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Maxwell-boltzmann speed distribution function - class-XI

Description: maxwell-boltzmann speed distribution function
Number of Questions: 20
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Tags: kinetic theory of matter physics kinetic theory kinetic theory of gases behavior of perfect gas and kinetic theory
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Hydrogen is a diatomic gas. Its molar specific heat at constant volume is very nearly

  1. $\frac { 3 R } { 2 }$

  2. $\frac { 5 R } { 2 }$

  3. $\frac { 7 R } { 2 }$

  4. (b) or (c) depending on the temperature.


Correct Option: D

$T _1$ is the temperature of oxygen enclosed in a cylinder. The temperature is increased to $T _2$ and Maxwellan distribution curves for $O _2$ at temperature $T _1$ and $T _2$ are plotted. If $A _1$ and $A _2$ are the areas under the curves and the speed axis, in both cases , then 

  1. $A _1 > A _2$

  2. $A _1 < A _2$

  3. A_1 = A_2$

  4. $A _1=\sqrt {A _2}$


Correct Option: A

let A and B the two gases and given :
$\frac{{T} _{A}}{{M} _{A}}$ = 4. $\frac{{T} _{B}}{{M} _{B}}$  Where T is the temperature and M is molecular mass. If ${C} _{A}$ and  ${C} _{B}$ are the r.m.s. speed, then the ratio $\frac{{C} _{A}}{{C} _{B}}$ will be equal to:

  1. 2

  2. 4

  3. 1

  4. 0.5


Correct Option: A

A mixture of ideal gases 7 kg of nitrogen and 11 Kg of $ CO _2 $ then (Take $\gamma$ for nitrogen and $CO _2$ as 1.4 and 1.3 respectively)

  1. Equivalent molecular weight of the mixture is 36.

  2. Equivalent molecular weight of the mixture is 18.

  3. $ \gamma $ for the mixture is 5/2

  4. $ \gamma $ for the mixture is 47/35


Correct Option: A,D

$3$ mole of gas ''X"  and $2$ moles of gas "Y" enters from end "P" and "Q" of the cylinder respectively. The cylinder has the area of cross section , shown as
under 
The length of the cylinder is $150cm$. The gas "X" intermixes with gas "Y" at the point . If the molecular weight of the gases X and Y is $20$ and $80$ respectively, then what will be the distance of point A from Q?

  1. $75cm$

  2. $50cm$

  3. $37.5$

  4. $90cm$


Correct Option: B
Explanation:

$\begin{array}{l} \frac { { rx } }{ { ry } } =\frac { { { w _{ x } } } }{ { { n _{ y } } } } \sqrt { \frac { { { M _{ y } } } }{ { { M _{ x } } } }  }  \ =\frac { 3 }{ 2 } \sqrt { \frac { { 80 } }{ { 20 } }  } =\frac { 3 }{ 1 } =3:1 \ \therefore \frac { { dis\tan  ce\, \, travelled\, \, by\, \, gas\, \, X } }{ { dis\tan  ce\, \, travelled\, \, by\, \, gas\, \, Y } } =3:1 \ \therefore dis\tan  ce\, \, of\, \, A\, \, from\, \, Q=\frac { { 150 } }{ 3 } =50\, \, cms \end{array}$

Hence, OPtion $B$ is correct.

The lowest pressure(the best Vaccum) that can be created in laboratory at 27 degree is $10^{-11} $ mm of Hg. At this pressure, the number of ideal gass molecules per $cm^{3}$ will be

  1. $3.22 \times 10 ^{12} $

  2. $1.61 \times 10 ^{12} $

  3. $3.21 \times 10 ^{6} $

  4. $3.22 \times 10 ^{5} $


Correct Option: A

If $P=10^6kT$, then the number of molecules per unit volume of the gas is:

  1. 1

  2. $10^2$

  3. $10^3$

  4. $10^6$


Correct Option: D

A sample of gas is at $0^{\circ}C$. To what temperature must it be raised in order to double the rms speed of its molecules?

  1. $102^{\circ}C$

  2. $273^{\circ}C$

  3. $819^{\circ}C$

  4. $1092^{\circ}C$


Correct Option: C

One mole of gas occupies 10 ml at 50 mm pressure. The volume of 3 moles of the gas at 100 mm pressure and same temperature is 

  1. 15 ml

  2. 100 ml

  3. 200 ml

  4. 500 ml


Correct Option: A

2 moles of an ideal monoatomic gas at temperature $T _0$ is mixed wth 4 moles of another ideal monoatomic gas at temperature $2T _0$ then  the temperature of the mixture is:

  1. $\frac{5}{3} T _0$

  2. $\frac{3}{2} T _0$

  3. $\frac{4}{3} T _0$

  4. $\frac{5}{4} T _0$


Correct Option: A

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


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.

  1. $-21^0$C

  2. $13^0$C

  3. $15^0$C

  4. $56.4^0$C


Correct Option: A

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


Correct Option: C

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


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

  1. proportional to $\sqrt{T}$

  2. proportional to T

  3. proportional to $T^{2}$

  4. equal to zero


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:

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

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

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

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


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?

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

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

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

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


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

  1. $1.46g\ per\ litre$

  2. $2.567 g \ per \ litre$

  3. $3.752 g \ per \ litre$

  4. $4.572 g \ per \ litre$


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

  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


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.

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