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Mean free path - class-XI

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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. $F _1=F _2=0$


Correct Option: A
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 $F _1 > F _2$

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


Correct Option: D
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:
$l.p=\dfrac{kT}{\sqrt{2}\pi d _m^2}$

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. $4m$


Correct Option: A
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 $\lambda =\dfrac{v _{rms}}{f}=\dfrac{10}{10}=1 m$

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


Correct Option: B
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.

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

  1. zero

  2. less

  3. more

  4. $\infty$


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 :

  1. T

  2. T$^{-1}$

  3. T$^2$

  4. T$^4$


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?

  1. $10^{-3}$

  2. $10^{-2}$

  3. $10^{-4}$

  4. $10^{-5}$


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 ?

  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


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 $d=0.24\ nm$.

  1. $1.56\times 10^{12}\ m$

  2. $2.56\times 10^{12}\ m$

  3. $3.56\times 10^{12}\ m$

  4. $4.56\times 10^{12}\ m$


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

  1. $2.25x{10^{ - 8}}m$

  2. $1.12x{10^{ - 7}}m$

  3. $11.2x{10^{ - 7}}m$

  4. $22.5x{10^{ - 8}}m$


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

  1. $\alpha $

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

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

  4. $3\alpha $


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

  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


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 :

  1. $0.2\mu$ $m$

  2. $0.3\mu$ $m$

  3. $0.2$mm

  4. $0.2nm$


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 

  1. the diamter of molecules

  2. molecular density of gas

  3. both'a' and 'b'

  4. neither 'a' nor 'b'


Correct Option: C

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

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

Estimate the mean free path of a cosmic ray proton in the atmosphere at sea level. Given $\sigma = 10 ^{-26} cm^2$

  1. $10^4 cm$

  2. $10^{-4} cm$

  3. $10^6 cm$

  4. $10^{-6} cm$


Correct Option: C
Explanation:

The mean free path is given by, $\lambda=\dfrac{1}{\pi \sigma n}$

Here the proton is scattered by the molecules of the atmosphere. 
The density of the molecules of the atmosphere is $n=\dfrac{N}{V}=\dfrac{P}{kT}=\dfrac{10^6}{(1.38\times 10^{-16})(300)}=2.4\times 10^{19}$  (all units are taken in CGS unit)
So, $\lambda=\dfrac{1}{\pi(10^{-26})(2.4\times 10^{19})}=1.3\times 10^6\sim 10^6 $  $cm$

Estimate the average number of collisions per second that each $N _2$ molecule undergoes in air at room temperature and at atmospheric pressure. The diameter of a $N _2$ molecule is $0.3\ mm$.

  1. $0.2\ \mu m$

  2. $0.1\ \mu m$

  3. $0.1\ mm$

  4. None of these


Correct Option: B

A container is divided into two equal parts I and II by a partition with a small hole of diameter d. The two partitions are filled with same ideal gas, but held at temperatures $T _I=150$K and $T _{II}=300$K by connecting to heat reservoirs. Let $\lambda _I$ and $\lambda _{II}$ be the mean free paths of the gas particles in the two parts such that $d > > \lambda _I$ and $d > > \lambda _{II}$. Then $\lambda _I/\lambda _{II}$ is close to.

  1. $0.25$

  2. $0.5$

  3. $0.7$

  4. $1.0$


Correct Option: C
Explanation:

Given,

Partition has hole of diameter $d$, Mean pressure between both sections is equal.

Boltzmann constant ${{K} _{B}}$

At constant pressure, Mean Free path $\lambda \ \alpha \ \sqrt{{{K} _{B}}T}$

Mean free path in ${{1}^{st}}$ section ${{\lambda } _{I}}=\sqrt{{{K} _{B}}\times 150}$

Mean free path in ${{2}^{nd}}$ section ${{\lambda } _{II}}=\sqrt{{{K} _{B}}\times 300}$

$\dfrac{{{\lambda } _{I}}}{{{\lambda } _{II}}}=\dfrac{\sqrt{{{K} _{B}}\times 150}}{\sqrt{{{K} _{B}}\times 300}}=0.707$

Hence, $\dfrac{{{\lambda } _{I}}}{{{\lambda } _{II}}}\cong 0.7$ 

Calculate the means free path of nitrogen molecule at $27^o$C when pressure is $1.0$ atm. Given, diameter of nitrogen molecule $=1.5\overset{o}{A}$, $k _B=1.38\times 10^{-23}$J $K^{-1}$. If the average speed of nitrogen molecule is $675$ $ms^{-1}$. The time taken by the molecule between two successive collisions is?

  1. $0.6$ns

  2. $0.4$ns

  3. $0.8$ns

  4. $0.3$ns


Correct Option: A
Explanation:
Here, $T=27^oC=27+273=300$K.
$P=1$atm $=1.01\times 10^5$N $m^{-2}$, d$=1.5\overset{o}{A}=1.5\times 10^{-10}$m,
$k _B=1.38\times 10^{-23}$J $K^{-1}, \lambda =?$
From $\lambda =\dfrac{k _BT}{\sqrt{2}\pi d^2p}=\dfrac{1.38\times 10^{-23}\times 300}{1.414\times 3.14(1.5\times 10^{-10})^2\times 1.01\times 10^5}$
$=4.1\times 10^{-7}$m
Time interval between two successive collisions
$t=\dfrac{distance}{speed}=\dfrac{\lambda}{v} = \dfrac{4.1\times 10^{-7}}{675}=0.6\times 10^{-9}s$

Ten small planes are flying at a speed of $150$km $h^{-1}$ in total darkness in an air space that is $20\times 20\times 1.5km^3$ in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius $10$m.

  1. $125$h

  2. $220$h

  3. $432$h

  4. $225$h


Correct Option: D
Explanation:
Here, $v=150$km $h^{-1}$
$N=10$
$V=20\times 20\times 1.5$ $km^3$
Diameter of plane, $d=2R=2\times 10$
$=20m=20\times 10^{-3}$km
$n=\dfrac{N}{V}=\dfrac{10}{20\times 20\times 1.5}=0.0167km^{-3}$
Mean free path of a plane
$\lambda =\dfrac{1}{\sqrt{2}\pi d^2n}$
Time elapse before collision of two planes randomly,
$t=\dfrac{\lambda}{v}=\dfrac{1}{\sqrt{2}\pi d^2nv}$
$=\dfrac{1}{1.414\times 3.14\times (20)^2\times 10^{-6}\times (0.0167)\times (150)}$
$=\dfrac{10^6}{4449.5}=224.74$h $=225$h

Estimate the mean free path for a water molecule in water vapor at $373K$,the diameter of the molecule is $2\ \times 10^{-10}\ m$ and at $STP$ number of molecular per unit volume is $2.7\ \times 10^{25}\ m^{-3}$ :

  1. $2.81 \times 10^{-7}\ m$

  2. $3 \times 10^{-7}\ m$

  3. $4 \times 10^{-7}\ m$

  4. $5 \times 10^{-7}\ m$


Correct Option: A

There are two vessels of same consisting same no of moles of two different gases at same temperature . One of the gas is $CH _{4}$ & the other is unknown X. Assuming that all the molecules of X are under random motion whereas in $CH _{4}$ except one all are stationary. Calculate $Z _{1}$ for X in terms of $Z _{1}$ of $CH _{4}$. Given that the collision diameter for both gases are same & $\displaystyle (U _{rms}) _{x}=\frac{1}{\sqrt{6}}(Uav) _{CH _{4}}$.

  1. $\displaystyle \frac{2\sqrt{2}}{3\sqrt{\pi }}Z _{1}$

  2. $\displaystyle \frac{3\sqrt{2}}{2\sqrt{\pi }}Z _{1}$

  3. $\displaystyle \frac{2\sqrt{3}}{2\sqrt{\pi }}Z _{1}$

  4. $\displaystyle \frac{4\sqrt{2}}{3\sqrt{\pi }}Z _{1}$


Correct Option: A
Explanation:

V, n, T $\rightarrow  same$(25) so $P\rightarrow $ also same ( P  5  25)
$\displaystyle \sigma \rightarrow same (25)$
given

$\displaystyle (v {rms})\times

x=\dfrac{1}{\sqrt{6}}(v _{avg.}) _{CH _{4}}$ &

$v _{rms}=\sqrt{\dfrac{3\pi }{8}}(v _{avg.})$ so
$\displaystyle \sqrt{\dfrac{3\pi }{8}}(v _{avg.}) _{CH _{4}}$
$\displaystyle \dfrac{(v _{avg.})x}{(v _{avg.}) _CH _{4}}=\sqrt{\dfrac{8}{3\pi }}.\frac{1}{\sqrt{6}}=\dfrac{2}{3\sqrt{\pi }}$
For X (9< ) : $\displaystyle Z _{1}=\sqrt{2}\pi \sigma ^{2}(v _{avg.}) _{x}N^{\ast }$
For CH
{4} (9< ) : $\displaystyle Z _{1}=\pi \sigma ^{2}(v _{avg.}) _{CH _{4}}N^{\ast }$
Since T, P, v, n are same, $N\ast $ will also be same.
$\displaystyle



\frac{Z _{1}X}{Z _{1}(CH _{4})}=\sqrt{2}\frac{(v _{avg.}) _{x}}{(v _{avg.}) _{CH _{4}}}=\sqrt{2}.\frac{2}{3\sqrt{\pi

}}$
$\displaystyle Z _{1}(X)=Z _{1}(CH _{4}).\frac{2\sqrt{2}}{3\sqrt{\pi }}$

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