NCERT Solutions Class 11th Physics. FREE Flip-BOOK by Study Innovations. 461 Pages

The P-T phase diagram for CO2 is shown in the following figure.

At 1 atm pressure and at –60°C, CO2 lies to the left of –56.6°C (triple point C). Hence, it
lies in the region of vaporous and solid phases.
Thus, CO2 condenses into the solid state directly, without going through the liquid state.
At 4 atm pressure, CO2 lies below 5.11 atm (triple point C). Hence, it lies in the region of
vaporous and solid phases. Thus, it condenses into the solid state directly, without passing
through the liquid state.
When the temperature of a mass of solid CO2 (at 10 atm pressure and at –65°C) is
increased, it changes to the liquid phase and then to the vaporous phase. It forms a line
parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the
intersection point where this parallel line cuts the fusion and vaporisation curves.
If CO2 is heated to 70°C and compressed isothermally, then it will not exhibit any
transition to the liquid state. This is because 70°C is higher than the critical temperature
of CO2. It will remain in the vapour state, but will depart from its ideal behaviour as
pressure increases.

Question 11.18:
A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers
fever) which causes an increase in the rate of evaporation of sweat from his body. If the
fever is brought down to 98 °F in 20 min, what is the average rate of extra evaporation
caused, by the drug? Assume the evaporation mechanism to be the only way by which
heat is lost. The mass of the child is 30 kg. The specific heat of human body is
approximately the same as that of water, and latent heat of evaporation of water at that
temperature is about 580 cal g–1.

Answer

Initial temperature of the body of the child, T1 = 101°F
Final temperature of the body of the child, T2 = 98°F

Change in temperature, ΔT = °C

Time taken to reduce the temperature, t = 20 min

Mass of the child, m = 30 kg = 30 × 103 g

Specific heat of the human body = Specific heat of water = c

= 1000 cal/kg/ °C

Latent heat of evaporation of water, L = 580 cal g–1

The heat lost by the child is given as:

Let m1 be the mass of the water evaporated from the child’s body in 20 min.
Loss of heat through water is given by:

∴Average rate of extra evaporation caused by the drug

Question 11.19:
A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of
cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of
5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h.
The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole
is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1]
Answer

Side of the given cubical ice box, s = 30 cm = 0.3 m
Thickness of the ice box, l = 5.0 cm = 0.05 m
Mass of ice kept in the ice box, m = 4 kg
Time gap, t = 6 h = 6 × 60 × 60 s
Outside temperature, T = 45°C
Coefficient of thermal conductivity of thermacole, K = 0.01 J s–1 m–1 K–1
Heat of fusion of water, L = 335 × 103 J kg–1
Let m’ be the total amount of ice that melts in 6 h.
The amount of heat lost by the food:

Where,
A = Surface area of the box = 6s2 = 6 × (0.3)2 = 0.54 m3

Mass of ice left = 4 – 0.313 = 3.687 kg
Hence, the amount of ice remaining after 6 h is 3.687 kg.

Question 11.20:
A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of
6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame
in contact with the boiler. Thermal conductivity of brass = 109 J s –1 m–1 K–1; Heat of
vaporisation of water = 2256 × 103 J kg–1.
Answer

Base area of the boiler, A = 0.15 m2
Thickness of the boiler, l = 1.0 cm = 0.01 m
Boiling rate of water, R = 6.0 kg/min
Mass, m = 6 kg
Time, t = 1 min = 60 s
Thermal conductivity of brass, K = 109 J s –1 m–1 K–1
Heat of vaporisation, L = 2256 × 103 J kg–1
The amount of heat flowing into water through the brass base of the boiler is given by:

Where,
T1 = Temperature of the flame in contact with the boiler
T2 = Boiling point of water = 100°C
Heat required for boiling the water:
θ = mL … (ii)
Equating equations (i) and (ii), we get:

Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.

Question 11.21:
Explain why:
a body with large reflectivity is a poor emitter
a brass tumbler feels much colder than a wooden tray on a chilly day
an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body
radiation gives too low a value for the temperature of a red hot iron piece in the open, but
gives a correct value for the temperature when the same piece is in the furnace
the earth without its atmosphere would be inhospitably cold
heating systems based on circulation of steam are more efficient in warming a building

than those based on circulation of hot water

Answer

A body with a large reflectivity is a poor absorber of light radiations. A poor absorber
will in turn be a poor emitter of radiations. Hence, a body with a large reflectivity is a
poor emitter.
Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted
from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a
lower value and one feels cooler.
Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is
conducted from the body to the wooden tray. Hence, there is only a negligible drop in the
temperature of the body and one does not feel cool.
Thus, a brass tumbler feels colder than a wooden tray on a chilly day.
An optical pyrometer calibrated for an ideal black body radiation gives too low a value
for temperature of a red hot iron piece kept in the open.
Black body radiation equation is given by:

Where,
E = Energy radiation
T = Temperature of optical pyrometer
To = Temperature of open space
σ = Constant
Hence, an increase in the temperature of open space reduces the radiation energy.
When the same piece of iron is placed in a furnace, the radiation energy, E = σ T4
Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric
gases, no extra heat will be trapped. All the heat would be radiated back from earth’s
surface.
A heating system based on the circulation of steam is more efficient in warming a
building than that based on the circulation of hot water. This is because steam contains

surplus heat in the form of latent heat (540 cal/g).

Question 11.22:
A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from
60 °C to 30 °C. The temperature of the surroundings is 20 °C.
Answer

According to Newton’s law of cooling, we have:

Where,
Temperature of the body = T
Temperature of the surroundings = T0 = 20°C
K is a constant
Temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s
Integrating equation (i), we get:

The temperature of the body falls from 60°C to 30°C in time = t’
Hence, we get:

Equating equations (ii) and (iii), we get:

Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.

Question 12.1:

A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If
the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat
of combustion is 4.0 × 104 J/g?

Answer

Water is flowing at a rate of 3.0 litre/min.
The geyser heats the water, raising the temperature from 27°C to 77°C.
Initial temperature, T1 = 27°C
Final temperature, T2 = 77°C
∴Rise in temperature, ΔT = T2 – T1
= 77 – 27= 50°C
Heat of combustion = 4 × 104 J/g
Specific heat of water, c = 4.2 J g–1 °C–1
Mass of flowing water, m = 3.0 litre/min = 3000 g/min
Total heat used, ΔQ = mc ΔT
= 3000 × 4.2 × 50
= 6.3 × 105 J/min

∴Rate of consumption = = 15.75 g/min

Question 12.2:

What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature)
to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R =
8.3 J mol–1 K–1.)

Answer

Mass of nitrogen, m = 2.0 × 10–2 kg = 20 g
Rise in temperature, ΔT = 45°C
Molecular mass of N2, M = 28
Universal gas constant, R = 8.3 J mol–1 K–1
Number of moles,

Molar specific heat at constant pressure for nitrogen,

The total amount of heat to be supplied is given by the relation:
ΔQ = nCP ΔT
= 0.714 × 29.05 × 45
= 933.38 J
Therefore, the amount of heat to be supplied is 933.38 J.

Question 12.3:
Explain why
Two bodies at different temperatures T1 and T2 if brought in thermal contact do not

necessarily settle to the mean temperature (T1 + T2)/2.

The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different
parts of a plant from getting too hot) should have high specific heat.

Air pressure in a car tyre increases during driving.

The climate of a harbour town is more temperate than that of a town in a desert at the
same latitude.

Answer

When two bodies at different temperatures T1 and T2 are brought in thermal contact, heat
flows from the body at the higher temperature to the body at the lower temperature till
equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The
equilibrium temperature is equal to the mean temperature (T1 + T2)/2 only when the
thermal capacities of both the bodies are equal.

The coolant in a chemical or nuclear plant should have a high specific heat. This is
because higher the specific heat of the coolant, higher is its heat-absorbing capacity and
vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a
nuclear or chemical plant. This would prevent different parts of the plant from getting too
hot.

When a car is in motion, the air temperature inside the car increases because of the
motion of the air molecules. According to Charles’ law, temperature is directly
proportional to pressure. Hence, if the temperature inside a tyre increases, then the air
pressure in it will also increase.

A harbour town has a more temperate climate (i.e., without the extremes of heat or cold)
than a town located in a desert at the same latitude. This is because the relative humidity
in a harbour town is more than it is in a desert town.

Question 12.4:

A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature
and pressure. The walls of the cylinder are made of a heat insulator, and the piston is
insulated by having a pile of sand on it. By what factor does the pressure of the gas
increase if the gas is compressed to half its original volume?

Answer

The cylinder is completely insulated from its surroundings. As a result, no heat is
exchanged between the system (cylinder) and its surroundings. Thus, the process is
adiabatic.
Initial pressure inside the cylinder = P1
Final pressure inside the cylinder = P2
Initial volume inside the cylinder = V1
Final volume inside the cylinder = V2
Ratio of specific heats, γ = 1.4
For an adiabatic process, we have:

The final volume is compressed to half of its initial volume.

∴

Hence, the pressure increases by a factor of 2.639.

Question 12.5:
In changing the state of a gas adiabatically from an equilibrium state A to another
equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is
taken from state A to B via a process in which the net heat absorbed by the system is 9.35
cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Answer

The work done (W) on the system while the gas changes from state A to state B is 22.3 J.
This is an adiabatic process. Hence, change in heat is zero.
∴ ΔQ = 0
ΔW = –22.3 J (Since the work is done on the system)
From the first law of thermodynamics, we have:
ΔQ = ΔU + ΔW
Where,
ΔU = Change in the internal energy of the gas
∴ ΔU = ΔQ – ΔW = – (– 22.3 J)
ΔU = + 22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the
system is:
ΔQ = 9.35 cal = 9.35 × 4.19 = 39.1765 J
Heat absorbed, ΔQ = ΔU + ΔQ
∴ΔW = ΔQ – ΔU
= 39.1765 – 22.3
= 16.8765 J
Therefore, 16.88 J of work is done by the system.

Question 12.6:
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A
contains a gas at standard temperature and pressure. B is completely evacuated. The entire
system is thermally insulated. The stopcock is suddenly opened. Answer the following:

What is the final pressure of the gas in A and B?
What is the change in internal energy of the gas?
What is the change in the temperature of the gas?
Do the intermediate states of the system (before settling to the final equilibrium state) lie
on its P-V-T surface?

Answer

Answer:
0.5 atm
Zero
Zero
No

Explanation:
The volume available to the gas is doubled as soon as the stopcock between cylinders A
and B is opened. Since volume is inversely proportional to pressure, the pressure will
decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the
pressure in each cylinder will be 0.5 atm.
The internal energy of the gas can change only when work is done by or on the gas. Since
in this case no work is done by or on the gas, the internal energy of the gas will not
change.
Since no work is being done by the gas during the expansion of the gas, the temperature
of the gas will not change at all.
The given process is a case of free expansion. It is rapid and cannot be controlled. The
intermediate states do not satisfy the gas equation and since they are in non-equilibrium
states, they do not lie on the P-V-T surface of the system.

Question 12.7:
A steam engine delivers 5.4×108 J of work per minute and services 3.6 × 109 J of heat per

minute from its boiler. What is the efficiency of the engine? How much heat is wasted per
minute?
Answer

Work done by the steam engine per minute, W = 5.4 × 108 J
Heat supplied from the boiler, H = 3.6 × 109 J

Efficiency of the engine =

Hence, the percentage efficiency of the engine is 15 %.
Amount of heat wasted = 3.6 × 109 – 5.4 × 108
= 30.6 × 108 = 3.06 × 109 J
Therefore, the amount of heat wasted per minute is 3.06 × 109 J.

Question 12.8:
An electric heater supplies heat to a system at a rate of 100W. If system performs work at
a rate of 75 Joules per second. At what rate is the internal energy increasing?
Answer

Heat is supplied to the system at a rate of 100 W.
∴Heat supplied, Q = 100 J/s
The system performs at a rate of 75 J/s.

∴Work done, W = 75 J/s
From the first law of thermodynamics, we have:
Q=U+W
Where,
U = Internal energy
∴U = Q – W
= 100 – 75
= 25 J/s
= 25 W
Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

Question 12.9:
A thermodynamic system is taken from an original state to an intermediate state by the
linear process shown in Fig. (12.13)

Its volume is then reduced to the original value from E to F by an isobaric process.
Calculate the total work done by the gas from D to E to F
Answer

Total work done by the gas from D to E to F = Area of ΔDEF

Area of ΔDEF =
Where,
DF = Change in pressure
= 600 N/m2 – 300 N/m2
= 300 N/m2
FE = Change in volume
= 5.0 m3 – 2.0 m3
= 3.0 m3

Area of ΔDEF = = 450 J

Therefore, the total work done by the gas from D to E to F is 450 J.

Question 12.10:

A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36° C,
calculate the coefficient of performance.

Answer

Temperature inside the refrigerator, T1 = 9°C = 282 K
Room temperature, T2 = 36°C = 309 K

Coefficient of performance =

Therefore, the coefficient of performance of the given refrigerator is 10.44.

Question 13.1:

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.

Answer

Diameter of an oxygen molecule, d = 3Å

Radius, r = 1.5 Å = 1.5 × 10–8 cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm3

Molecular volume of oxygen gas,
Where, N is Avogadro’s number = 6.023 × 1023 molecules/mole

Ratio of the molecular volume to the actual volume of oxygen
= 3.8 × 10–4

Question 13.2:
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

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

PV = nRT
Where,
R is the universal gas constant = 8.314 J mol–1 K–1
n = Number of moles = 1
T = Standard temperature = 273 K
P = Standard pressure = 1 atm = 1.013 × 105 Nm–2

= 0.0224 m3
= 22.4 litres
Hence, the molar volume of a gas at STP is 22.4 litres.

Question 13.3:
Figure 13.8 shows plot of PV/T versus Pfor 1.00×10–3 kg of oxygen gas at two different
temperatures.

What does the dotted plot signify?
Which is true: T1 > T2 or T1 < T2?
What is the value of PV/T where the curves meet on the y-axis?
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

The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio is
equal. μR (μ is the number of moles and R is the universal gas constant) is a constant
quality. It is not dependent on the pressure of the gas.
The dotted plot in the given graph represents an ideal gas. The curve of the gas at
temperature T1 is closer to the dotted plot than the curve of the gas at temperature T2. A
real gas approaches the behaviour of an ideal gas when its temperature increases.
Therefore, T1 > T2 is true for the given plot.
The value of the ratio PV/T, where the two curves meet, is μR. This is because the ideal
gas equation is given as:
PV = μRT

Where,
P is the pressure
T is the temperature
V is the volume
μ is the number of moles
R is the universal constant
Molecular mass of oxygen = 32.0 g
Mass of oxygen = 1 × 10–3 kg = 1 g
R = 8.314 J mole–1 K–1

∴

= 0.26 J K–1
Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is
0.26 J K–1.
If we obtain similar plots for 1.00 × 10–3 kg of hydrogen, then we will not get the same
value of PV/T at the point where the curves meet the y-axis. This is because the molecular
mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).
We have:

R = 8.314 J mole–1 K–1
Molecular mass (M) of H2 = 2.02 u

m = Mass of H2

∴

= 6.3 × 10–2 g = 6.3 × 10–5 kg
Hence, 6.3 × 10–5 kg of H2 will yield the same value of PV/T.

Question 13.4:
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

Volume of oxygen, V1 = 30 litres = 30 × 10–3 m3
Gauge pressure, P1 = 15 atm = 15 × 1.013 × 105 Pa
Temperature, T1 = 27°C = 300 K
Universal gas constant, R = 8.314 J mole–1 K–1
Let the initial number of moles of oxygen gas in the cylinder be n1.
The gas equation is given as:
P1V1 = n1RT1

= 18.276

But,
Where,
m1 = Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
∴m1 = n1M = 18.276 × 32 = 584.84 g
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume, V2 = 30 litres = 30 × 10–3 m3
Gauge pressure, P2 = 11 atm = 11 × 1.013 × 105 Pa
Temperature, T2 = 17°C = 290 K
Let n2 be the number of moles of oxygen left in the cylinder.

The gas equation is given as:
P2V2 = n2RT2

= 13.86

But,
Where,
m2 is the mass of oxygen remaining in the cylinder
∴m2 = n2M = 13.86 × 32 = 453.1 g
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
= 584.84 g – 453.1 g
= 131.74 g
= 0.131 kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.

Question 13.5:
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

Volume of the air bubble, V1 = 1.0 cm3 = 1.0 × 10–6 m3
Bubble rises to height, d = 40 m
Temperature at a depth of 40 m, T1 = 12°C = 285 K
Temperature at the surface of the lake, T2 = 35°C = 308 K
The pressure on the surface of the lake:
P2 = 1 atm = 1 ×1.013 × 105 Pa
The pressure at the depth of 40 m:
P1 = 1 atm + dρg
Where,
ρ is the density of water = 103 kg/m3
g is the acceleration due to gravity = 9.8 m/s2
∴P1 = 1.013 × 105 + 40 × 103 × 9.8 = 493300 Pa

We have:
Where, V2 is the volume of the air bubble when it reaches the surface

= 5.263 × 10–6 m3 or 5.263 cm3
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3.

Question 13.6:
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

Volume of the room, V = 25.0 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 relating pressure (P), Volume (V), and absolute temperature (T)
can be written as:
PV = kBNT
Where,
KB is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1
N is the number of air molecules in the room

∴

= 6.11 × 1026 molecules
Therefore, the total number of air molecules in the given room is 6.11 × 1026.

Question 13.7:
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

At room temperature, T = 27°C = 300 K
Average thermal energy
Where k is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1
∴
= 6.21 × 10–21J
Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 ×
10–21 J.
On the surface of the sun, T = 6000 K
Average thermal energy

= 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.
At temperature, T = 107 K
Average thermal energy

= 2.07 × 10–16 J
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10–16 J.

Question 13.8:

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

Yes. All contain the same number of the respective molecules.
No. 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, N = 6.023 × 1023.
The root mean square speed (vrms) of a gas of mass m, and temperature T, is given by the
relation:

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

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

Question 13.9:

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

Temperature of the helium atom, THe = –20°C= 253 K
Atomic mass of argon, MAr = 39.9 u
Atomic mass of helium, MHe = 4.0 u
Let, (vrms)Ar be the rms speed of argon.
Let (vrms)He be the rms speed of helium.
The rms speed of argon is given by:

… (i)
Where,
R is the universal gas constant
TAr is temperature of argon gas
The rms speed of helium is given by:

… (ii)
It is given that:
(vrms)Ar = (vrms)He

= 2523.675 = 2.52 × 103 K
Therefore, the temperature of the argon atom is 2.52 × 103 K.

Question 13.10:
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

Mean free path = 1.11 × 10–7 m
Collision frequency = 4.58 × 109 s–1
Successive collision time ≈ 500 × (Collision time)
Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × 105 Pa
Temperature inside the cylinder, T = 17°C =290 K
Radius of a nitrogen molecule, r = 1.0 Å = 1 × 1010 m
Diameter, d = 2 × 1 × 1010 = 2 × 1010 m
Molecular mass of nitrogen, M = 28.0 g = 28 × 10–3 kg
The root mean square speed of nitrogen is given by the relation:

Where,
R is the universal gas constant = 8.314 J mole–1 K–1

= 508.26 m/s
The mean free path (l) is given by the relation:

Where,
k is the Boltzmann constant = 1.38 × 10–23 kg m2 s–2K–1

= 1.11 × 10–7 m
Collision frequency

= 4.58 × 109 s–1
Collision time is given as:

= 3.93 × 10–13 s
Time taken between successive collisions:

= 2.18 × 10–10 s

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

Question 13.11:
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

Length of the narrow bore, L = 1 m = 100 cm
Length of the mercury thread, l = 76 cm
Length of the air column between mercury and the closed end, la = 15 cm
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) = 9 cm
Hence, the total length of the air column = 15 + 9 = 24 cm
Let h cm of mercury flow out as a result of atmospheric pressure.
∴Length of the air column in the bore = 24 + h cm
And, length of the mercury column = 76 – h cm
Initial pressure, P1 = 76 cm of mercury
Initial volume, V1 = 15 cm3
Final pressure, P2 = 76 – (76 – h) = h cm of mercury

Final volume, V2 = (24 + h) cm3
Temperature remains constant throughout the process.
∴P1V1 = P2V2
76 × 15 = h (24 + h)
h2 + 24h – 1140 = 0

= 23.8 cm or –47.8 cm
Height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore and
52.2 cm of mercury will remain in it. The length of the air column will be 24 + 23.8 =
47.8 cm.

Question 13.12:
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.
[Hint: Use Graham’s law of diffusion: R1/R2 = (M2/M1)1/2, where R1, R2 are diffusion
rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a
simple consequence of kinetic theory.]
Answer

Rate of diffusion of hydrogen, R1 = 28.7 cm3 s–1
Rate of diffusion of another gas, R2 = 7.2 cm3 s–1
According to Graham’s Law of diffusion, we have:

Where,
M1 is the molecular mass of hydrogen = 2.020 g
M2 is the molecular mass of the unknown gas

= 32.09 g
32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

Question 13.13:
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)/ (ρRT)]
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.] [Hint: Use Archimedes
principle to find the apparent weight of the suspended particle.]
Answer

According to the law of atmospheres, we have:
n2 = n1 exp [-mg (h2 – h1)/ kBT] … (i)

Where,
n1 is the number density at height h1, 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, the
effective weight of the suspended particle is given as:
Weight of the medium displaced – Weight of the suspended particle
= mg – m'g

Gas constant, R = kBN

… (iii)
Substituting equation (ii) in place of mg in equation (i) and then using equation (iii), we
get:
n2 = n1 exp [-mg (h2 – h1)/ kBT]

= n1 exp [- (h2 – h1) ]
= n1 exp [-
(h2 – h1) ]

Question 13.14:

Given below are densities of some solids and liquids. Give rough estimates of the size of
their atoms:

Substance Atomic Mass (u) Density (103 Kg m-3)
Carbon (diamond) 12.01 2.22
Gold 197.00 19.32
Nitrogen (liquid) 14.01 1.00
Lithium 6.94 0.53
Fluorine (liquid) 19.00 1.14

[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

Substance Radius (Å)
Carbon (diamond) 1.29
1.59
Gold 1.77
Nitrogen (liquid) 1.73
1.88
Lithium
Fluorine (liquid)

Atomic mass of a substance = M
Density of the substance = ρ

Avogadro’s number = N = 6.023 × 1023

Volume of each atom

Volume of N number of molecules N … (i)

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

N=

For carbon:
M = 12.01 × 10–3 kg,
ρ = 2.22 × 103 kg m–3

= 1.29 Å
Hence, the radius of a carbon atom is 1.29 Å.
For gold:
M = 197.00 × 10–3 kg
ρ = 19.32 × 103 kg m–3

= 1.59 Å
Hence, the radius of a gold atom is 1.59 Å.
For liquid nitrogen:
M = 14.01 × 10–3 kg

ρ = 1.00 × 103 kg m–3

= 1.77 Å
Hence, the radius of a liquid nitrogen atom is 1.77 Å.
For lithium:
M = 6.94 × 10–3 kg
ρ = 0.53 × 103 kg m–3

= 1.73 Å
Hence, the radius of a lithium atom is 1.73 Å.
For liquid fluorine:
M = 19.00 × 10–3 kg
ρ = 1.14 × 103 kg m–3

= 1.88 Å
Hence, the radius of a liquid fluorine atom is 1.88 Å.

Question 14.1:
Which of the following examples represent periodic motion?
A swimmer completing one (return) trip from one bank of a river to the other and back.
A freely suspended bar magnet displaced from its N-S direction and released.
A hydrogen molecule rotating about its center of mass.
An arrow released from a bow.

Answer

Answer: (b) and (c)
The swimmer’s motion is not periodic. The motion of the swimmer between the banks of
a river is back and forth. However, it does not have a definite period. This is because the
time taken by the swimmer during his back and forth journey may not be the same.
The motion of a freely-suspended magnet, if displaced from its N-S direction and
released, is periodic. This is because the magnet oscillates about its position with a
definite period of time.
When a hydrogen molecule rotates about its centre of mass, it comes to the same position
again and again after an equal interval of time. Such motion is periodic.
An arrow released from a bow moves only in the forward direction. It does not come
backward. Hence, this motion is not a periodic.

Question 14.2:
Which of the following examples represent (nearly) simple harmonic motion and which
represent periodic but not simple harmonic motion?
the rotation of earth about its axis.
motion of an oscillating mercury column in a U-tube.
motion of a ball bearing inside a smooth curved bowl, when released from a point slightly
above the lower most point.

general vibrations of a polyatomic molecule about its equilibrium position.
Answer

Answer: (b) and (c) are SHMs
and (d) are periodic, but not SHMs
During its rotation about its axis, earth comes to the same position again and again in
equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple
harmonic. This is because earth does not have a to and fro motion about its axis.
An oscillating mercury column in a U-tube is simple harmonic. This is because the
mercury moves to and fro on the same path, about the fixed position, with a certain period
of time.
The ball moves to and fro about the lowermost point of the bowl when released. Also, the
ball comes back to its initial position in the same period of time, again and again. Hence,
its motion is periodic as well as simple harmonic.
A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the
superposition of individual simple harmonic motions of a number of different molecules.
Hence, it is not simple harmonic, but periodic.

Question 14.3:
Figure 14.27 depicts four x-t plots for linear motion of a particle. Which of the plots
represent periodic motion? What is the period of motion (in case of periodic motion)?
(a)

(b)

(c)

(d)

Answer

Answer: (b) and (d) are periodic
It is not a periodic motion. This represents a unidirectional, linear uniform motion. There
is no repetition of motion in this case.
In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic
motion, having a period of 2 s.
It is not a periodic motion. This is because the particle repeats the motion in one position
only. For a periodic motion, the entire motion of the particle must be repeated in equal
intervals of time.
In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic
motion, having a period of 2 s.

Question 14.4:
Which of the following functions of time represent (a) simple harmonic, (b) periodic but
not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic
motion (ω is any positive constant):
sin ωt – cos ωt
sin3 ωt
3 cos (π/4 – 2ωt)
cos ωt + cos 3ωt + cos 5ωt
exp (–ω2t2)
1 + ωt + ω2t2
Answer

SHM
The given function is:

This function represents SHM as it can be written in the form:

Its period is:
Periodic, but not SHM
The given function is:

The terms sin ωt and sin ωt individually represent simple harmonic motion (SHM).
However, the superposition of two SHM is periodic and not simple harmonic.
SHM
The given function is:

This function represents simple harmonic motion because it can be written in the form:

Its period is:

Periodic, but not SHM

The given function is . Each individual cosine function

represents SHM. However, the superposition of three simple harmonic motions is

periodic, but not simple harmonic.

Non-periodic motion

The given function is an exponential function. Exponential functions do not

repeat themselves. Therefore, it is a non-periodic motion.

The given function 1 + ωt + ω2t2 is non-periodic.

Question 14.5:
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart.
Take the direction from A to B as the positive direction and give the signs of velocity,
acceleration and force on the particle when it is
at the end A,
at the end B,
at the mid-point of AB going towards A,
at 2 cm away from B going towards A,
at 3 cm away from A going towards B, and
at 4 cm away from B going towards A.
Answer

Zero, Positive, Positive
Zero, Negative, Negative
Negative, Zero, Zero
Negative, Negative, Negative
Zero, Positive, Positive
Negative, Negative, Negative
Explanation:

The given situation is shown in the following figure. Points A and B are the two end
points, with AB = 10 cm. O is the midpoint of the path.

A particle is in linear simple harmonic motion between the end points
At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at

this point.
Its acceleration is positive as it is directed along AO.
Force is also positive in this case as the particle is directed rightward.
At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at
this point.
Its acceleration is negative as it is directed along B.
Force is also negative in this case as the particle is directed leftward.
(c)

The particle is executing a simple harmonic motion. O is the mean position of the
particle. Its velocity at the mean position O is the maximum. The value for velocity is
negative as the particle is directed leftward. The acceleration and force of a particle
executing SHM is zero at the mean position.
(d)

The particle is moving toward point O from the end B. This direction of motion is
opposite to the conventional positive direction, which is from A to B. Hence, the
particle’s velocity and acceleration, and the force on it are all negative.
(e)

The particle is moving toward point O from the end A. This direction of motion is from A
to B, which is the conventional positive direction. Hence, the values for velocity,
acceleration, and force are all positive.
(f)

This case is similar to the one given in (d).

Question 14.6:
Which of the following relationships between the acceleration a and the displacement x of
a particle involve simple harmonic motion?
a = 0.7x
a = –200x2
a = –10x
a = 100x3
Answer

A motion represents simple harmonic motion if it is governed by the force law:
F = –kx
ma = –k

Where,
F is the force
m is the mass (a constant for a body)
x is the displacement
a is the acceleration
k is a constant
Among the given equations, only equation a = –10 x is written in the above form with

Hence, this relation represents SHM.

Question 14.7:
The motion of a particle executing simple harmonic motion is described by the
displacement function,
x (t) = A cos (ωt + φ).
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what
are its amplitude and initial phase angle? The angular frequency of the particle is π s–1. If
instead of the cosine function, we choose the sine function to describe the SHM: x = B sin
(ωt + α), what are the amplitude and initial phase of the particle with the above initial
conditions.
Answer

Initially, at t = 0:
Displacement, x = 1 cm
Initial velocity, v = ω cm/sec.
Angular frequency, ω = π rad/s–1
It is given that:

Squaring and adding equations (i) and (ii), we get:

Dividing equation (ii) by equation (i), we get:

SHM is given as:
Putting the given values in this equation, we get:
Velocity,
Substituting the given values, we get:
Squaring and adding equations (iii) and (iv), we get:

Dividing equation (iii) by equation (iv), we get:

Question 14.8:
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm.
A body suspended from this balance, when displaced and released, oscillates with a
period of 0.6 s. What is the weight of the body?
Answer

Maximum mass that the scale can read, M = 50 kg
Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m
Time period, T = 0.6 s
Maximum force exerted on the spring, F = Mg
Where,
g = acceleration due to gravity = 9.8 m/s2
F = 50 × 9.8 = 490

∴Spring constant,
Mass m, is suspended from the balance.

Time period,

∴Weight of the body = mg = 22.36 × 9.8 = 219.167 N
Hence, the weight of the body is about 219 N.

Question 14.9:
A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as
shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then
pulled sideways to a distance of 2.0 cm and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and
(iii) the maximum speed of the mass.
Answer

Spring constant, k = 1200 N m–1
Mass, m = 3 kg
Displacement, A = 2.0 cm = 0.02 cm
Frequency of oscillation v,is given by the relation:

Where, T is the time period