Tag: physics

Questions Related to physics

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


Correct Option: C
Explanation:

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

Mean free path does not depend on

  1. $\rho $

  2. T

  3. d

  4. b


Correct Option: D
Explanation:

Mean free path is the average distance between collisions for a gas and is given as $\lambda=\dfrac{RT}{\sqrt2\pi d^2N _AP}$

where d is molecule diameter.
Density $\rho=\dfrac{N _AP}{RT}$

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


Correct Option: A
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
$He > H _2 > O _2 > N _2 > CO _2$

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) $10.612\times 10^{2}:Pa$


Correct Option: A
Explanation:

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

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. $0.1m$


Correct Option: A
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 = $\dfrac {speed} {frequency}$

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


Correct Option: A
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

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. $0.25m$


Correct Option: A
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, $\lambda = \displaystyle \frac{1}{\sqrt{2}\pi \times 10\times 0.1^2} \approx 2.25 \textrm{ m}$

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


Correct Option: A
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.


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


Correct Option: C
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

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


Correct Option: A
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

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