In this question paper, some of the important one mark, two and five marks questions from the chapter Kinetic Theory are covered. The questions are prepared from the book back and previous year questions.

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Questions:

Generic 2

14 x 2 = 28

Explain qualitatively, how the extent of Brownian motion is affected by the
(a) size of the Brownian particle,
(b) density of the medium,
(c) temperature of the medium,
(d) viscosity of the medium?

What is basic law followed by equipartition of energy?

If there are f degrees of freedom with n moles of a gas, then find the internal energy possessed at a temperature T.

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters \(1\overset { 0 }{ A } \) and \(2\overset { 0 }{ A } \). The gases may be considered under identical conditions of temperature, pressure and volume.

A gas is contained in a closed vessel. How pressure due to the gas will be affected if force of attraction between the molecules disseppear sudenly?

How is mean free path depends on number density of the gas?

The specific heat of argon at constant volume is 0.075kg^-1K^-1, then what will be its atomic weight?[Given, R = 2 cal mol^-1K^-1]

What is the number of degree of freedom of a bee flying in a room?

What would be the effect on rms velocity of gas molecules, if the temperature of the gas is increased by a factor 4 ?

10.

What is the minimum possible temperature on the basis of Charle's law ?

11.

The volume of a given mass of a gas at 27⁰C, 1 atm is 100 cc. What will be its volume at 327⁰C?

12.

Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197g mol^-1.

13.

What is the rms speed of hydrogen gas molecules at STP. Given, density is 0.09 kg m^-3.

14.

What is the value of \(\gamma\) to monoatomic gas ?

Generic 5

2 x 5 = 10

15.

A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity,e.g. does not have uniform density(and pressure). As you might except, its density decreases with height. The precise dependence is given by the so called law of atmosphere.
n₂= n₁ exp [- mg(h₂- h₁)/ k_BT]
Where, n₂ and n₁ refer to number density at heights h₂andh_1,respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column.
\({ n }_{ 2 }={ n }_{ 1 }\quad exp\left[ -mg\quad { N }_{ A }(\rho -{ \rho }^{ ' })({ h }_{ 2 }-{ h }_{ 1 })/(\rho RT) \right] \)
Where, \(\rho \) is the density of the suspended particle and \({ \rho }^{ ' }\), that of surrounding medium.
[\(\because \) N_A is Avogadro's number and R is the universal gas constant.]

16.

An electric bulb of volume 250 cm³ was sealed off during manufacture at a pressure of 10^-3 mm of mercury at 27⁰C. Compute the number of air molecule contained in the bulb.Given that, molecules contained in the bulb. Given that, R = 8.31 J/mol/K N_A= \(6.02\times { 10 }^{ 23 }{ mol }^{ -1 }\).

Generic 3

5 x 3 = 15

17.

Explain,
(i) why there is no atmosphere on moon.
(ii) there is fall in temperature with altitude

18.

Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).

19.

Calculate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2 atm and temperature 17^oC.Take the radius of a nitrogen molecule to be roughly \(1.0\mathring { A } \) Compare the collision time with the time,the molecule moves freely between two successive collisions.Molecular mass of \({ N }_{ 2 }=28.0u\)

20.

The molecules of a given mass of a gas have root mean square speeds of 100 ms^-1 at 27⁰ C and 1.00 atm pressure. What will be the root mean square speeds of the molecules of the gas at 127⁰C and 2.0 atm pressure?

21.

An oxygen cylinder of volume 30 L has an initial gauge pressure of 15 atm and a temperature of 27⁰ C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17⁰ C.Estimate the mass of oxygen taken out of the cylinder (R = 8.31 mol^-1K^-1 , molecular mass of O₂= 32 u).
Whenever masses are taken out of the closed system, no more it is a closed removed should be calculated and then by multiplying with molecular mass of the gas, the same can be converted into mass of the gas removed.

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Answers Key:

Generic 2

14 x 2 = 28

The effect of the various factors on the Brownian motion is as follows

Factors	Effects
(a) Decrease in the size of Brownian particle.	Increase of Brownian motion
(b) Decrease in the density of the medium	Increase of Brownian motion
(c) Increase in temperature of the medium	Increase of Brownian motion
(d) Increase in viscosity of the medium	Decrease of Brownian motion

The law of equipartition of energy for any dynamical system in thermal equilibrium, the total energy is distributed q = equally amongst all the degrees of freedom.
The energy associated with each molecule per degree of freedom is \(\frac { 1 }{ 2 } { k }_{ B }T\), where K_B is Boltzmann's constant and T is temperature of the system.

For 1 mole with f degrees of freedom,
Internal energy, U = 1 x C_vx T = f²/RT
For n moles, U = nC_vT = nf²/RT

As, we know, mean free path,
λ ∝ 1/d²
Given, d₁= 1A^o and d₂ = 2A^o⇒ λ₁ : λ₂=4:1

As force of attraction between molecules disappears, then the molecules will hit the wall with more speeds, hence , F = \(\frac{\Delta p}{\Delta t}\) , where F is average force on the wall due to the molecules.
\(\Delta\) p is change in momentum and \( \Delta \)t is the time duration . Due to increase in \(\Delta\) p, force F will also increase, hence pressure, p = \(\frac{F}{A}\) will increase. Here, A is area of one wall.

The mean free path is inversely proportional to the number density of the gas.

Argon is a monoatomic gas,
\( C_v=\frac{3}{2} R=\frac{3}{2} \times 2=3 \mathrm{calmol}^{-1} K^{-1} \)
\(C_v=M c_v\)
\(M=\frac{C_v}{c_v}=\frac{3}{0.075}=40 \)

Three, because bee is free to move along x-direction or y-direction or z-direction.

As, V_rms ∝ √T
If temperature of the gas is increased 4 times, then v_rms will be doubled.

10.

The minimum possible temperature on the basis of Charles' law is - 273.15⁰C.

11.

Keeping p constant, we have
V²/2 = \(\frac{V_1T_2}{T_1}\)
= \(\frac{100×600}{300}\)
= 200cc

12.

Molar mass of gold is 197 g mol^-1 , the number of atoms
= 6 x 10²³
∴ Number of atoms in 39.4 g
= \(\frac{6.0×10^{23} ×39.4}{197}\)
= 1.2 x 10²³

13.

1.8 x 10³ ms^-1

14.

For monoatomic gas, N =1
The total degree of freedom = 3
\( C_V=\frac{3 R}{2} \)
\( \text { Since } C_P=C_V+R \)
\( C_P=\frac{3 R}{2}+R \)
\( C_P=\frac{5 R}{2}\)
\( Y=\frac{\frac{5 R}{2}}{\frac{3 R}{2}} \)
\( Y=\frac{5}{3}=1.66 \)

Generic 5

2 x 5 = 10

15.

According to the law of atmospheres.
\({ n }_{ 2 }={ n }_{ 1 }\quad exp.\left[ -\frac { mg }{ { K }_{ B }T } \left( h_{ 2 }-{ h }_{ 1 } \right) \right] --\quad (i)\)
where, n₂ and n₁refer to number density of particles at heights h₂ and h_1, respectively.
If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we will have to take effective weight of the suspended particles.
Let, V = average volume of a suspended particle,
\(\rho \) = density of suspended particle, \({ \rho }^{ ' }\)= density of liquid, m = mass of one suspended particle, \({ m }^{ ' }\)= mass of equal volume of liquid displaced.
According to Archimedes' priciple, effective weight of one suspended particle
= Actual weight-weight of liquid displaced = mg-m^'g
\(=mg-V{ \rho }^{ ' }g=mg-\left( \frac { m }{ \rho } \right) { \rho }^{ ' }g=mg\left( 1-\frac { { \rho }^{ ' } }{ \rho } \right) \)
\(\\ Also,\ Boltzmann\ constant,{ K }_{ B }=\frac { R }{ { N }_{ A } } \)
where, R is gas constant and N_A is Avogardro's number.
\(putting,\ mg\left( 1-\frac { { \rho }^{ ' } }{ \rho } \right) in\ place\ of\ mg\ and\ value\ of\ { K }_{ B }\quad in\)
\( Eq.(i)\ we\ get\)
\( { n }_{ 2 }={ n }_{ 1 }exp\left[ -\frac { mg{ N }_{ A } }{ RT } \left( 1-\frac { { \rho }^{ ' } }{ \rho } \right) \left( { h }_{ 2 }-{ h }_{ 1 } \right) \right] ,\ which\ id\ required\ relation.\)

16.

\( V=250 \mathrm{cc}=250 \times 10^{-6} \mathrm{~m}^3 \)
\( \mathrm{P}=10^{-3 \mathrm{~mm}}=10^{-3} \times 10^{-3} \mathrm{~m} \)
\( =\left(10^{-6} \times 13600 \times 10\right) \)
\(=136 \times 10^{-3} \text { Pascal } \)
\( \mathrm{T}=27^0 \mathrm{C}=300 \mathrm{k} \)
\( \mathrm{n}=\frac{\mathrm{PV}}{R T} \)
\( =\frac{136 \times 10^{-3} \times 250 \times 10^{-6}}{8.3 \times 300}=1.36 \times 10^{-8}\)
No. of molecules
\(=1.36 \times 10^{-8} \times 6 \times 10^{23} \)
\(=8.17 \times 10^{15}\)

Generic 3

5 x 3 = 15

17.

(i) The moon has small gravitational; force and hence the escape velocity is small .As the moon is in tyhe proximity of the earth as seen from the sun, the moon has the same amount of heat per unit area as that of the earth , The air molecules have l;arge range of speeds.
Even though the rms speed of the air molecules is smaller than the escape velocity on the moon, a significant number of molecules have speed greater than escape velocity and they escape.
Now, rest of the molecules arrange the speed distribution for the equilibrium temperature. Again, a significant number of molecules escape as their speeds exceed escape sppeed. Hence, over a long time the moon has lost most of its atmosphere.
(ii) As the molecules move higher , their potential energy increases and hence kinetic energy decreases and hence temperature reduces.
At greater height, more volume is available and gas expands and hencde some cooling takes place.

18.

(i) Given, T = 27 \(^0\)C
= ( 273.15 + 27 )
= 300.15K
Average thermal energy , E = \(\frac{3}{2}\)k_BT
( where, kB = Boltzman constant
= 1.38\(\times\)10^-23 JK^-1)
E = \(\frac{3}{2}\)\(\times\)1.38\(\times\)10^-23\(\times\)300.15
= 6.21 \(\times\)10^-21 J
(ii) At the temperatures , T = 107 K
Average thermal energy , E = \( \frac{3}{2}\)k_BT
= \( \frac{3}{2}\)\(\times\)1.38\(\times\)10^-23\(\times\)6000
= 1.241\(\times\)10^-19 J
(iii) At temperature , T = 10⁷K
Average thermal energy,
E = \(\frac{3}{2}\)k_BT
= \(\frac{3}{2}\)\(\times\)1.38\(\times\)10^-23\(\times\)10⁷
= 2.07\(\times\)10^-16 J

19.

500 times

20.

According to ideal gas equation, we get
\(\frac { { p }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\frac { { p }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \)
\(\\ \frac { { V }_{ 1 } }{ { V }_{ 2 } } =\frac { { p }_{ 2 }{ T }_{ 1 } }{ { p }_{ 1 }{ T }_{ 2 } } =\frac { 2\times 300 }{ 1\times 400 } =\frac { 3 }{ 2 } \)
\(\\ { p }_{ 1 }=\frac { 1 }{ 3 } \frac { M }{ { V }_{ 1 } } \left( { v }_{ rms } \right) _{ 1 }^{ 2 },\quad { p }_{ 2 }=\frac { 1 }{ 3 } \frac { M }{ { V }_{ 2 } } \left( { v }_{ rms } \right) _{ 2 }^{ 2 }\)
\(\\ \left( { v }_{ rms } \right) _{ 2 }^{ 2 }=\left( { v }_{ rms } \right) _{ 1 }^{ 2 }\times \frac { { V }_{ 2 } }{ { V }_{ 1 } } \times \frac { { p }_{ 2 } }{ { p }_{ 1 } }\)
\( \\=\left( 100 \right) ^{ 2 }\times \frac { 2 }{ 3 } \times 2\)
\(\\ or\ \left( { v }_{ rms } \right) _{ 2 }=\frac { 200 }{ \sqrt { 3 } } \ { ms }^{ -1 }\)

21.

Given, absolute pressure. P₁ = (15 + 1) atm
\(\left[ \because \ absolute\ pressure\ =gauge\ pressure+1atm \right]\)
\( \\ =16\times 1.013\times { 10 }^{ 5 }{ P }_{ a }\)
\(\\ { V }_{ 1 }=30\ L=30\times { 10 }^{ -3 }{ m }^{ 3 }\)
\(\\ { T }_{ 1 }=273.15+27=300.15\ K\)
\(\\ Using\ ideal\ gas\ equation,\ pV=nRT\)
\(\\ or\ n=\frac { pV }{ RT } =\frac { { p }_{ 1 }{ V }_{ 1 } }{ { RT }_{ 1 } } =\frac { 16\times 1.013\times { 10 }^{ 5 }\times 30\times { 10 }^{ -3 } }{ 8.314\times 300.15 } \)
\(=19.48\)
\(\\ Final\ { p }_{ 2 }=(11+1)=12\ atm=12\times 1.013\times { 10 }^{ 5 }Pa\)
\(\\ { V }_{ 2 }=30\ L=30\times { 10 }^{ -3 }m^{ 3 }\)
\(\\ { T }_{ 2 }=273.15+17=290.15\ k\)
\(\\ Number\ of\ moles\)
\(\\ =\frac { { p }_{ 2 }{ V }_{ 2 } }{ { RT }_{ 2 } } =\frac { 12\times 1.013\times { 10 }^{ 5 }\times 30\times { 10 }^{ -3 } }{ 8.314\times 290.15 } \)
\(\\ Hence,moles\ removed\ =19.48-15.12=4.36\)
\(\\ Mass\ removed=4.36\times 32\ g\)
\(\\ =139.52\ g=0.1395\ kg\)