Tag: behavior of perfect gas and kinetic theory

Questions Related to 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.

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. Only II

  2. II & III

  3. II & IV

  4. I, II & IV

Show answer
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)

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

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

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

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

Show answer
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?

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. 2.25m

  2. 2m

  3. 3m

  4. 1m

Show answer
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 $\sigma = 10 ^{-26} cm^2$

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. $10^4 cm$

  2. $10^{-4} cm$

  3. $10^6 cm$

  4. $10^{-6} cm$

Show answer
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 $0.3\ mm$.

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. $0.1\ \mu m$

  3. $0.1\ mm$

  4. None of these

Show answer
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.

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. $0.25$

  2. $0.5$

  3. $0.7$

  4. $1.0$

Show answer
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?

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. $0.6$ns

  2. $0.4$ns

  3. $0.8$ns

  4. $0.3$ns

Show answer
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.

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. $125$h

  2. $220$h

  3. $432$h

  4. $225$h

Show answer
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 $2.7\ \times 10^{25}\ m^{-3}$ :

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

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

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

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

Show answer
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 & $\displaystyle (U _{rms}) _{x}=\frac{1}{\sqrt{6}}(Uav) _{CH _{4}}$.

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

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

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

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

Show answer
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

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