Tag: behavior of perfect gas and kinetic theory

Questions Related to behavior of perfect gas and kinetic theory

In two vessels of the same volume, atomic hydrogen and helium with pressure 1 atm and 2 atm are filled. If temperature of both the same is the same, then the average speed of hydrogen atom $v _H$ will be related to helium $v _{He}$ as

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $v _{H}$ $= \sqrt{2}$ $v _{He}$ 

  2. $v _H$ $=$ $v _{He}$

  3. $v _H$ $=$ 2$v _{He}$

  4. $v _H$ $=$ $\dfrac{v _{He}}{2}$

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Correct Option: C
Explanation:

By Maxwell's speed distribution, $<v>\alpha \sqrt { \dfrac { RT }{ M }  } $. Since the temperature of two gases is same, hence


$<v>\alpha \sqrt { \dfrac { 1 }{ M }  } $

Also, ${ M } _{ He }=4{ M } _{ H }$

Hence, $<{ v } _{ H }>=2<{ v } _{ He }>$

Answer is option C.

The molecular weights of $O _2$ and $N _2$ are 32 and 28 respectively. At $15^0$C, the pressure of 1 gm will be the same as that of 1 gm in the same bottle at the temperature.

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $-21^0$C

  2. $13^0$C

  3. $15^0$C

  4. $56.4^0$C

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Correct Option: A

Average kinetic energy of a gas molecule is

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. Inversely proportional to the square of its absolute temperature

  2. Directly proportional to the square root of its absolute temperature

  3. Directly proportional to its absolute temperature

  4. Directly proportional to square of absolute temperature

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Correct Option: C

Maxwell's laws of distribution of velocities shows that

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. the number of molecules with most probable velocity is very large

  2. the number of molecules with most probable velocity is small

  3. the number of molecules with most probable velocity is zero

  4. the number of molecules with most probable velocity is exactly equal to 1

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Correct Option: A
Explanation:

The form of Maxwell's velocity distribution function is gaussian type. So the maximum of this function represents the speed at which most of the molecules travel. This speed is known as most probable speed.

The average velocity of the molecules in a gas in equilibrium is

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. proportional to $\sqrt{T}$

  2. proportional to T

  3. proportional to $T^{2}$

  4. equal to zero

Show answer
Correct Option: A
Explanation:

the average velocity of the gas molecules = $\sqrt{\frac{8RT}{\pi M}}$
so clearly, the average velocity $\alpha \sqrt{T}$
So, A is the correct answer.
Note that T is the temperature in Kelvins

The average kinetic energy of a gas molecule at ${27}^{o}C$ is $6.21\times {10}^{-21}J$, then its average kinetic energy at ${227}^{o}C$ is:

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $10.35\times {10}^{-21}J$

  2. ${11.35}\times {10}^{-21}J$

  3. $52.2\times {10}^{-21}J$

  4. $5.22\times {10}^{-21}J$

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Correct Option: A
Explanation:

Average kinetic energy of gas molecules $\propto$ Temperature (Absolute)
$\cfrac{K.E(at\quad {227}^{o}C)}{K.E (at\quad {27}^{o}C)}=\cfrac{273+227}{273+27}=\cfrac{500}{300}=\cfrac{5}{3}$
$K.E({227}^{o})=\cfrac{5}{3}\times 6.21\times {10}^{-21}J=10.35\times {10}^{-21}J$

For a given gas, which of the following relationships is correct at a given temp?

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $u _{rms} > u _{av} > u _{mp}$

  2. $u _{rms} < u _{av} < u _{mp}$

  3. $u _{rms} > u _{av} < u _{mp}$

  4. $u _{rms} < u _{av} > u _{mp}$

Show answer
Correct Option: A
Explanation:

$u _{rms} = $ Root mean square velocity
$u _{ar}= $Average velocity 
$u _{mp} $= Most probable velocity
$u _{mp}:u _{ar}:u _{rms}= 1: 1.128: 1224$
$\therefore u _{rms} > u _{ar} > u _{mp}$

A vessel contains a mixture consisting of m$ _{1}$ - 7 g of nitrogen (M$ _{1}$ = 28) and m$ _{2}$ = 11 g of carbon dioxide (M$ _{2}$ = 44) at temperature T - 300 K and pressure P$ _{0}$ = 1 atm. The density of the mixture is

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $1.46g\ per\ litre$

  2. $2.567 g \ per \ litre$

  3. $3.752 g \ per \ litre$

  4. $4.572 g \ per \ litre$

Show answer
Correct Option: A
Explanation:

Let the volume occupied $=V$

By Dalton's law of partial pressure
$\cfrac{P _{nit}}{P _0}=\cfrac{n _{nit}}{n _{nit}+n _{carb}}$
No. of moles of Nitrogen $\eta _{nit}=\cfrac{M _1}{M _{nit}}=\cfrac{7}{28}=0.25 mol$
No. of moles of carbon $\eta _{carb}=\cfrac{M _2}{M _{carb}}=\cfrac{11}{4}=0.25 mol$
Thus,
$P _{nit}=P _0\times\cfrac{0.25}{0.25\times0.25}=P _0/2=0.5atm$
From ideal gas equation
$V=\cfrac{nRT}{P}=\cfrac{0.25\times8.314\times290}{0.5\times101325}=0.0119mole$
Total mixture $m=(7+11)\times 10^{-11}kg$
Thus density s $P=\cfrac{m}{V}=\cfrac{(7+11)\times 10^{-3}}{0.0119}\approx1.46kg/m^3$
Option A is correct.

A vessel of volume V contains a mixture of $1$mole of hydrogen and $1$ mole of oxygen(both considered as ideal). Let $f _1(v)dv$ denote the fraction of molecules with speed between v and $(v+dv)$ with $f _2(v)dv$, similarly for oxygen. then

physics kinetic theory maxwell-boltzmann speed distribution function behavior of perfect gas and kinetic theory kinetic theory of matter

  1. $f _1(v)+f _2(v)=f(v)$ obeys the Maxwell's distribution law

  2. $f _1(v), f _2(v)$ will obey the Maxwell's distribution law separately

  3. Neither $f _1(v)$ nor $f _2(v)$ will obey the Maxwell's distribution law

  4. $f _2(v)$ and $f _1(v)$ will be the same

Show answer
Correct Option: B
Explanation:

The Maxwell-Boltzmann speed distribution function $\left(N _v=\dfrac{dN}{dv}\right)$ depends on the mass of the gas molecule. [Here, dN is the number of molecules with speeds between v and $(v+dv)$]. The masses of hydrogen and oxygen molecules are different.

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

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. $F _1 > F _2$

  2. $F _1 < F _2$

  3. $F _1=F _2$, both non-zero

  4. $F _1=F _2=0$

Show answer
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 $F _1 > F _2$