Tag: behaviour of perfect gas and kinetic theory of gases

Quizzes (2) Questions (105)

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

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 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 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The mean free path in 3D is lambda = 1 / (sqrt(2) * pi * n * d^2). When considering motion along a single coordinate axis, the effective mean free path is reduced by a factor of sqrt(3) due to the projection of the velocity components.

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 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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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. $0.2nm$
Reveal answer Fill a bubble to check yourself
C Correct answer
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
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 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'
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The mean free path formula lambda = 1 / (sqrt(2) * pi * n * d^2) shows that it depends on both the molecular diameter (d) and the molecular number density (n).

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