Tag: behaviour of perfect gas and kinetic theory of gases

Questions Related to behaviour of perfect gas and kinetic theory of gases

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

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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. $0.2nm$

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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

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

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. 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

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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

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. $\rho $

  2. T

  3. d

  4. b

Show answer
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
$He > H _2 > O _2 > N _2 > CO _2$

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. True

  2. False

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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.

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

Show answer
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?

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. $1m$

  2. $2m$

  3. $3m$

  4. $0.1m$

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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?

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. $1m$

  2. $0.1m$

  3. $2m$

  4. None of the above

Show answer
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