular mass
            $p$ is the pressure

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 in a 1 $m^3$ container has a molecular diameter of 0.1 m. There are 10 molecules. What is its mean free path?

  1. 2.25m

  2. 2m

  3. 3m

  4. 1m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The mean free path of a molecule is given by the formula ($\lambda $)  $\cfrac{1}{\sqrt{2\pi d^2 n}}$ 

where d is the diameter of the molecules; n - number of molecules

$\cfrac{1}{\sqrt{2\pi \times 0.1 \times 0.1 \times 10}} = 2.25m$