NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Physics
Chapter : 13
Chapter Name : Kinetic Theory

Q13.1 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 3Å.

Diameter of an oxygen molecule, d = 3A

d 3 ^ −8
= = = 1.5A = 1.5 × 10 cm
2 2

3
Actual volume occupied by 1 mole of oxygen gas at STP = 22400cm

4 3
V = πr ⋅ N
3
23
Where, N is Avogadro's number = 6.023 × 10 molecules/mole

3
4 −8 23 3
∴ V = × 3.14 × (1.5 × 10 ) × 6.023 × 10 = 8.51cm
3

8.51
Ratio of the molecular volume to the actual volume of oxygen =
22400

−4
= 3.8 × 10

Page : 338 , Block Name : Exercise

Q13.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature
and pressure (STP : 1 atmospheric pressure, O C). Show that it is 22.4 litres.
∘

The ideal gas equation relating pressure (P), volume (V ), and absolute temperature (T)

is given as:

P V = nRT

−1 −1
R is the universal gas constant = 8.314Jmol K

n = Number of moles = 1

T = Standard temperature = 273K

5 −2
P = Standard pressure = 1atm = 1.013 × 10 Nm
nRT
∴ V =
P

1×8.314×273
=
5
1.013×10

3
= 0.0224m

= 22.4 litres

Page : 338 , Block Name : Exercise

Q13.3
−3
Figure 13.8 shows plot of P V /T versus For 1.00 × 10 kg of oxygen gas at two different

temperatures.

(a) does the dotted plot signify?
(b) Which is true. Tl > T2 or Tl < 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 xlO-3 kg of hydrogen, would we get the same value of
PV/T at the point where the curves meet on the v-axis? If not, what mass of
hydrogen yields the same value of p V/ T (for pressure high temperature region of the plot)?
(Molecular mass of H = 2.02u, of O = 32.0u, R = 8.31Jmol K )
−1 −1
2 2

(a) The dotted plot in the graph signifies the ideal behaviour of the gas, l.e., the ratio

PV
is equal. μR(μ is the number of moles and R is the universal gas constant) is a
T

constant quality. It is not dependent on the pressure of the gas.

(b) The dotted plot in the given graph represents an ideal gas. The curve of the gas at
temperature Tl is closer to the dotted plot than the curve of the gas at temperature T . A real gas
2

approaches the behaviour of an ideal gas when its temperature increases. Therefore, T > T is
1 2

true for the given plot.

(c) The value of the ratio P V /T , where the two curves meet, is μR . This is because the

ideal gas equation is given as:

P V = μRT

PV
= μR
T

P is the pressure
T is the temperature
V is the volume
p is the number of moles
R is the universal constant
Molecular mass Of oxygen = 32.0 g

−3
Mass of oxygen = 1 × 10 kg = 1g

−1 −1
R = 8.314mole K

PV 1
= × 8.314
T 32

−1
= 0.26JK

Therefore, the value of the ratio PV/T , where the curves meet on the y -axis, is

−1
0.26JK

(d) If We obtain similar plots for 1.00 × 10 kg Of hydrogen, then we will not get the same
−3

value of PV/T at the point where the curves meet the V axis. This is because the molecular mass
of hydrogen (2.02 u) is different from that of oxygen (32.0 u) We have:
PY −1
= 0.26JK
T

−1 −1
R = 8.314Jmole K

Molecular mass (M ) of H2 = 2.02u

PV
= μR at constant temperature
T

m
Where, μ =
M

m = Mass of H2
PV M
m = ×
T R

0.26×2.02
=
8.31

−2 −5
= 6.3 × 10 g = 6.3 × 10 kg

−5
Hence, 6.3 × 10 kg of H2 will yield the same value of P V /T

Page : 338 , Block Name : Exercise

Q13.4 An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a
temperature of 27 oc. After some oxygen is withdrawn from the cylinder, the gauge pressure
drops to 11 atm and its temperature drops to 17 oc. Estimate the mass of oxygen taken out of the
cylinder (R = 8.31 J mole K , molecular mass of 02 = 32 u).
−1 −1

−3 3
Volume of oxygen, V1 = 30 litres = 30 × 10 m

5
Gauge pressure, P1 = 15atm = 15 × 1.013 × 10 Pa
∘
Temperature, T1 = 27 C = 300K

−1 −1
Universal gas constant, R = 8.314Jmole K

Let the initial number of moles of oxygen gas in the cylinder be n1

The gas equation is given as:

P1 V1 = n1 RT1
P V1
∴ n1 =
RT1

5 −3
15.195×10 ×30×10
= = 18.276
(8.314)×300

m1 = Initial mass of oxygen

M = Molecular mass of oxygen = 32g

∴ m1 = n1 M = 18.276 × 32 = 584.84g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces
−3 3
Volume, V2 = 30 litres = 30 × 10 m

5
Gauge pressure, P2 = 11 atm = 11 × 1.013 × 10 Pa

∘
Temperature, T2 = 17 C = 290K

P2 V2 = n2 RT2

P2 V2
∴ n2 =
RT2

5 −3
11.143×10 ×30×10
= = 13.86
8.314×290
m2
n2 =
M

m2 is the mass of oxygen remaining in the cylinder

∴ m2 = r2 M = 13.86 × 32 = 453.1g

The mass of oxygen taken out of the cylinder is given by the relation:

Initial mass of oxygen in the cylinder - Final mass of oxygen in the cylinder

= m1 − m2

= 584.84g − 453.1g

= 131.74g

= 0.131kg

Therefore, 0.131kg of oxygen is taken out of the cylinder.

Page : 339 , Block Name : Exercise

Q13.5 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 ?

3 −6 3
Volume of the air bubble, V1 = 1.0cm = 1.0 × 10 m

Bubble rises to height, d = 40m

∘
Temperature at a depth of 40m, T1 = 12 C = 285K

∘
Temperature at the surface of the lake, T2 = 35 C = 308K

The pressure on the surface of the lake:

5
P2 = 1 atm = 1 × 1.013 × 10 Pa

The pressure at the depth of 40m :

P1 = 1atm + dρg

3 3
ρ is the density of water = 10 kg/m

2
∴ P1 is the acceleration due to gravity = 9.8m/s

5 3
∴ P1 = 1.013 × 10 + 40 × 10 × 9.8 = 493300Pa
P1 V1 P2 V2
=
T1 T2

Where, V is the volume Of the air bubble when it reaches the surface
2
P1 V1 T2
V2 =
T1 P2

−6
(493300)(1.0×10 )308
=
5
285×1.013×10

−6 3 3
= 5.263 × 10 m or 5.263cm

Therefore, where, the air bubble reaches the surface, its Volume becomes 5.263 crn³.

Page : 339 , Block Name : Exercise

Q13.6 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.

3
Volume of the room, V = 25.0m
∘
Temperature of the room, T = 27 C = 300K

5
Pressure in the room, P = 1atm = 1 × 1.013 × 10 Pa

The ideal gas equation relating pressure (p), Volume (V), and absolute temperature (T) can be
written as:
P V = ks N T

−23 2 −2 −1
KB is Boltzmann constant = 1.38 × 10 m kgs K
−23 2 −2 −1
KB is Boltzmann constant = 1.38 × 10 m kgs K

N is the number of air molecules in the room
PV
N =
kB T

5
1.013×10 ×25 26
= = 6.11 × 10 molecules
−23
1.38×10 ×300

26
Therefore, the total number of air molecules in the given room is 6.11 × 10 .

Page : 339 , Block Name : Exercise

Q13.7 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).

∘
(i) At room temperature, T = 27 C = 300K

3
Average thermal energy = kT
2

−23 2 −2 −1
Where k is Boltzmann constant = 1.38 × 10 m kgs K
3 3 −18
∴ kT = × 1.38 × 10 × 300
2 2

−21
= 6.21 × 10 J
∘
Hence, the average thermal energy of a helium atom at room temperature (27 C) is

−21
6.21 × 10 J

(ii) On the surface of the sun, T = 6000K

3
Average thermal energy = kT
2

3 −38
= × 1.38 × 10 × 6000
2

−19
= 1.241 × 10 J
7
(iii) At temperature, T = 10 K

3
Average thermal energy = kT
2
3 −23 7
= × 1.38 × 10 × 10
2

−16
= 2.07 × 10 J

−16
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10

J.

Page : 339 , Block Name : Exercise

Q13.8 Three vessels Of equal capacity have gases at the same temperature and pressure. The rst
vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third
contains uranium hexa uoride (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 the largest ?

Answer. Yes. All contain the same number of the respective molecules. The root mean square
speed Of neon is the largest. Since the three vessels have the same capacity, they have the same
volume. Hence, 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. This
number is equal to Avogadro's number,
The root mean square speed (vrm,) of a gas of mass m, and temperature T, is
23
N = 6.023 × 10

given by the relation:
3kT
vrms = √
m

Where, k is Boltzmann constant
For the given gases, k and T are constants.
Hence v depends only on the mass of the atorns, i.e. ,

1
vrms ∝ √
m

Therefore, the root mean square speed of the molecules in the three cases is not the same.
Among neon, chlorine, and uranium hexa uoride, the mass of neon is the
smallest. Hence, neon has the largest root mean square speed among the given gases.

Page : 339 , Block Name : Exercise

Q13.9 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).

∘
Temperature of the helium atom, THe = −20 C = 253K

Atomic mass of argon, MAr = 39.9u

Atomic mass of helium, MHe = 4.0u

Let, (v rms ) be the rms speed of argon.
ar

Let (V rms ) be the rms speed of helium.
He

The rms speed of argon is given by:

The rms speed of argon is given by:

3RTAr
(vmat ) = √
Nx MN r

R is the universal gas constant

T Ar is temperature of argon gas

The rms speed of helium is given by:

3RTHe
(vms ) = √
He MHe

3RTN 3RTHe
√ = √
MN MHe

253
= × 39.9
4

3
= 2523.675 = 2.52 × 10 K

3
Therefore, the temperature of the argon atom is 2.52 × 10 K .

Page : 339 , Block Name : Exercise

Q13.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder
containing nitrogen at 2.0 atm and temperature 17 0C. 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 N = 28.0 u).
2

−7
Mean free path = 1.11 × 10 m

9 −1
Collision frequency = 4.58 × 10 s

Successive collision time ≈ 500 × ( Collision time )
5
Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × 10 Pa
∘
Temperature inside the cylinder, T = 17 C = 290K

10
Radius of a nitrogen molecule, r = 1.0A = 1 × 10 m
10 10
Diameter, d = 2 × 1 × 10 = 2 × 10 m

−3
Molecular mass of nitrogen, M = 28.0g = 28 × 10 kg

The root mean square speed of nitrogen is given by the relation:

3RT
vm= √
M

−1 −1
R is the universal gas constant = 8.314 J mole K

3×8.314×290
∴ vms = √ = 508.26m/s
−3
28×10

∴ l the mean free path (I ) is given by the relation:

kT
l =
√2×d 2 ×P

−23 2 −2 −1
k is the Boltzmann constant = 1.38 × 10 kgm s K
−23
1.38×10 ×290
∴ I =
−10 2 5
√2×3.14×(2×10 ) ×2.026×10

−7
= 1.11 × 10 m
v rme
Collision frequency =
l

508.26 9 −1
= −7
= 4.58 × 10 s
1.11×10

Collision time is given as:
d
T =
vmin

−10
2×10 −13
= = 3.93 × 10 s
508.26

Time taken between successive collisions:

′ l
T =
v mys

−7
1.11×10 m −10
= = 2.18 × 10 s
508.26m/s

′ −10
T 2.18×10
∴ = −13
= 500
T 3.93×10

Hence, the time taken between successive collisions is 500 times the time taken for a collision.

Page : 339 , Block Name : Exercise

Q13.11 A metre long narrow bore held horizontally (and closed at one and) 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 ?

Length of the narrow bore, L = 1m = 100cm

Length of the mercury thread, / = 76cm

Length of the air column between mercury and the closed end, la = 15cm

since the bore is held vertically in air with the open end at the bottom, the mercury

length that occupies the air space is: 100 − (76 + 15) = 9cm

Hence, the total length of the air column = 15 + 9 = 24cm

Let hcm of mercury flow out as a result of atmospheric pressure.

∴ Length of the air column in the bore = 24 + hcm

And, length of the mercury column = 76 − hcm

Initial pressure, P1 = 76cm of mercury

3
Initial volume, V1 = 15cm

Final pressure, P2 = 76 − (76 − h) = hcm of mercury

3
Final volume, V2 = (24 + h)cm

Temperature remains constant throughout the process.

∴ P1 V 1 = P2 V 2

76 × 15 = h(24 + h)

2
h + 24h − 1140 = 0
2
−24±√(24) +4×1×1140
∴ h =
2×1

= 23.8cm or − 47.8cm

Height cannot be negative. Hence, 23.8 cm of mercury will ow out from the bore and
522 cm of mercury will remain in it. The length of the air column Will be 24 + 23.8 — 47.8 cm

Page : 340 , Block Name : Additional Exercise

Q13.12 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 ?

3 −1
Rate of diffusion of hydrogen, R1 = 28.7cm s

3 −1
Rate of diffusion of another gas, R2 = 7.2cm s

According to Graham's Law of diffusion, we have:

R1 M2
= √
R2 M1

M1 is the molecular mass of hydrogen = 2.020g

M2 is the molecular mass of the unknown gas

2
R1
∴ M2 = M1 ( )
R2

2
28.7
= 2.02( )
7.2

32g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

Page : 340 , Block Name : Additional Exercise

Q13.13 A gas in equilibrium has uniform density and pressure throughout its volume. This is
strictly true only if there are no external in uences. 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 ) /kB T ]

Where n n , refer to number density at heights h and hl respectively. use this relation to derive
2, 1
2

the equation for sedimentation equilibrium of a suspension in a liquid column:
n = n exp[−mgN (ρ − P ) (h − h ) /(ρRT )]Where p is the density of the suspended
′
2 1 A 2 1

particle, and p' that of surrounding medium. CNA s Avogadro's number, and R the universal gas
constant.] [Hint: use Archimedes principle to nd the apparent weight of the suspended
particle.]

According to the law of atmospheres, we have:

n2 = n1 exp[−mg (h2 − h1 ) /kBT ] … (i)

n1 is the number density at height h11 , and n2 is the number density at height h2

mg is the weight of the particle suspended in the gas column
′
Density of the medium = ρ

Density of the suspended particle = ρ

′
Mass of one suspended particle = m

Mass of the medium displaced = m

Volume of a suspended particle = v

According to Archimedes' principle for a particle suspended in a liquid column,

effective weight of the suspended particle is given as:

Weight of the medium displaced- Weight of the suspended particle
′
= mg − m g

m
′ ′
= mg − V ρ g = mg − ( )ρ g
ρ
′
ρ
=mg (1 − )
ρ

Gas constant, R = kB N

R
kB = … (iii)
N

n2 = n1 exp[−mg (h2 − h1 ) /kB T ]
′
ρ N
mg(1 − )
ρ RT
(h2 −h1 )

N
′
mg(ρ−ρ ) RT ρ
= n1 exp[− (h2 − h1 ) ]

Page : 340 , Block Name : Additional Exercise

Q13.14 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].

Atomic mass of a substance = M

Density of the substance = ρ

23
Avogadro's number = N = 6.023 × 10

4 3
Volume of each atom = πr
3
4 3
Volume of N number of molecules = πr
3

M
Volume of one mole of a substance = … (ii)
ρ

4 3
πr
3

3
3M
∴ r = √
4πρN

For carbon:

−3
M =12.01×10 kg

3 −3
ρ = 2.22 × 10 kgm
1
−3 3
3×12.01×10
∴ r = ( ) = 1.29A
3 23
4π×2.22×10 ×6.023×10

Hence, the radius of a carbon atom is 1.29 A.

For Gold
−3
M = 197.00 × 10 kg

3 −3
ρ = 19.32 × 10 kgm
1
−3 3
3 × 12.01 × 10
∴ r = ( ) = 1.29A
3 23
4π × 2.22 × 10 × 6.023 × 10

For liquid nitrogen :
−3
M = 14.01 × 10 kg

3 −3
ρ = 1.00 × 10 kgm
1
−3 3
3×14.01×10
∴ r = ( 3 25
) = 1.77A
4π×1.00×10 ×6.23×10

For lithium :
−3
M = 6.94 × 10 kg

3 −3
ρ = 0.53 × 10 kgm
1
−3 3
3×6.94×10
∴ r = ( 3 23
) = 1.73A
4π×0.53×10 ×6.23×10

For liquid Fluorine :
−3
M = 19.00 × 10 kg

3 −3
ρ = 1.14 × 10 kgm
1
−3 3
3×19×10
∴ r = ( ) = 1.88A
3 23
4π×1.14×10 ×6.023×10

Page : 340 , Block Name : Additional Exercise