Kinetic Theory
Class 11th Physics Part II CBSE Solution
Exercise
Question 1.Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
Answer:
Given:
Diameter of an oxygen molecule (D) = 3 Å
∴ the radius of an oxygen molecule (r) = 1.5 Å (∵ r = D/2)
Fraction = ?
Explanation:We know that the volume occupied by 1 mole of oxygen gas at STP =22.4 litres. The molecular volume occupied 1 mole of oxygen molecules can be found by multiplying the volume of a single oxygen molecule (considering it to be a sphere) with the
Avogadro's constant. Then dividing the latter by the former the required result is obtained.
Volume of 1 mole of oxygen gas at S.T.P(Vmolecular)
Vmolecular = 
πr3 × NA ---(1)
The actual volume occupied by oxygen gas at STP (Vactual)
Vactual = 22.4 litres = 0.0224 m3
Vactual = 0.0224 m3 ---(2)
(Where NA = 6.02214086 × 1023 mol-1 is Avogadro's constant.)
Substituting for ‘r’ and ‘NA’ in equation (1) we get
Vmolecular = πr3 × NA = π(1.5 × 10-10)3 × ( 6.02214086 × 1023)
= 0.000008513≈ 8.51 × 10-6 m3
(∵ 1 Å = 1 × 10-10 m)
∴ Vmolecular = 8.52 × 10-6 m3 ---(3)
Fraction = 
= 
= 0.00038007 ≈ 3.8 × 10-4
Fraction = 3.8 × 10-4
The required ratio/fraction is3.8 × 10-4.
Question 2.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.
Answer:
Given:
P = 1 atm = 101 kPa
T = 273.15 K
n = 1 mole
V = ?
Explanation:We know that the volume occupied by 1 mole of oxygen gas at STP =22.4 litres. By using the Ideal Gas Law (PV = nRT) we can obtain the required result.
Wkt…
PV = nRT ---( Ideal Gas Law )
∴ V = 
⇒ V = 
≈ 0.0224 m3 = 22.4 litres
⇒ V = 22.4 litres
The required volume is22.4 litres.
Question 3.
Figure 13.8 shows plot of PV/T versus P for 1.00 × 10–3 kg of oxygen gas at two different temperatures.
(a) What does the dotted plot signify?
(b) Which is true: T1 > T2 or T1 < T2?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00 × 10–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.)
Answer:
(a) Seeing that the dotted line doesn’t vary with pressure and by using theIdeal Gas Law (PV = nRT ) we can say that it implies the ideal behaviour of Gas.
Wkt…
PV = nRT
⇒ nR = PV/T = constant
The dotted plot impliesPV/T = constant.
(b) T2is heavily deviates from ideal behaviour than T1.
⇒ T1 > T2 is true.
T1>T2is true.
(c) Using PV = nRT
⇒ PV/T = nR
Given: Mass of the gas = 1 × 10-3 kg = 1 g
Molecular mass of Oxygen = 32g/mole
So, number of moles = given weight/molecular weight
⇒ number of moles = 1/32
So, nR = 1/32 × 8.314 J / mol /K = 0.26 J/K
Hence, value of PV/T = 0.26 J/K.
(d) 1g of Hydrogen doesn't represent the same number of mole. molecular mass of H2 = 2g/mol
Hence, number of moles of Hydrogen required is 1/32
So, mass of H2 required = no of moles of H2 × molecular mass of H2
⇒ mass of H2 required = 1/32mole × 2 g/mol = 1/16 g = 0.0625 g
⇒Hence themass of H2 required = 0.0625 g
Question 4.
An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K-1, molecular mass of O2 = 32 u).
Answer:
Explanation: We have to consider 2 cases here and find the masses of O2in each case and then by calculating the difference of those masses gives thethe mass of oxygen taken out of the cylinder.
Given:
Initially, Case1
V1 = 30 litres = 0.03 m3
P1 = 15 atm = 15 × 1.01 × 105 Pa
T1 = 27+273 = 300 K
If n1 is the moles O2 in the cylinder, then by using
P1V1 = n1RT1 ---( Ideal Gas Law )
⇒ n1 = 
= 
⇒ n1 = 18.276 moles
Molecular weight of O2 (M) = 32 g
Initial mass of cylinder (m1) = n1M
⇒ m1 = n1M
⇒ m1 = (18.276 moles) × ( 32 g)
⇒ m1 = 584.84 g
Now,
Let n2 be the moles of O2 left in the cylinder
V2 = 30 litres = 0.03 m3
P2 = 11 atm = 11 × 1.01 × 105 Pa
T2 = 17+273 = 290 K
Thus n2 = 
= 
⇒ n1 = 13.86 moles
Therefore, m2 be the final mass of O2 in the cylinder
⇒ m2 = n2M
⇒ m2 = (13.86 moles) × ( 32 g)
⇒ m2 = 453.1 g
∴ the mass of oxygen taken out of the cylinder
Δm = m1-m2
⇒ Δm = 584.84 g - 453.1 g
⇒ Δm = 131.74 g
The mass of oxygen taken out of the cylinder is 131.74 g.
Question 5.
An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?
Answer:
This problem can by solved using the fact that ‘the number of moles/molecules of the air bubble remains unchanged as the bubbles rises from bottom to the surface’.
Given:
Volume of the air bubble (V1) = 1.0 cm3 = 1.0 × 10 –6 m3
Bubble rises to height (h) = 40 m
Temperature at a depth of 40 m (T1) = 12°C = 285 K
Temperature at the surface (T2) = 35°C = 308 K
The pressure on the surface (P2) = 1 atm = 1 × 1.01 × 105 Pa
The pressure at the depth of 40 m (P1) = 1 atm + ρgh
(Where, ρ is the density of water = 103 kg/m3
g is the acceleration due to gravity = 9.8 m/s2 )
∴ P1 = (1.013 × 10 5 pa)+ (40 m × 10 3 kg/m3 × 9.8 m/s2 )
⇒ P1 = 493300 Pa
We have 
= 
( = number of moles/molecules of the air bubble)
(Where, V2 is the volume of the air bubble when it reaches the surface )
∴ V2 = 
= 
⇒ V2 = 5.263 × 10 –6 m3
⇒ V2 = 5.263 cm3
When the air bubble reaches the surface, its volume is 5.263 cm3
Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.
Answer:
We have to consider the Ideal Gas Law with theBoltzmann constant to solve this problem.
Given:
Volume of the room (V) = 25 m3
Temperature of the room (T) = 27°C = 300 K
Pressure in the room (P) = 1 atm = 1 × 1.013 × 105 Pa
The ideal gas equation also be written as
PV = kBNT
(Where, KB is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1 and N is the number of air molecules in the room)
∴ N = 
= 
⇒ N = 6.11 × 10 26 molecules
The total number of air molecules in the given room is 6.11 × 1026
Question 7.
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).
Answer:
This problem can be solved using the relation between kinetic energy and temperature i.e. averagethermal energy = (3/2)kBTper atom.
(i) Average thermal energy of a helium atom at room temperature (27 °C)
At room temperature (T) = 27°C = 300 K
Average thermal energy (ETh) = (3/2)kBT
(Where kB is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1 )
∴ ETh = (3/2)kBT
⇒ ETh = (3/2) 1.38 × 10–23 × 300
⇒ ETh = 6.21 × 10–21J
The average thermal energy of a helium atom at room temperature is 6.21 × 10–21 J.
(ii) The surface of the sun has T = 6000 K
∴ ETh = (3/2)kBT
⇒ ETh = (3/2) × 1.38 × 10-23 × 6000
⇒ ETh = 1.241 × 10 -19 J
Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19 J.
(iii) The temperature of 10 million
kelvin Average thermal energy = (3/2)kBT
⇒ ETh = (3/2) × 1.38 × 10-23 × 107
⇒ ETh = 2.07 × 10-16 J
Hence, the average thermal energy of a helium atom(the typical core temperature in the case of a star)is 2.07 × 10–16 J.
Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?
Answer:
All the three vessels have the same capacity, which implies they have the same volume. Thus, each gas has the same pressure, volume, and temperature. According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules.
Which is equal to Avogadro’s number, NA = 6.023 × 1023.
The root mean square speed (Vrms) of a gas of mass m, and temperature T, is given by the relation:
mVrms2 = 
kBT
⇒ Vrms = 
where, kB is Boltzmann constant For the given gases, k and T are constants. Hence Vrms depends only on the mass of the atoms, i.e., Vrms∝ 
The root mean square speed of the molecules in the vessels is not the same. The mass of neon is the smallest. Hence, neon has the largest root mean square speed among the given gases.
Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer:
All the three vessels have the same capacity, which implies they all have the same volume. Thus, each gas has the same pressure, volume, and temperature. According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules.
Given:
Temperature of the helium atom(Th)
Th = –20°C = 253 K
Atomic mass of argon(Ma) = 39.9 u
Atomic mass of helium(Mh) = 4u
Let, Va be the rms speed of argon and let Vh be the rms speed of helium.
Since, MV2 = kBT (let kB be simply k)
∴ V = 
The rms speed of argon is given by-
Va = 
---(1)
The rms speed of helium is given by-
Vh = 
---(2)
∴ since Va = Vh
⇒ =
⇒ 
= 
⇒ Ta = Ma
⇒ Ta = 
× 39.9 u
⇒ Ta = 2.52 × 103 K
Therefore, the temperature of the argon atom is 2.52 × 103 K.
Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).
Answer:
First we have to find the rms speed of the nitrogen molecule, then by finding the collision time and the time the between two successive collisions, and comparing the two by dividing the latter by former we get the required result.
Given:
Mean free path = 1.11 × 10–7 m
Collision frequency = 4.58 × 109 s–1
Pressure inside the cylinder containing nitrogen (P)
P = 2 atm = 2.026 × 105 Pa
Temperature inside the cylinder (T) = 17°C = 290 K
Radius of a nitrogen molecule (r) = 1 Å = 1 × 1010 m
Diameter (d) = 2r
d = 2 × 1010 m
Molecular mass of nitrogen (M) = 28 g
M = 28 × 10–3 kg
The root mean square speed of nitrogen is given by
Vrms = 
⇒ Vrms = 
⇒ Vrms = 508.26 m/s
Let the mean free path be L, given as
L = 
⇒ L = 
⇒ L = 1.11 × 10-7 m
Collision frequency (f) = Vrms /L
⇒ f = 508.26 / (1.11 × 10-7 )
⇒ f = 4.58 × 109 s-1
The Collision time is given as (T)
T = d / Vrms
⇒ T = (2 × 10 -10 ) / (508.26)
⇒ T = 3.93 × 10-13 s
Time taken between successive collisions be Tc
Tc = l / Vrms
⇒ Tc = (1.11 × 10-7 ) / (508.26)
⇒ Tc = 2.18 × 10-10 s
∴ Tc / T = (2.18 × 10-10) / (3.93 × 10-13)
⇒ Tc / T = 500
Hence, the time taken between successive collisions is 500 times the time taken for a collision.
Question 11.
A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Answer:
If the temperature is constant, then the Boyle’s law state that,
P1 V1 = P2 V2
Where,
‘P1’ is the initial pressure.
‘V1’ is the initial pressure.
‘P2’ is the final pressure.
‘V2’ is the final pressure.
Given,
The length of the tube = 1 m = 100 cm
The length of the mercury thread = 76 cm
The length of the air column = 15 cm
Therefore,
The length of 9 cm is left in the open end when the tube is kept in a horizontal position. We suppose ‘a’ cm2 as the cross section area of the tube.
⇒ When the tube is held horizontally,
Initial pressure, P1 = 76 cm (of mercury)
Initial Volume, V1 = 15 a cm3
⇒ When the tube is held vertically,
The length of the air column = 15 cm + 9 cm = 24 cm
(The two air column on the both sides of the mercury thread adds up)
The length of the mercury thread = 76 cm
(at the open end of the tube)
Since, the pressure of the mercury and the air adds up to be greater than the atmospheric pressure. Some mercury flows out of the open end until the total pressure inside reaches the atmospheric pressure.
Suppose, The length of the mercury thread that is reduced = ‘h’ cm
The length of the air column now = ( 24 + h ) cm
The length of the mercury thread now = ( 76 – h ) cm
The new final pressure, P2 = 76 cm – (76 – h ) cm
⇒ P2 = h cm ( of mercury)
The new final Volume, V2 = (24 + h) a cm3
By Boyle’s law, we have
P1V1 = P2V2
⇒ 76 × 15 a = h × (24 + h) a
⇒ 1140 = 24h + h2
⇒ h2 + 24h – 1140 = 0
Solving the quadratic equation we get,
⇒ 
cm
⇒ 
cm
⇒ 
⇒ h = 23.83 cm or h = -47.83
Since, the height cannot be negative, h = 23.83 cm
Thus,if the tube is held vertically with the open end at the bottom the mercury will flow out of the tube to equalize the pressure with the atmospheric pressure and the thread of mercury will reduce by h = 23.83 cm.
Question 12.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s–1. Identify the gas.
Answer:
The Graham’s Law of Diffusion states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. We have,
Where,
M is the molecular mass of the first gas,
M’ is the molecular mass of the second gas
R is the diffusion/ effusion rate of the first gas,
R’ is the diffusion/ effusion rate of the second gas
Given,
Molecular mass of hydrogen, M = 2.020 g
Rate of diffusion of hydrogen, R = 28.7 cm3 /s
Rate of diffusion of unknown gas, R’ = 7.2 cm3 /s
Thus, From the Graham’s law of diffusion, we get:
⇒ 
⇒ 
⇒ M’ = 32.09 g
32 g is the molecular mass of Oxygen .
The unknown gas is identified to be Oxygen.
Question 13.
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, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
n2 = n1 exp [ -mg (h2 – h1)/ kBT]
Where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n2 = n1 exp [ -mg NA(ρ - P′) (h2 – h1)/ (ρ RT)]
Where ρ is the density of the suspended particle, and ρ’ that of surrounding medium.
[NA is Avogadro’s number, and R the universal gas constant.]
Answer:
Let us suppose:
The density of the suspended particle is ρ
The density of the medium is ρ’
Mass of the medium which is displaced by the suspended particle is m
Mass of the suspended particle is m’
Volume of the suspended particle is V
Since, the equation for sedimentation equilibrium of a suspension in a liquid column is asked we will use the Archimedes’ principle.
The Archimedes’ Principle for a suspended particle in a liquid column, the effective weight W’ of the suspended particle in the liquid is given as:
W’ = Weight of the displaced medium – Weight of the suspended particle
⇒ W’ = mg – m’g
⇒ W’ = mg – Vρ’g = 
⇒ 
……..(i)
And we have,
Gas Constant, R = kBN
⇒ kB = 
………………..(ii)
Given,
The law of atmosphere states that:
n2 = n1 exp [ -mg (h2 – h1)/ kBT] ……………(iii)
where,
n1 is the number density at the height ‘h1’
n2 is the number density at the height ‘h2’
‘mg’ is the apparent weight of the particle suspended in the gas column.
Replacing the values from equation (i) and (ii) in equation (iii) we get:
⇒ 
(h2 – h1) 
⇒
(h2 – h1) 
This is the equation for sedimentation equilibrium of a suspension in a liquid column.
Question 14.
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].
Answer:
Let us suppose that the atoms are ‘tightly packed’ in a solid or liquid phase and
The Density of the substance = D
Atomic mass of the substance = M
Avogadro’s Number, N = 6.022 × 1023
Radius of an atom of the substance = r
Volume of an atom = 
Therefore, the volume of the N molecules = 
……(1)
Also, the volume of one mole of the substance = 
…….(2)
Equating the equation (1) and (2) we get :
=
⇒ 
Thus,
(i) Carbon
Given,
Atomic mass of carbon, M = 12.01× 10-3 kg
The Density of the carbon, D = 2.22× 103 kg / m3
⇒ 
⇒ r = 1.29× 10-10 m = 1.29 Å
The radius of the carbon atom is 1.29Å.
(ii) Gold
Given,
Atomic mass of Gold atom, M = 197.0× 10-3 kg
The Density of the Gold atom, D = 19.32× 103 kg / m3
⇒ 
⇒ r = 1.59× 10-10 m = 1.59 Å
The radius of the Gold atom is 1.59Å.
(iii) Liquid Nitrogen
Given,
Atomic mass of Liquid nitrogen, M = 14.01× 10-3 kg
The Density of the Liquid nitrogen,D = 1.0× 103 kg / m3
⇒ 
⇒ r = 1.77× 10-10 m = 1.77 Å
The radius of the Liquid nitrogen atom is 1.77 Å.
(iv) Lithium
Given,
Atomic mass of Lithium, M = 6.94× 10-3 kg
The Density of the Lithium, D = 0.53× 103 kg / m3
⇒ 
⇒ r = 1.73× 10-10 m = 1.73 Å
The radius of the Lithium atom is 1.73 Å.
(v) Liquid Fluorine
Given,
Atomic mass of Liquid flourine, M = 19.0× 10-3 kg
The Density of the Liquid flourine,D = 1.14× 103 kg / m3
⇒ 
⇒ r = 1.88× 10-10 m = 1.88 Å
The radius of the Liquid flourine atom is 1.88 Å.
The Rough estimates of the size of the atoms thus are:
