Tag: maxwell-boltzmann speed distribution function

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Questions Related to maxwell-boltzmann speed distribution function

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

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

  1. $v _{H}$ $= \sqrt{2}$ $v _{He}$ 

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

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

  4. $v _H$ $=$ $\dfrac{v _{He}}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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.

  1. $-21^0$C

  2. $13^0$C

  3. $15^0$C

  4. $56.4^0$C
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a fixed mass of gas in a constant volume, P is proportional to T/M. P1 = P2 implies T1/M1 = T2/M2. T1 = 15C = 288K. M1 = 32 (O2). M2 = 28 (N2). 288/32 = T2/28. 9 = T2/28. T2 = 252K. In Celsius, 252 - 273 = -21C.

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

Average kinetic energy of a gas molecule is

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

The average kinetic energy of a gas molecule is given by (3/2)kT, where k is the Boltzmann constant and T is the absolute temperature. Thus, it is directly proportional to the absolute temperature.

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

Maxwell's laws of distribution of velocities shows that

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

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

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

  1. proportional to $\sqrt{T}$

  2. proportional to T

  3. proportional to $T^{2}$

  4. equal to zero
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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:

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

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

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

  4. $5.22\times {10}^{-21}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. $u _{rms} &gt; u _{av} &gt; u _{mp}$

  2. $u _{rms} &lt; u _{av} &lt; u _{mp}$

  3. $u _{rms} &gt; u _{av} &lt; u _{mp}$

  4. $u _{rms} &lt; u _{av} &gt; u _{mp}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

  1. $1.46g\ per\ litre$

  2. $2.567 g \ per \ litre$

  3. $3.752 g \ per \ litre$

  4. $4.572 g \ per \ litre$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.