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Table of Contents
Heat and Calorimetry ...................................................................................................................................... 6
Heat is the Energy in Transit ..................................................................................................................................... 6
Specific Heat .................................................................................................................................................................. 7
Specific Heat of Gas ..................................................................................................................................................... 8
Molar specific heat at constant volume Cv ............................................................................................................ 8
Molar specific heat at constant pressure Cp .......................................................................................................... 9
Relation between Cv and Cp: Mayer’s formula ..................................................................................................... 9
Dulong and Petit’s Law ................................................................................................................................ 10
Variation of specific heat of solid with temperature ..........................................................................................11
Heat Capacity or Thermal Capacity .......................................................................................................... 11
Water Equivalent .........................................................................................................................................................12
Change in phase ...........................................................................................................................................................12
Latent Heat ....................................................................................................................................................................13
Latent heat of fusion ...................................................................................................................................................13
Latent heat of Vaporization ......................................................................................................................................13
Some definitions ............................................................................................................................................ 14
Regelation......................................................................................................................................................................14
Super cooled Water .....................................................................................................................................................14
Superheated steam .....................................................................................................................................................14
Dry ice .............................................................................................................................................................................14
Vapour & gas .................................................................................................................................................................14
Law of Mixture or Law of Calorimety .....................................................................................................................16
Heat Transmission ......................................................................................................................................... 27
Conduction.....................................................................................................................................................................27
Convection .....................................................................................................................................................................27
Radiation ........................................................................................................................................................................28
Conduction of Heat in a Conducting Rod ................................................................................................ 28
Isothermal Surface and Temperature Gradient...................................................................................................29
Temperature Gradient ................................................................................................................................................29
Coefficient of Thermal Conductivity .......................................................................................................................30
Thermal Resistance .....................................................................................................................................................31
Thermal Resistance of the combined conductor .................................................................................................32
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Equivalent Thermal Conductivity .............................................................................................................................33
Equivalent Thermal Conductivity .............................................................................................................................35
Thermal Resistance ....................................................................................................................................... 35
Combination of Metallic Rods ...................................................................................................................................36
Combinations Formula on Electricity Analogy ....................................................................................... 39
Rods in Series ...............................................................................................................................................................39
Rods in Parallel .............................................................................................................................................................39
Thermal Expansion ........................................................................................................................................ 42
Linear Expansion of Solids.........................................................................................................................................42
Application of Linear Expansion...............................................................................................................................43
Radius of Bimetallic strip ...........................................................................................................................................44
Example based on the Differential Expansion of two solid rods .....................................................................45
Thermal Stress ................................................................................................................................................ 46
Example based on Thermal Stress ..........................................................................................................................47
Effect of temperature on Pendulum Clock ............................................................................................................48
Example based on Pendulum clock .........................................................................................................................49
Superficial Expansion of Solid .................................................................................................................... 50
Volume Expansion of Solid .......................................................................................................................... 50
Change in Density of Solid with Temperature......................................................................................................52
Convection ....................................................................................................................................................... 53
Experiment 1 .................................................................................................................................................................53
Experiment 2 .................................................................................................................................................................53
Natural convection ......................................................................................................................................................54
Forced convection ........................................................................................................................................................54
Phenomenon Based on Convection .........................................................................................................................54
Radiation .......................................................................................................................................................... 56
Properties of thermal radiation ................................................................................................................................56
Reflectance, absorptance and transmittance ........................................................................................ 56
Spectral Absorptive Power .......................................................................................................................... 57
Spectral Emissive Power .............................................................................................................................. 58
Emissivity ......................................................................................................................................................... 58
Black Body ....................................................................................................................................................... 59
Fery's Black Body ........................................................................................................................................... 59
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Kirchhoff's Law ............................................................................................................................................... 59
Thermos flask.................................................................................................................................................. 61
Prevost Theory of Heat Exchange (1792) ............................................................................................... 61
Stefan-Boltzmann Law ................................................................................................................................. 62
Newton's Law of Cooling ............................................................................................................................. 62
WIEN'S DISPLACEMENT LAW ..................................................................................................................... 74
Solar Constant ................................................................................................................................................ 75
Temperature of the sun ............................................................................................................................... 75
Solar spectrum.............................................................................................................................................. 109
Gas Laws ........................................................................................................................................................ 110
Boyle's law in terms of molecules per unit volume ..........................................................................................111
Gay Lussac's Law (or Pressure Law) ...................................................................................................... 113
Perfect Gas Equation .................................................................................................................................. 113
Perfect Gas Equation .................................................................................................................................. 114
Postulates of Kinetic theory of gases ..................................................................................................... 115
Expression for the Pressure of a Gas...................................................................................................... 115
Relation between pressure and kinetic theory of gas......................................................................................118
Kinetic interpretation of Temperature and Absolute Temperature ..............................................................119
Absolute temperature ................................................................................................................................. 120
(i) Boyle's law. ............................................................................................................................................................120
(ii) Charle's Law. ........................................................................................................................................................120
(iii) Avogadro's hypothesis. ....................................................................................................................................121
(iv) Graham's Law of Diffusion of Gases. ............................................................................................................121
(v) Regnault's or Gay Lussac's Law. .....................................................................................................................122
(vi) Gas Equation or Equation of state .................................................................................................................122
(vii) Dalton's Law of partial pressure. ..................................................................................................................123
Mean, Root Mean Square and Most Probable Speeds ....................................................................... 123
Specimen Numerical ................................................................................................................................... 126
Degrees of freedom ..................................................................................................................................... 126
Illustrations .................................................................................................................................................................127
Degree of freedom of a diatomic gas molecule .................................................................................................129
Degree of freedom of a solid...................................................................................................................................129
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The law of equipartition of Energy ......................................................................................................... 130
Definition of law of equipartition of energy .......................................................................................................131
Total Energy of a System ........................................................................................................................... 131
Specific Heats of Gases in Terms of Energy ......................................................................................... 131
Triatomic gas ..............................................................................................................................................................133
Concept of Mean Free Path ....................................................................................................................... 135
Expression for mean free path ...............................................................................................................................135
Numerical ....................................................................................................................................................... 137
Brownian Motion .......................................................................................................................................... 138
Reason of Zig-Zag motion .......................................................................................................................................138
Factors affecting the Brownian motion ...............................................................................................................138
Isothermal and Adiabatic Processes ...................................................................................................... 141
1. Isothermal Process ...............................................................................................................................................141
2. Adiabatic Process ..................................................................................................................................................141
3. Isothermal and Adiabatic Curves ......................................................................................................................143
4. Work done by an Ideal Gas in Isothermal Expansion..................................................................................146
5. Work done by an Ideal Gas in Adiabatic Expansion .....................................................................................146
6. Volume Elasticities of Gases ...............................................................................................................................148
Isothermal Modulus of Elasticity ...........................................................................................................................148
Adiabatic Modulus of Elasticity ................................................................................................................ 149
Example ........................................................................................................................................................................151
Thermodynamics .......................................................................................................................................... 160
Thermodynamic System ...........................................................................................................................................160
Thermodynamic Variables .......................................................................................................................................161
Thermal Equilibrium ..................................................................................................................................................161
Heat, Work and Internal Energy .............................................................................................................. 163
Sign conventions for heat ........................................................................................................................................163
Work Done ...................................................................................................................................................................163
Internal Energy ..........................................................................................................................................................165
Some other processes in Thermodynamics .........................................................................................................166
Applications of First law of Thermodynamics ...................................................................................... 168
Reversible Process.....................................................................................................................................................174
First Law of Thermodynamics .................................................................................................................. 178
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Limitations of First Law of Thermodynamics .....................................................................................................179
Free expansion ............................................................................................................................................. 180
Zeroth Law of Thermodynamics .............................................................................................................. 181
Temperature .................................................................................................................................................. 182
Thermal equilibrium .................................................................................................................................... 183
Different types of temperature scales ................................................................................................... 183
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Heat and Calorimetry
Heat is the Energy in Transit
When two bodies at different temperature make in contact, something is
transferred between them. The word heat is meaningful only when energy
is being transferred. The expressions like heat of a body or heat in a body
are meaningless. So the heat can be defined as energy in transit that flows
from one body to another due to difference in temperature between them.
Once heat is transferred to a body, it becomes the part of its internal
energy.
Unit of Heat
CGS unit of Heat: CGS unit of heat is calorie.
Definition of Calorie: It is the amount of heat energy required to raise
the temperature of one gram of water through 1C (from 14.5C to
15.5C).
S.I. Unit of Heat: The SI unit of heat is Joule (J).
Joule’s mechanical equivalent of heat (J):
From experiments, he proved that heat is a form of energy.
He showed that if an amount of work w (or any other form of energy) is
converted into heat, the equal amount of heat is produced. Thus
wQ
or
w = JQ
or
J=
If Q = I, then J = w
The proportionality constant J is called Joules‟s mechanical equivalent of
heat.
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The value of J = 4.186 J/Cal
Note: J is not a physical quantity. It is just a conversion factor.
British thermal unit: It is the amount of heat required to raise the
temperature of 1 pound of water through 1F.
1 BT = 252 Calorie
Specific Heat
It is defined as the amount of heat required to raise the temperature of
unit mass of a substance, the rise in temperature of substance is T, then
specific heat is given by
c=
or

Q = mcT
Units of Specific Heat
(i) In CGS system, the unit of Q is calorie, m is gram T in C. Therefore
unit of c in this system becomes
a. c = cal/g-C
(ii) In SI system, the unit of Q is Joule, m in Kg and T in kelvin.
Therefore in this system unit of specific heat is J/Kg-K.
1.
2.
3.
4.
Specific heat of water is 4200 J/Kg-K.
Specific heat of ice is 0.5 cal/g-C or 2100 j/Kg-K.
The maximum value of specific heat is 3.5cal/gC for H.
The minimum value of specific heat is 0.022 cal/gC for radon.
The specific heat of a substance is not constant at all temperature.
Therefore specific heat used in the above formula is the mean value of
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specific heats. When c varies considerably with temperature, then for small
change in temperature dT, we can write
dQ = mcdT

Q=
Here T1 and T2 are the initial and final temperatures.
Specific Heat of Gas
Limit of specific heat of gas
Consider a gas of mass m and volume V at a pressure P.
Suppose gas is compressed suddenly without supplying heat:
Let temperature of the gas rises by T.

c=


From zero to infinity, it may have any positive or negative value. The exact
value depends on the conditions of pressure and volume when heat is
being supplied. Out of many specific heats of a gas, two are of prime
significance.
Molar specific heat at constant volume Cv
It s the amount of heat required to raise the temperature of 1 mole of gas
through 1 K (1C) at constant volume. If Qv is the heat given to n moles of
a gas at constant volume and change in temperature be T, then
Cv =

Qv = nCvT
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Molar specific heat at constant pressure Cp
It s the amount of heat required to raise the temperature of 1 mole of gas
through 1 K (1C) at constant pressure. If Qp is the heat given to n moles
of a gas at constant pressure and change in temperature be T, then
Cp =

Qp = nCpT
Relation between Cv and Cp: Mayer’s formula
Heat supplied to a gas at constant volume entirely used to raise its
temperature. When a gas is heated at constant pressure, it expands to
keep pressure constant and therefore some mechanical work is to be done
in addition to raise the temperature of the gas. Hence more heat is
required at constant pressure than that at constant volume. Thus for one
mole of a gas, we have
Cp – Cv = work done
= PV
……………………………(1)
At constant pressure, we have
PV1 = RT
and
……………………..(2)
PV2 = R(T + 1) ………………………(3)
where V1 is the volume of gas at temperature T and V2 is the volume of
gas at temperature (T + 1).
Subtracting equation (2) from (3), we get
P(V2 – V1) = R

PV = R
Substituting this value in equation (1), we get
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CP – CV = R
Mayer‟s formula
Note:
1.
Substance which expand on heating, PV is positive and therefore, CP
– CV = +ve or CP > CV. If any substance contracts on heating, PV
will be negative and therefore, CP – CV = -ve or CP < CV.
2.
CP – CV = R holds good for all ideal gases.
3.
For one gram of a gas, we have cV and cP we can write CV = McV
and CP = McP.
Also, cP – cV = r
Here r =
which is different for different gases.
Dulong and Petit’s Law
At near about room temperature the molar specific heat of most of the
solids is equal to 3R or 6 cal/mol-K at constant volume.
In case of solids, the significant motions of atoms are vibratory motion.
During vibration, the kinetic energy (EK) of an atom changes periodically
into potential energy (EP) nad vice versa. So the average values of EK and
EP are equal.
For each form of energy there are three degrees of freedom. Therefore a
molecule has six degrees of freedom (3 for kinetic + 3 for potential).
According to law of equi-partition of energy, each degree of freedom
possesses energy
per atom. Therefore total energy associated with one
mole of a substance at a temperature T is given by;
Ek = 3 
=
kT
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Ek = 3 
=
kT
 Average vibrational energy per atom =Ek + EP = 3kT
The internal energy (due to vibration) of one mole of an atom of the solid
is given by
U = (3kT)  N
= 3(kN)T
= 3RT
Also, we have
(kN = R)
CV =
=
or
CV = 3R
Variation of specific heat of solid with temperature
At higher temperature the molar specific heat of all solids is close to a
value 3R.
Heat Capacity or Thermal Capacity
It is the amount of heat required to raise the temperature of whole amount
of substance through 1 K(1C).
Heat Capacity = mass  specific heat
S = mc
The CGS unit of heat capacity is cal/C and SI unit is J/K.
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Water Equivalent
The water equivalent of a body is defined as the mass of water which
requires the same amount of heat as is required by the given body for the
equal rise of temperature. Let the mass of substance be m, specific heat c
and rise in temperature is T, then
Q = mcT
If water equivalent is w, then
Q = w  1  T

w  1  T = mcT
or
w = mc
(c of water is 1cal/g-C)
Note: Water equivalent numerically equal to the heat capacity but the
unit of water is gm or kg and that of heat capacity is cal/C or J/C
Change in phase
A substance can exist in three possible phases i.e. solid, liquid & gas.
Transition from one phase to another are accompanied by the absorption
or liberation of heat & usually by change in volume, even when the
transition occurs at constant temperature. As an example take small piece
of ice in a container at -20 & heat is supplied to a container at a uniform
rate. The temperature starts increasing steadily, as shown by the segment
a to b in figure until the temperature rises to 0 .There after ice starts
melting. The melting process is a change in phase, from the solid phase to
the liquid phase(b c) where temp remains constant .When the whole ice
has melted(point c) the temp of water now rises at uniform rate (c
)
although this rate is slower than that from a to b. When temperature of
water reaches 100 ,it begins to boil. The temperature remain constant
until whole water has converted into water vapour.
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If heating has still continued (from e to f), the temperature of vapour starts
rising. The gaseous state would now be called superheated steam.
Latent Heat
The heat when is used to change the phase of substance at constant
temperature is called latent heat or hidden heat.
Latent heat of fusion
The amount of heat required to convert unit mass of solid into liquid as its
melting point is called latent heat of fusion.
Latent heat of Vaporization
The amount of heat required to convert unit mass of liquid into vapor at its
boiling point is called latent heat of vaporization.
If Q is the amount of heat absorbed or liberated by m amount of
substance, then latent heat L is defined as:
L
or Q = m
Unit of L
The CGS unit of L is cal/g
The SI unit of L is J/Kg
Some values
for ice = 80 cal/gm or 336 KJ/kg
of water = 540 cal/gm = 2259 KJ/Kg at 1 atm pressure
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Some definitions
Regelation
The phenomenon of melting of ice & its resolidification is called regelation.
Super cooled Water
Water below
is known as super cooled water. Both can be possible as
a pressure greater than atmosphere pressure.
Superheated steam
Steam at temperature greater than 100
is known as super heated steam.
Dry ice
Solid carbon dioxide is called dry ice. Carbon dioxide at -78 remains in
solid state. Solid carbon dioxide does not melt when exposed to air. It
directly evaporates & form vapour.
Vapour & gas
These are the gaseous states of the substance. The gaseous state of a
substance below the critical tamp is called vapour & above critical temp is
called gas. A gas cannot be liquefied by mere application of pressure,
however high it may be. A vapour can be liquefied by applying pressure.
Thus to liquefy a gas, first it bring below critical temperature.
A solid material is supplied heat at a constant rate. The temperature of the
material is changing with the heat input as shown in figure.
Study the graph carefully & answer the following questions:
(a)
(b)
(c)
What do the horizontal regions AB & CD represent?
If CD=2AB, what do you infer?
The slope of OA > the slope of BC, what does this indicate?
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(d)
What does the slope DE represent?
Sol:
(a)
(b)
(c)
In the region AB & CD the temp of the material remain constant
so AB represents fusion & CD represents vaporisation.
CD = 2AB or Q4 = 2Q2, it shows that latent heat of vaporization is
twice than that of latent heat of fusion.
If c1 & c2 are the specific heats of solid & liquid sates represents
then
Q1 = mc1ΔT1
or c1 =
=
Similarily, c2=
(d)
Since Slope of OA > slope of BC
 c1 < c 2
If c3 is the specific heat of vapour state(region DE) then
c3 =
or slope of DE =
=
15
Law of Mixture or Law of Calorimety
When two or more non reacting substances are placed in contact, the heat
lost by the hot substance is equal to the heat gained by cold substance.
That is Heat lost = Heat gained
Two substances at temperatures T1 & T2 are taken together in a
calorimeter. Let T is the equilibrium temp of the given system
S1(T1 - T)=
S2(T - T2)
T 1 > T2
T=
T2 < T < T1
T=
Ex: 1g ice at 0 is raplacedin a calorimrter having 1g water at
40 , find equilibrium temp & final contents. Assuming heat
capacity of calorimeter is negligibly small.
Sol: the heat required to melt the ice completely = mL = 1×80 = 80
cal
The heat available on water = mcΔT = 1×1×(40-0) = 40 cal

Entire heat of water is utilized to melt ice & its temp falls to 0 .
Ice still at 0 .so equilibrium temp of contents remain 0 . Let m is
the amount of ice melt due to 40 cal heat then
m×80 = 40 or m=1/2g final
constant ice =
g, water =
g
16
Ex: 1g ice at -400 is placed in a container having 1g water at 1 Find
equilibrium temp. Assume heat capacity of container is negligibly
small.
Sol: The heat available on water to cool from 10
to 0
= mcΔT = 1×1×(10 - 0)= 10 cal
Let temp of ice becomes T after taking this heat

mice cΔT = 10
Or
1×0.5[T- (- 40)] = 10
T + 40 = 20
T = -20
Now the system have 1g ice at -20
get freeze to bring the ice from -20
& 1g water at 0 . Let m gm water
to 0
Heat gained by ice = heat lost by water
mice c[0 - (-20)] = m × 80
m=
g
Thus equilibrium temp becomes 0 , as both ice & water change into 0
Final contents:
Ice = 1g +
Water = 1 -
g=
=
g
g
17
Ex.
1g of steam at 100 is passed in a insulating vessel having 1g ice at
0 . Find the equilibrium temp of the mixture, neglecting heat
capacity of the vessel.
Sol. Heat available on steam
(changes into steam to water) = mL = 1×540 = 540 cal
Heat gained by ice to change into water and then rise its temperature
to 100
=
L+
c T
=1×80+1×1×(100-0)
=180 cal
The above calculations show that some part of steam will condense
to change the ice into water at 100 . Let m is the mass of steam
considered then
m × 540 = 180
m=
=
Final Contents
g
ice = 0g
Water = 1 +
=
g
Steam = 1 -
=
g
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Ex: The temperature of equal masses of three different liquids A,B,C
are 12 ,19 and 28 respectively.The temperature when A and B
are mixed is 16
. What
should be +ve temperature when A and B are mixed?
Sol:
A(16 )
m
B(19 )
(16-12) = m
=
---------(1)
B(19 )
m
(19 - 16)
C(28 )
(23-19)=m
=
(28-23)
……………………(2)
From (1) \and (2) we get
=
m
(T - 12) = m (28 - T)
m
(T - 12)=m
(28 - T)
31T = 628
T = 20.26
19
Q. A 5g piece of ice at -20˚is put into 10g of water at 30˚. Assuming that
heat is exchanged only between the ice and the water, find the final
temperature of the mixture.
Specific heat capacity of ice = 2100 J/kg-˚C.
Specific heat capacity of water = 4200J/Kg-˚C
& Latent heat of fusion of ice = 3.36×
J/Kg
Sol:
30˚C water = 0˚C water
Heat given = 0.01 × 4200 × 30
=1260 J
5 g ice at -20˚C  0˚C ice
0.005×2100(20)
=210 J
Heat required to melt 5 g of ice to 0˚C water
=0.005×3.36×
=1680 J
Ice melted =
=3.125 g
Mixture = (10g + 3.125g) water + 1.875 g ice
Final mixture content = 13.125 g of water + 1.875 g of ice.
20
Q. Calculate the amount of heat required to convert 1.00 kg of ice at -10˚C
into steam at 100˚C at normal pressure.
Specific heat capacity of ice = 2100J/Kg-K,
Latent heat of fusion of ice = 3.36×
J/kg
Specific heat of capacity of water = 4200 J/kg-K
Latent heat of vaporization of water = 2.25×
J/kg
Sol:
Q1 = 1×2100×10 = 21000J
Q2 = 1×3.38×
J
Q3 = 1×4200×100 = 420000J
Heat required to convert 1 kg of water at 100˚C into steam
=1×2.25×
J
 total heat required =3.03×
J
Q. An aluminium container of mass 100 g contains 200g of ice at -20˚C.
Heat is added to the system at a rate of 100 cal/s. What is the temp of the
system after 4 minutes?
Draw a rough sketch showing the variations in the temperature of the
system as a function of time.
Specific heat capacity of ice = 0.5 cal/g-˚C
Specific heat capacity of aluminium = 0.2 cal/g-˚C
Specific heat capacity of water = 1 cal/g-˚C
Latent heat of fusion of ice = 80cal/g
21
Sol:
Total heat supplied to the system in 4 minutes is Q = 100cal/s×240
=2.4×
cal
The heat requires to take the system from -20˚C to 0˚C
=100×0.2×20+200×0.5×20
=400+2000 cal
=2400cal
The time taken in this process
=
s
=24 s
The heat required to melt the ice at 0˚C = 200×80
= 16000 cal
The time taken in this process
=
s
= 160s
If the final temperature is , then heat required to take the system to the
final temperature is
=100 × 0.2× + 200 × 1 × 
So, 2.4×
cal = 2400 + 16000 + 220 × 
=
= 25.5˚C
22
Q. An aluminium vessel of mass 0.5Kg contains 0.2Kg of water at 20 . A
block of iron of mass 0.2 kg at 100 is gently put into the water. Find the
equilibrium temperature of the mixture.
Specific heat capacities of aluminium is 910 J/Kg-K
Specific heat capacities of iron is 470 J/Kg-K
Specific heat capacities of water is 42000 J/Kg-K respectively
Heat gain = 0.5×910×(-293)+0.2×4200(-293)
Heat Lost = 0.2×470×(373 - )
 = 25
Q. A piece of iron of mass 100g is kept inside a furnace for a long time and
then put in a calorimeter of water equivalent 10g containing 240g of water
at 20 .The mixture attains an equilibrium temperature of 60 .
Find the temperature of the furnace.
Specific heat capacity of iron=470J/kg×470×( - 65) =
×42000(60-20)
 = 953.61
23
Q. A thermally isolated vessel contains 100g of water at 0 . When air
above the water is pumped out, some of the water freezes and some
evaporates at 0 itself. Calculate the mass of ice formed if no water is left
in the vessel.
Latent heat of vaporization of water at 0 =2.1×
Latent heat of fusion of ice=3.36×
J/kg
J/kg
Sol: Let the mass of the ice formed = m
Then the mass of water evaporated = M - m
Heat taken by the water to evaporate = (M-m)
Heat taken by the water in freezing=m
m
= (M-m)
m=
=
= 86g
24
Q. A calorimeter of water equivalent 15g contains 165g of water at 25 .
Steam at 100 is passed through the water for some time .The
temperature is increased to 30 and the mass of the calorimeter and the
contents is increased by 1.5g.
Calculate the specific latent heat of vaporization of water.
Specific heat capacity of water is 1 cal/gSol. Heat lost in condensation by steam
=1.5L
Heat lost in cooling condensed water from 100
=1.5×1×100.30
=105 cal
Heat supplied to the calorimeter and the cold water during the rise in temp
from 25 to 30 .
=900cal
1.5 L +105 = 900
L = 530 cal/g
25
Q. A lead ball at 25 is dropped from a height of 2 Km. It is heated due to
air resistance and it is heated due to air resistance and it is assumed that
all of its Kinetic Energy is used in increasing the temperature of ball. If the
specific heat of lead is s = 126J/kg- find the final temperature of ball?
Sol.
V=
=200m/s (in the absence of air resistance)
KE =
J
m(126)×  = 2×

m
158.73
26
Heat Transmission
The flow of heat from one body to another or from one place to another, is
called transmission of heat. There are three methods of heat transmission :
conduction, convection and radiation.
Conduction
Whenever there is a temperature difference between parts of a
body then particles of the body at higher temperature give heat by
mutual contact to the particles at a lower temperature. Thus heat
flows from the part at higher temperature to the part at lower
temperature. This process of heat transmission in which the
particles of the body do not leave their positions is called
conduction. For example, when one end of a metallic bar is kept in
fire, then heat flows by conduction from the hot end of the bar to
the cold end. As a result, the other also becomes hot. In solids
(and in mercury) the transmission of heat takes place by
conduction.
Convection
Whenever the temperature of the lower region in a fluid(liquid or
gas) becomes higher than the temperature of the upper region,
the density of the fluid in the lower region becomes less than that
in the upper region. Hence particles of the fluid in the lower region
begin to rise and their places are occupied by the particles in the
upper region. The process continues until the temperature of the
whole fluid becomes the same. The process of heat transmission
in which the particles of the fluid move is called convection. The
water in a vessel placed on fire becomes hot by convection.
Convection takes place only in liquid and gases (not in solids).
In solids (and in mercury) transmission of heat takes place only by
conduction, while in liquids and gases it mainly takes place by
convection(in liquid and gases it is possible by convection also).
27
Radiation
Heat (energy) is also transmitted in the form of
electromagnetic waves from a hot body. This is called radiation. It
does not require any medium, if present then the medium does
not become hot. Heat from the sun reaches to the earth by
radiation.
Heat transmission by conduction and convection takes place
slowly but by radiation heat is transmitted with speed of light.
Further more in conduction and convection the path of heat
transmission may be zig-zag, but the path of radiation is a straight
line. This is why we can protect our self from the heat rays of the
sun by using umbrella.
Conduction of Heat in a Conducting Rod
If we heat one end of a conducting (metallic) rod, heat flows by conduction
from the hot end to the cold end. In the process of conduction, each cross
section of the rod receives heat from the adjacent cross section towards
the heated end. A part of this heat is absorbed by the cross section
towards the heated end. A part of this heat is absorbed by the cross
section itself whose temperature increases, another part goes in
atmosphere by convection and radiation from the sides of the cross section
and the rest is conducted away to the next cross section. This state of the
rod is collective “variable state”, because in this state the temperature of
each cross section of the rod is increasing with time.
After sometime, however, a state is reached when the temperature of each
section of the rod becomes constant. In this state, no heat is absorbed by
the rod. The heat that reaches any section is transferred to the next except
that some heat goes out from the sides by convection and radiation.
This state of the rod in which no part of the rod absorbs heat is called the
“steady state”.
28
By steady state it does not mean that the temperature of the whole rod is
the same. In steady state, the temperatures of different parts of the rod
are different (as we go away from the hot end, the temperature falls), but
the temperature of each part whatever it may be, remains constant.
If the escape of heat into the atmosphere is prevented by covering the rod
by some non conducting material, then in the steady state, the heat
received by any section will be totally transferred to the next. In this state
the rate of conduction of heat in the rod depends only upon the
conductivity of the rod.
Isothermal Surface and Temperature Gradient
Any surface of a material, whose all points are at the same temperature, is
called an “isothermal surface”. The direction of heat conduction at a point
in a material is normal to the isothermal surface passing through that
point.
Above figure shows a long and thin metallic rod in which heat is flowing
from one end to the other.
Temperature Gradient
The rate of change of temperature with distance between two isothermal
surfaces is called “temperature gradient”.
If the temperature of two isothermal surfaces be  and  -  and the 
distance between them be x, then the temperature gradient between
them is
=

=




=-


The negative sign indicates that the temp  decreases as distance x
increases in the direction of heat flow.
The unit of temp gradient is C per meter.
29
Coefficient of Thermal Conductivity
Let a long conducting rod of uniform area of cross section be in a steady
state. Let  and  -  be the temperatures of two isothermal plane
surfaces of the rod and x be the  distance between these surfaces. It
has been found experimentally that the amount of heat Q flowing normally
to isothermal surface in time „t‟ is
(i) Directly proportional to the area „A‟ of each surface
(ii) Directly proportional to the temp gradient -


between surfaces
(iii) Directly proportional to the time „t‟.
Q-A
Thus,


Q = - kA
t


t
where k is a constant whose value depends upon the material of the rod
and is known as “coefficient of thermal conductivity” of the material.
If 1 and 2 be the temperatures of the ends of the rod and l be the
distance between them.


=

Q = kA






=


t
If the area of cross section of each end A = 1 m2, distance between the
ends, l = 1m, temp difference (
 ) = 1C and time t = 1 sec
Then Q =
=k
Hence the coefficient of thermal conductivity of a material is the amount of
heat flowing in 1 sec through a rod of that material of 1 m length & 1 m2
30
area of cross section in the steady state, when the difference of
temperatures between two ends of the rod is 1C & the flow of heat is
normal to the end phases of the rod.
S.I unit is J/m-sec-C [MLT-3-1]
Higher the value of k, greater will be the conductivity of material. For an
ideal or perfect heat conductor, the value of k will be , & for a perfect
heat insulator (or non conductor), the value k will be zero.
Thermal Resistance
Just as charge flows in an electric circuit due to a potential difference
between two points of the circuit, in the same way heat flows in a
conductor due to a temp difference between two points of the conductor.
Hence like electric resistance, there is also a thermal resistance in a
material.
H=
= kA





=
sec-C/K-Cal
[M-1L-2T3]
Rate of Flow of Heat, Interface temperature, Equivalent Thermal
Resistance, Equivalent Thermal Conductivity for Conductors
connected in Series
Let us consider two conductors of same area of cross section A and joined
in series. Let d2 be their thickness and K1 and K2 their thermal
conductivities. Let heat be allowed to flow through this combination. After
the steady state is reached, let 1 be the temperature of the open face of
the first conductor and 2 be the temperature of the open face of the
second conductor. Let  be the steady temperature of the interface. In the
steady state, the rate of flow of heat in both the conductors will be same.
Hence
H=
= k1 A


= k2 A
 
31

) =
(


+
 )
(
=


=
=


……………….(i)
This is the equation for the temperature of the interface. Substituting this
value of  in equation, then the flow of heat in the combined conductor is


H=

=

=



=


…………………………(ii)
Thermal Resistance of the combined conductor
If R be the equivalent thermal resistance of the combined conductor, then
we have
R=


=
Thermal resistances of the individual conductors are
R1 =
and R2 =
Comparing the above equation, we get
R = R1 + R2
32
Equivalent Thermal Conductivity
If K be the equivalent thermal conductivity of the combined conductor,
then
H=


……………………(iii)
Comparing it with equation (ii), we get
K=
Since thickness of the two conductors are equal,
(say),
Then
K=
=
=
Conductors of different lengths are joined in series, then the equivalent
thermal conductivity would be
K=
Rate of Flow of Heat, Equivalent Thermal Resistance and
Equivalent Thermal Conductivity of Conductors connected in
Parallel
Let two conductors of thickness d be connected in parallel. Let A1 and A2
be their areas of cross section and K1 and K2 their thermal conductivities.
Let heat be allowed to the flow from left to right and in the steady state,
the temperatures of left and right surfaces of the conductors be 1 and 2
33
The rate of flow of heat in the first conductor is
= k1 A1


The flow of heat in the second conductor is
= k2 A2


The flow of heat in the combined conductor is
=
Where
+
= k1 A1

= (
 )
= (
 )
= (
 )
and

+ k2 A2


are thermal resistances of the conductors respectively.
If the equivalent thermal resistance of the conductors be R, then
=


Comparing equation (i) and (ii), we get
=
34
Equivalent Thermal Conductivity
If the the thermal conductivity of the conductors be k, then

=

Comparing it with equation (i), we have
= K1A1 + K2A2
K=
If the area of cross sections of the two conductors be the same, then
K=
Thermal Resistance
We know that heat current
H = KA
H=
…………………(1)
In electricity, electric current is given by
i=
………………..(2)
where V1 – V2 is the potential difference across the resistor R.
If we compare equation (1) and (2), we find
is a type of resistance
which is called as thermal resistance RH.
So
RH =
Temperature difference is analogous to the potential difference in
electricity.
SI unit of RH =
=
35
Combination of Metallic Rods
1. Series Combination: Suppose n number of rods, each of cross
sectional area A and lengths l1, l2,…….., ln and conductivities
K1, K2, …………., Kn are placed in series.
(i) Equivalent Thermal Resistance: If R1, R2, ……… are the
thermal resistances of the rods, then equivalent resistance
RH = R1 + R2 +………….+ Rn
(ii) Heat Current: The heat current remains same for all rods.
H=
= H1 = H2 = ………. =
(iii) Equivalent Thermal Conductivity:
We know that RH = R1 + R2 +………….+ Rn
or
or
=
+
K=
+ …………+
=
(iv) For two rods of same length and same cross sectional
area:
K=
2. Parallel Combination: Suppose n number of rods or slabs each of
length l and area of cross sections A1, A2, ……………An and thermal
conductivities K1, K2,……………, Kn are placed in contact.
(i) Equivalent Thermal Resistance:
If R1 + R2 +………….+ Rn are the resistances of the rods, then
equivalent resistance
=
(ii)
+
+ …………..+
Heat Current:
H1 =
, H2 =
,…………., Hn =
36
and H = H1 + H2 +………….+ Hn =
(iii) Equivalent Thermal Conductivity:
We know that
=
or
or
+
+ …………..+
=
+
+ ………….+
K=
=
(iv)
For two slabs:
K=
37
Example:
An electric heater is used in a room of total wall area 137 m2 to
maintain a temperature of +20C inside it, when the outside
temperature is - 10C. The walls have three different layers
materials. The innermost layer of wood of thickness 25.0 cm. Find
the power of the electric heater. Assume that there is no heat loss
through the floor and the ceiling. The thermal conductivities of wood,
cement and brick are 0.125, 1.5 and 1.0 W/m-C respectively.
Sol:
Equivalent thermal conductivity of the wall
K=
=
=
= 0.624 W/m-C
The rate of flow of heat is given by
H = KA
= 0.624  137 
=


= 9000 W
38
Combinations Formula on Electricity Analogy
Rods in Series
TH – T = ir1
T – TC = ir2
TH – TC = i(r1 + r2)
i=
T = TC + ir2  TC +
. r2
T=
TH – TC = ir
i(r1 + r2) = ir
r = r1 + r2
=
+
K=
Rods in Parallel
i = i1 + i2
TH – TC = ir1
TH – TC = ir2
i1r1 = i2r2
i1 =
.i
i2 =
.i
TH – TC = ir
=
=
+
+
 r=
39
(A1 + A2) =
+
K=
Example:
Above figure shows two conducting rod AB and CD made up of same
material. C is the midpoint. The temperature of A, B and D are
100C, 0C and 25C respectively. Find the rate of heat flow in CD.
Sol. Given, AC = BD =
As resistance of the rod is proportional to the length, so
RAC = RCB =
=
= 2.5 K/W
Let H1 is the heat current in CD and H2 in CB, then
H1 =
=
H2 =
=
……………..(1)
……………..(2)
If H is the heat current through AC, then
H=
=
We know that H = H1 + H2

Or
=
+
TC = 45C
From (i)
H1 =
= 4.0 W
Ans.
40
Example:
Consider the situation shown in fig. The frame is made of the same
material and has a uniform cross sectional area everywhere.
Calculate the amount of heat flowing per second through a cross
section of the bent part if the total heat taken out per second from
the end at 100C is 130 J.
Sol. Suppose resistance of 10 cm length of frame is R, then
RAB = REF = 2 R, RBE = 7 R
Resistance of bent part (length = 60 + 5 + 5 = 70 cm) R = 7 R.
The equivalent network of resistor is shown in figure.
Let H1 and H2 are the heat current in straight and bent parts of the
frame, then
H1  6R = H2  7R
Or 6H1 = 7H2 ………….(i)
Given H1 + H2 = 130 ………….(ii)
H1 = 70 J/s
H2 = 60 J/s at 100C is 130 J
41
Thermal Expansion
Solid are made up of atoms and molecules. At a given temperature,
the atoms and molecules are placed at some equilibrium distance.
When heat is supplied to solid , the inter atomic separation
increases by which there is an expansion of solids. This expansion
can be in terms of length / area / volume.
From this tree, it is clear that the thermal expansion of solids is
classified into three categories.
1. Linear expansion of solids.
2. Superficial expansion of solids.
3. Cubical expansion of solids.
Linear Expansion of Solids
1) Almost all solids expand on heating. On increasing the temperature
of solids, their length increases. This change in length of a solid on
heating is called linear expansion.
2) Coefficient of linear expansion is defined as fractional increase in
length per °C rise in temperature. If l is the length of the rod at T
K and as the temperature is changed to T + ΔT its length becomes l
+ l, so the coefficient of linear expansion is given by
=
=
3) Unit of is per Kelvin or per °C. It is positive for metals except
carbon. The value of is negative for plastic because in plastic when
the temperature increases, length decreases.
4) The numerical value of is same in both the units i.e. in per Kelvin
or per °C.
42
5) If
is the coefficient of linear expansion at t1°C.
l1= length of rod at t1°C
l2 = length of rod at t2 °C
l2 = l1[1+ (t2-t1)]
t may be in any unit °C or K because in the formula there is a
difference of temperature which remains same for °C or ˚K.
(6) If length of the rod is lo at 0°C and l1 at t°C then
l1 = lo[1+ (t - 0)]
l1= lo(1 + t)
Where is temperature coefficient of linear expansion at 0°C.
Here t should be in °C only because initial temperature is taken as 0
°C.
Application of Linear Expansion
Differential expansion of two solid rods
1) Suppose there are two rods of length l1 and l2. The first rod is kept on
the other such that the initial separation between the free ends of
the rods is
S= l2 - l1
Both the rods are initially at a temperature of t1°C.
On heating the entire system, the temperature increases to t2 such
that the length of both the rods increases. So, if the new length of
the rod be l1 and l2 then
l1‟=l1[1+ (t2-t1)]
l‟2=l1[1+
(t2-t1)]
Here
and
are the coefficient of linear expansion at t1 and t2 °C.
Now the separation between the free ends of the rod also changes
such that
S‟=l‟2-l1‟
= l2[1+
(t2-t1)] - l1[1+ (t2-t1)]
43


= (l2-l1) + (
l2- l1) (t2-t1)
‟
If the new separation S of the composite rod is equal to the original
separation then
S‟ = S
(l2-l1) + (
l2- l1) (t2-t1) = l2-l1
(
l2 - l1) (t2-t1) = 0
(
l2 - l1) =0
l2 = l1
Radius of Bimetallic strip
If two strip of different metals are welded together to form a
bimetallic strip, when heated uniformly it bends in the form of an arc,
the metal with greater coefficient of linear expansion lies on convex
side. The radius of arc thus formed by bimetal is:
R=
R=
Where t = temperature difference between the two ends
d= thickness of each strip
and  Coefficients of linear expansion.
44
Example based on the Differential Expansion of two solid rods
Exercise
The length of the steel rod which would have the same difference in
length with a copper rod of length 24 cm at all temperatures .
(acopper = 1810-6 K-1 ,
= 12  10-6 K-1)
(a) 20 cm
(b) 18 cm
(c) 24 cm
(d) 48 cm
Sol: (d) By linear expansion of solids, we have
l =l2
=
So , l2 - l1= 24
1.5l1 – l1= 2
l1= 48 cm
l2 = 72 cm
45
Thermal Stress
1) When a rod is heated or cooled, it expands or contracts. It is turned
as free expansion of the rod.
Actually no strain is being developed because on increasing the
temperature the length of the rod increase so at t2 °C, l2 because
natural length of rod.
2) Now if the ends of the rods are rigidly fixed so as to prevent it from
expansion or contraction than stress is produced in the rod.
By the virtue of this thermal stress the rod exerts a large force on the
supports.
The first figure indicates the rod kept at a room temperature t1°C
having l1. The rod is between two rigid supports. If the supports were
not there then on increasing the temperature the free expansion
occurs by which length becomes l2 at higher temperature t2°C .The
third figure indicates that on increasing the temperature to t2°C the
length remains the same but a compressive strain is induced in the
rod. If l1 is the length of the rod at t1°C and on increasing to t2°C its
length becomes l2 then
l2 = l1[1+ (t2-t1)]
l2 = l1+ l1 (t2-t1)
= (t2-t1)
The above relationship indicates the thermal strain developed in the
rod.
i.e. Thermal strain =
= (t2-t1)
In elasticity, Young‟s modulus =
Stress = Y (t2-t1)
As Force=Stress  Area
Force = YA (t2-t1)
46
3) If the rod is in its natural length at t1°C while at t2°C it is in
compressed state, then
Strain =
In this case first figure represents the rods of length l at t1°C when
the temperature is lowered to at t2 the length of the rod remains
same but a tensile strain is developed in the rod .
=
=
=
=
Strain =
4) When the temperature of the rod is increased, the compressive stress
is developed while on decreasing the temperature of the rod the
tensile stress is developed.
Example based on Thermal Stress
Exercise:
A steel rod of length 1 m rests on a smooth horizontal base. If it is
heated from 0°C to 100°C, what is the longitudinal strain developed?
Sol: In this case, rod rest on a horizontal base which is the free expansion
on heating .Hence no strain is developed in the rod.
Thus,
Strain = 0
47
Exercise:
A steel rod of length 50 cm has a cross sectional area of 0.4 cm2
.What force would be required to stretch this rod by the same
amount as the expansion produced by heating it through 10°C.
(
Y = 2  1011 N/m2)
Sol: We have
Force= YA
= 2  1011 0.410-410-510
=0.8  103
=800N
Effect of temperature on Pendulum Clock
A pendulum clock consists of a metal rod or wire with the bob at one
end. Let l1 be the length of the simple pendulum at θ1 °C than time
period T1 is given by
T1=
Now when the temperature increases to
becomes l2 so that
the effective length
T2=
Now on dividing
But
l2=l1[1+ ( 2- 1)]
1/2
=1+½
48
–
=
Change in time Period
T2 – T1
=½
The above expression represents the time lost per oscillation. Thus, a
pendulum clock losses time in summer and gains time in winter.
Note:
If a pendulum clock is giving correct time when time period is T
Than
If T increases , clock becomes slow.
If T decreases , clock becomes fast.
Example based on Pendulum clock
Exercise:
A pendulum clock with a pendulum made of invar
(

) has a period of 0.5 s and is accurate at 25 .
If the clock is used in a country where the temperature averages
35 , what correction in necessary at the end of a month (30 days)
to the time given by the clock –
Sol:
In the time interval t, the clock will become slow by
= 

30  86400(35 - 25)
= 9.1 sec
49
Superficial Expansion of Solid
1) On increasing the temperature of solid, its area increases. This
change in area is referred as superficial expansion of solids.
2) If A0 is the area of solid at °C . On heating the rod t1°C, the area
becomes At so that
At=A0[1+ ]
Where
is coefficient of superficial expansion at 0°C and t should
be in °C only.
3) If the area of solid at temperature t1 is A1 and on heating the rod ,
the area becomes A2 at t2 then
A2=A1[1+ (t2-t1)]
Where β is the coefficient expansion at t1°C.
4) Coefficient of superficial expansion is defined as fractional increase in
area per
rise in temperature.
β=
5) Unit of
=
is per
or per Kelvin.
Volume Expansion of Solid
1) On increasing the temperature of rod, its volume changes.
2) If V0 is the volume of solid at 0°C and on increasing the temperature
, volume becomes Vt then, Vt= Vo[1+ t]
Where V0 is volume of solid at 0°C.
Here also t should be in °C only.
3) If V1 is the volume of solid at t1°C and on increasing the temperature
to t2°C the volume becomes V2 then,
V2= V1[1+ (t1 – t2]
Where is coefficient of volume expansion at t1°C.
4) Coefficient of volume expansion is defined as the fractional increase
in volume per
rise in temperature.
50
=
5) Unit of
=
is per °C Kelvin.
Relation Between Coefficient of Linear Expansion ( ), Coefficient
of Superficial Expansion ( ) and Coefficient of Cubical Expansion
( )
Exercise:
Consider the following statements
(a)The coefficient of linear expansion has dimension K-1
(b)The coefficient of volume expansion has dimension K-1.
(1) Both a and b are correct.
(2) a is correct but b is wrong.
(3) b is correct but a is wrong.
(4) a and b are both wrong
Sol: (1)
Coefficient of Linear expansion is given by =
and coefficient of
volume is given by
=
So from above formula it is clear that both have units of per Kelvin.
51
Change in Density of Solid with Temperature
Suppose m is the mass of a solid which at a given temperatures
occupies a volume V so that density at 0°C is
d0 =
Now if the temperature is increased by t , mass will remain
unchanged but due to thermal expansion volume increases so that,
V‟= V(1+ )
Now density
dt = =
dt =
=
dt =
Here
is coefficient of cubical expansion at 0°C.
52
Convection
It is the process by which heat flows from the region of
higher temperature to the region of lower temperature by
the actual movement of the particles of the medium. There
is no simple equation for convective heat transfer as there
is for conduction. Convective heat transfer depends on
many factors, such as the shape and orientation of the
surface, the mechanical and thermal properties of the fluid.
For practical calculations, one can define heat current due
to convection is;
H = hAT
where h is called convection coefficient. A is the surface
area and T is the temperature difference between the
surface and the main body of the fluid.
Experiment 1
Take a glass tube, filled with water and put some charcoal
powder in the tube, and starts heating it from one of its
sides (see Fig. 8.23). The heated water molecules rise up
(due to low density) and cold water molecules take their
places, coming from the side tube and therefore a
convection current is set-up in the whole tube. The
direction of movement of water molecules is indicated by
the movement of the charcoal particles.
Experiment 2
Take a flask containing water and put some crystals of
potassium permagnet. When flask starts heating, coloured
streaks of water rise up due to lower density. The denser
cold water takes its place by moving downwards. Thus
convection current is setup in the water.
53
Natural convection
If the material of the medium moves due to difference in
density caused by difference in temperature, the process of
heat transfer it called natural or free convection. Natural
convection currents always move upward due to difference
in density and gravity. Different types of winds in the
atmosphere are originated due to natural convection.
Forced convection
If the heated material is forced to move by a machine like
a blower or a pump, the process of heat transfer is called
forced convection. Het convector, hair drier, airconditioning are the examples of forced convection.
Phenomenon Based on Convection
(i)
Monsoon: In summer, the surface of the earth of the
Indian subcontinent becomes hotter than the Indian Ocean.
This sets up convection current with hot air from the land
rising and moving towards the Indian Ocean, while the
moisture laden air from the ocean moves towards the land.
When obstructed by mountains, the moist air rushes
upwards to great height and gets cooled. In this process
moisture condenses and causes rains in all over India.
(ii)
Trade winds: The surface of the earth gets heated more
at the equator than at the poles. Warm air at the equator
moves up and cold air from the poles moves towards the
equator. In the northern hemisphere, it is coming from the
north and due to the rotation of the earth from west to
east, the wind appears to come from north-east. In the
southern hemisphere, the wind appears to be from south54
west. These winds are called trade winds because they
were used by traders for sailing their vessels in ancient
days.
(iii)
Land and sea breezes: Sun shines almost uniformly on
the land mass near coastal regions, giving equal amount of
heat energy. However the temperature of land rises more
rapidly as compared to sea, because specific heat capacity
of land is much smaller than that of water. Thus the air
above the land becomes hot and light and hence rises up.
This results decrease in pressure over landmass. So the
colder air starts blowing from sea towards land and thereby
setting up sea breeze.
During night the land as well as sea water radiate out heat
energy. However the temperature of land decreases more
rapidly as compared to sea water due to higher specific
heat of water. Thus at night the temperature of sea water
becomes more than land. The air above sea water become
warm and light and rises up. The cold air from land takes
its place. This set-up land breeze.
55
Radiation
Radiation is the process by which heat is transmitted from
one place to another without heating and transferring the
intervening medium.
Properties of thermal radiation
(i)
These are electromagnetic waves having wavelength range
from 1µm to 100 µm. There are also called infrared waves.
(ii)
Thermal radiations travel in straight line with the speed of
light.
(iii)
They obey the laws of reflection and refraction like light
does.
(iv)
They show the phenomenon of interference, diffraction and
polarisation.
NOTE: Word 'radiation' uses for process and energy both.
Reflectance, absorptance and transmittance
When thermal radiations falls on a body, they are partly
reflected, absorbed and rest get transmitted. Lt Q amount
of thermal energy is incident on a body. Suppose the part
R is reflected, A is absorbed ant T is transmitted, then
R+A+T=Q
…(1)
Dividing both sides of equation (1), by Q we have
R A T
 
1
Q Q Q
where
…(2)
R
 r, is called reflectance
Q
56
A
 a, is called absorptance and
Q
T
 t, is called transmittance.
Q
Thus equation (2) takes the form
r+a+t=1
for any specific wavelength , we can write,
r + a + t = 1
Special cases
(i)
If a body does not transmit the radiations, t = 0, then r + a
= 1.
It shows that if r is more, a is less and vice versa. That is
good reflectors will be bad absorbers and vice versa.
(ii)
If a body neither reflects nor transmits any radiation, r = 0
and t = 0, then a = 1, such a body is called a black body.
Spectral Absorptive Power
The absorptive power of any body for a given wavelength 
is defined as the ratio of amount of heat energy absorbed
by certain surface area of the body in a given time to the
total heat energy incident on that area and in same time
within a unit wavelength range around the wavelength . It
can be denoted by a.
57
If dQ is the quantity of heat radiations incident on the
surface in one second and Q1 is the quantity of heat
absorbed by the surface in a wavelength range  to  + d,
then
a 
Q1
or Q1  a dQ
dQ
Spectral Emissive Power
The emissive power of a body at a given temperature and
for a given wavelength  is defined as the amount of
radiant energy emitted by unit surface are of the body per
unit time within a unit wavelength range around the
wavelength . If e is the emissive power of the body, then
the radiant energy emitted by it in one second = e (d).
The SI unit of emissive power is W/m2–Å.
Emissivity
It is defined as the ratio of the heat energy radiated per
unit surface area per second by the given body to the
amount of the heat energy radiated per unit area per
second by a black body of the same temperature. If e and
E are the emissive powers of any body and black body
respectively, then emissivity

e
E
It is a dimensionless quantity. Its value ranges from 0 to 1.
For a black body, it is 1.
58
Black Body
A perfectly black body is one which absorbs all the heat
radiation incident on it. When such a body is placed inside
an isothermal enclosure, it will emit all the radiations of the
enclosure after it is in equilibrium with the enclosure.
Fery's Black Body
It is not possible to construct a perfectly black body, but a
body showing close approximation to a perfectly black body
can be constructed. Fery constructed such a black body. It
has a hollow copper sphere and coated with lamp black on
its inner surface. A find hole is made and a pointed
projection is made just in front of the hole. When the
radiations enter the hole, they suffer multiple reflections
and are completely absorbed. This body behaves as a black
body.
When this body is heated, the heat radiations come out of
the hole. It should be remembered that only the hole, not
the walls of the body, acts as the black body emitter.
Kirchhoff's Law
It states that at any temperature, the ratio of the emissive
power to the absorptive power of any body is a constant,
and equal to the emissive power of the perfectly black body
at the same temperature.
Consider any body (not black body which is suspended in a
hollow enclosure maintained at a constant temperature.
59
The amount of energy absorbed per unit area per unit time
= a (dQ)
The amount of energy emitted per unit area per second =
e (d)
As the body is in thermal equilibrium,
so e (d) = a (dQ)
…(1)
Suppose at black body at the same temperature is
suspended in the enclosure. For this boy, a = 1 and e =
E, so equation (1) becomes
E d   dQ
…(2)
Dividing equation (1) by (2), we get
e
 a
E
or
…(3)
e
 E  constant 
a
The above equation can be stated as; good emitter is a
good absorber.
As
e
 ,
E

from equation (3), we have
a  
60
Thermos flask
A thermos flask is constructed by a double walled glass
bottle. The space in between the two walls is evacuated
and sealed. by doing this, heat cannot go out by
conduction and convection. The inner surface of the outer
wall and outer surface of inner wall are highly polished, to
prevent heat loss by radiation. When a hot liquid is kept in
the bottle, it remains hot for a long time. Similarly ice kept
inside the flask will not melt for a long time.
Prevost Theory of Heat Exchange (1792)
According to this theory, all bodies radiate thermal
radiation at all temperature. Besides, body also absorbs
radiations from its surroundings. This is known as theory of
heat exchange. According to this theory:
(i)
All bodies at temperature above 0 K emit radiation to the
surroundings and gain from the surroundings at all the
time.
(ii)
The amount of heat radiated per second depends on the
nature of the surface, its area and its temperature, and it
does affect by the presence of surroundings bodies.
(iii)
The rise or fall in temperature of a body is the net result of
exchange of heat radiations between body and the
surroundings.
61
Stefan-Boltzmann Law
According to this law, the rate of emission of heat energy
by unit area of a perfectly black body is proportional to the
fourth power of its absolute temperature of its surface.
Thus
E  T4
or
E  T 4 W / m2
Here  is a universal constant called Stefan's constant. Its
values is 5.67 × 108 W/m2-K4.
If H is the rate of energy radiated by a blackbody of
surface area A, then
H = EA = AT4
If T0 is the temperature of the surrounding and  is the
emissivity of the body, then the net rate of loss of energy
per unit area will be;


E   T 4  T04 W / m2
net
Newton's Law of Cooling
It states that the rate of cooling of the body is directly
proportional to the temperature difference between body
and the surrounding, provided it to be small.
If T and T0 are the temperatures of the body and the
surrounding respectively and A is the surface area of the
body, then by Stefan-Boltzmann law, the rate of loss of
heat due to radiation
62

H1    A T 4  T04

If T is the temperature difference between body and
surroundings, then we can write
T  T  T0
or T  T0  T
4
 H1    A  T0  T   T04 


4
 

T 
4
4
 T0 
=   A  T0 1 
T0 
 

If T is small, then

T 
1


T0 
4
1 4
T
T0
Thus, we have
 

4T 
4
H1    A  T04 1 

T
0 
T0 
 

=   A  T04  4T03 T  T04 
or H1  4AT03  T 
or H1  k1A  T  T0 
…(1)
where k1 is a constant. The body may loose heat due to
convection also The rate of loss of heat by convection can
be written as
H2  k2A  T  T0 
…(2)
63
The net rate of loss of heat energy
H = H1 + H2
= (k1 + k2) A (T – T0)
…(3)
If c is the specific heat capacity of the body and m its
mass, then
 dT 
H = mc  
 dt 
or 
=
or
dT
H

dt mc
k1  k2  A  T  T0 
mc
dT
  k  T  T0 
dt
…(4)
…(5)
The constant k depends on the surface involved and the
surrounding conditions.
NOTE:
1.
From equation (3), it is clear that the rate of loss of heat
H  A. Thus if two bodies of equal surface areas, one solid
and other hollow are kept t same temperature difference,
then their ratio of loss of heat will be equal.
2.
For spherical body, A = 4r2.
 H  r2
3.
dT
A

 T  T0  . Thus two bodies of
dt
mc
same material an equal surface areas , one solid other
From equation (4), 
64
hollow (ms > mH), then rate of cooling of solid body will be
less.
4.
For spherical body,


2

A
4r   1

   .
mc  4 3   r 
r c
3

 Rate of cooling of body is inversely proportional to its
radius, provided other things remain constant.
5.
If temperature of the body falls fromT1 and T2 in time t,
T  T2
then we can write T = 1
2
and
6.
T1  T2  T1  T2


 T0 
t
 2

We have,

dT
 k  T  T0 
dt
T
or

Ti
t
dT
   kdt
0
T  T0
ln  T  T0 
Tf
Ti
  kt
 T  T0 
  kt
or ln  f
 Ti  T0 
or  Tf  T0    Ti  T0  e kt
or Tf  T0   Ti  T0  e kt
65
Ex:
A spherical body with radius 12 cm radiates 450 W power
at 500 K. if the radius were halved and the temperature
doubled, what would be the power radiated?
Sol:
By Stefan's law, power radiated
E = AT 4


=  4r2 T 4
When radius is halved and temperature is doubled, power
radiates
  r 2
4
E' =  4    2T   4E
 2 


= 4 × 450 = 1800 W
Ex.
A thin brass rectangular sheet of sides 15.0 and 12.0 cm is heated
in a furnace to 600°C, and taken out. How much electric power is
needed to maintain the sheet at this temperature, given that its
emissivity is 0.250? Neglect heat loss due to convection (Stefan–8
W/m2-k4).
Sol:
Area of the both side of the plate
A = 2 × (15.0) × (12.0) × 10–4 m2
= 3.60 × 10–2 m2
The energy radiated by the plate
=  AT4
= 0.250 × 5.67 × 10–8 × 3.60 × 10–2
× (600 + 273)
4
= 5.10 × 10–12 × 8734 = 296.4 W Ans.
66
Ex.
A body cools in 7 minute from 60°C to 40°C. what will be
its temperature after the next 7 minute? The temperature
of the surroundings is 10°C. Assume that Newton's law of
cooling holds good throughout the process.
Sol:
Newton's law of cooling can be written as:
T1  T2
 T  T2

k 1
 T0 
t
 2

In first case;
T1 = 60°C, T2 = 40°C, T0 = 10°C
and t = 7 minute

60  40
 60  70

k
 10
7
2


or k =
1
14
In second case;
T1 = 40°C and T2 = ?
Next t = 7 minute

40  T2
1  40  T2


 10

7
14  2

or 80  2T2  20 
T2
 10
2
or T2 = 28°C Ans.
67
Ex.
A hot body placed in air is cooled down according to
Newton's law of cooling, the rate of decrease of
temperature being k times the temperature difference from
the surrounding. Starting from t = 0, find the time in which
the body will lose half the maximum heat it can lose.
Sol:
We have,

dT
 k  T  T0 
dt
where T0 is the temperature of the surrounding. If T1 is the
initial temperature and T is the temperature at any time t,
then
T
t
dT
 T  T0    k 0 dt
T1 
ln  T  T0 
T
T1
  kt
 T  T0 
or ln  f
   kt
T

T
0 
 i
or T = T0 + (T1 – T0) e–kt
…(1)
The body continues to lose heat till its temperature
becomes equal to that of the surrounding. the loss of heat
Q = mc(T1 – T0)
If the body loss half of the maximum lose that it can, then
decrease in temperature
Q
 T  T0 
 mc  1
 2 
2
68
If body loses this heat in time, t then its temperature at
time t' will be
 T  T0  T1  T0
T1   1

 2 
2
Putting these values in equation (i), we have
T1  T0
 T0   T1  T0  e kt '
2
or
T1  T0
  T1  T0  ekt '
2
or ekt ' 
or t' =
Ex.
1
2
ln 2
Ans.
k
Fig. shows water in a container having 2.0 mm thick walls
made of a material of thermal conductivity 0.50 W/m-°C.
The container is kept in a melting ice bath at 0°C. The total
surface area in contact with water is 0.05 m2. A wheel is
clamped inside the water and is coupled to a block of mass
M as shown in the figure. As the block goes down, the
wheel rotates. It is found that after some time a steady
state is reached in which the block goes down with a
constant speed of 10 cm/s and the temperature of the
water remain constant at 1.0°C. Find the mass M of the
block. Assume that the heat flows out of the water only
through the walls in contact. Take g = 10 m/s2.
69
Sol:
At steady state,
Rate of doing work by gravity on block M
= Rate of energy produced
or
d
d 
1

Mgh 
Q

Mv2 


dt
dt 
2

dv
 dh dQ M
or Mg   
…(i)
  2v
 dt 
dt 2
dt
As block goes down with constant velocity,
so
dh
dv
 v and
0
dt
dt
 Mgv 
Here
dQ
dt
…(ii)
dQ
T
 KA
dt
L
 Mgv  KA
T
L
or M × 10 × (0.1) = 0.50  0.50
1.0  0
2  103
or M = 12.5 kg Ans.
70
Ex.
A copper sphere is suspended in an evacuated chamber
maintained at 300 K. The sphere is maintained at a
constant temperature of 500 K by heating it electrically. A
total of 210 W of electric power is needed to do it. When
the surface of the copper sphere is completely blackened,
700 W is needed to maintain the same temperature of the
sphere. Calculate the emissivity of copper.
Sol:
Let  is the emissivity of the sphere then by Stefan's law

E = A T 4  T04



or 210 = A 5004  3004 …(i)
When sphere is blackened, it behaves like a perfectly
blackbody, so we have

700 = A 5004  3004

…(ii)
Dividing equation (i) and (ii), we get
 = 0.3 Ans.
Ex.
Sol:
A solid copper sphere (density  and specific heat c) of
radius r at an initial temperature 200 K is suspended inside
a chamber whose walls are almost at 0 K. Calculate the
time required for the temperature of sphere to drop t 100
K.
 dT 
Let T is the temperature at any time and 
is the rate
 dt 
of fall of temperature. Then the rate of loss of heat
 dT 
= mc  
 dt 
…(i)
71
The rate of heat lost by sphere due to radiation only
E = AT4
…(ii)
Equating equations (i) and (ii), we have
 dT 
mc  
 A T 4  04

 dt 


= AT 4
or
dT  A

dt
T4
mc
Integrating both sides of above equation
100

t
T
4
200
A
dT  
dt
mc 0
100
T 3
A

t
3 200
mc
100
mc  1 
t 
3A  T 3 200
4 3 
 r 
3
=
3  4r2
1 
 1

1003 2003 


=
rc
  9  106
1

1


8 

=
7 rc
 106 s Ans.
72 
72
Ex.
One end of a rod of length L and cross-sectional area A is
kept in a furnace of temperature T1.
The other and of the rod is kept at a temperature T2. The
thermal conductivity of the material of the rod is K an
emissivity is . It is given that T2 = TS + T, where T <<
TS, TS being the temperative of surrounding. If T  (T1 –
TS), find the proportionality constant. Consider that heat it
lost only by radiation at the end where the temperature of
the rod is T2.
Sol:
At steady state rate of heat gained by left end of the rod
= Rate of heat radiates by right end of the rod
or KA
 T1  T2   A
L
T
4
2

 TS4 …(i)
Given T2 = TS + T
 T24   TS  T 

T 
= TS4 1 
TS 

4
4
As T << TS, so by Binomial theorem, we get
T24

T 
TS4 1  4
TS 

or T24  TS4  4TS3 T
Also T2  TS  T
Substituting these values in equation (i), we have
73
 T1   TS  T  
KA
 A 4TS3 T
L

or K
or K
T
1
 TS 
L
T
1
or T 
 TS 
L

K
T  4TS3 T
L

K

  4TS3   T

L
K  T1  TS 
 4LT
3
S
K

…(ii)
It is given that
T = C(T1 – TS)


K
 Constant of proportionality C = 

3
 4LTS  K 
Ans.
WIEN'S DISPLACEMENT LAW
The intensity of energy radiated by a black body is not
uniformly distributed over all the wavelengths but it is
maximum for a particular wavelength m. The value of m
decreases with the increase of temperature.
According to Wien's law the product of the wavelength
corresponding to maximum intensity and absolute
temperature is a constant i.e.,

m T = b (constant)
where b is Wien's constant.
Its value is 2.9 × 10–3 m-K.
74
Solar Constant
It is the amount of radiant energy that a unit area of a
perfectly black body placed at a mean distance of the earth
from the sun would receive per second in the absence of
the atmosphere with its surface held perpendicular to the
sun rays.
Temperature of the sun
Let RS be the radius and T be the temperature of sun, then
the solar energy radiated per second
E = AT4
= 4R2S T 4
If r is the mean distance of the earth from the sun then
surface are over which solar energy will spread
= 4r2
Let S is the solar constant, then by the definition
4R 2ST 4
E

S=
4r2
4r2
R 
or S = T  S 
 r 
2
4
1/4
 r2S 
or T =  2 
RS  
By this formula the surface temperature of sun is found to
be 5742 K.
75
Ex.
The spectral energy distribution of the sun has a maximum
at 4753 Å. If the temperature of the sun is 6050 K, what is
the temperature of a star for which this maximum is at
9506 Å?
Sol:
Given m = 4753 Å, T = 6050 K
m = 9506 Å
If T' is the temperature of star, then
m T = m T'
or T' =
=
mT
 'm
4753  6050
 3025 K Ans.
9506
Ex.
The thickness of ice in a lake is 5 cm and the temperature
of air is –10°C. Calculate the time required for the
thickness of ice be doubled. Constant for ice are :
conductivity = 0.004 cal cm–1 sec–1 °C–1 density = 0.92
gm/cm3, latent heat = 80 cal/gm.
Sol:
The cold air (below 0°C) above the water in a lake takes
heat from the water of the lake and hence, the water
begins to freeze into ice layer. Here, we shall obtain an
expression for the rate of growth of this layer.
As shown in fig. (16), let a layer of ice x cm thick has
already formed on a lake at 0°C. The air above lake is at –
T°C. Let A be the area of the layer, L the latent heat of
fusion of ice and  its density. The heat given up when the
ice layer increases in thickness by dx
76
= mass × latent heat
= (A. dx. ) × L calories
Let this quantity of heat is conducted upwards through the
layer in dt seconds. Then
A.dx. × L = KA
0   T 
dt
x
where K = Thermal conductivity of ice.

dx
KT

dt
Lx
or dt =
Lx
dx
KT
…(1)
Now the time in which the thickness of ice will increase
from x1 to x2 can be obtained by integrating eq. (1) within
the limits of x from x1 to x2. Hence
x2
L x2
L  x2 
x dx 
t=
KT x1
KT  2  x
1
or t =
L
x22  x12
2KT


….(2)
In the given problem,
K = 0.004 cal.cm–1 sec–1 °C–1,
L = 80 cal/gm, –T = – 10°C,
x1 = 5 cm, x2 = 10 cm (doubled).
Substituting these values, we get
77
t=
0.92  80
102  52
2  0.004  10


= 69000 seconds
= 19.1 hours.
Ex.
A rubber tube to length 20 cm, through which steam at
100°C is passing is immersed in a calorimeter whose water
equivalent is 15 gm and which contains 300 gm of water at
16°C. The temperature of water rises at the rate of 2°C per
minute. If the outer and inner diameters of the tube are
1.0 cm and 0.6 cm respectively, calculate the thermal
conductivity of rubber.
Sol:
We know that the radial rate of heat flow through a rubber
tube is
Q=
K.2l  T1  T2 
cal / sec.
 r2 
loge  
 r1 
…(1)
where the symbols have their usual meaning.
If this heat rises the temperature of m gm of water
contained in a calorimeter of water equivalent w by dT,
then
Q = (m + w) dT cal/sec.
…(2)
Equating eqs. (1) and (2), we get
K.2l  T1  T2 
 m  w dT
 r2 
loge  
 r1 
78
r 
or K =
m  w dT.loge  r2 
2l  T1  T2 
 1
Substituting the given values, we have
K=
=
2
300  15  60

 1.0 
C / sec  2.3log10 

 0.6 
2  3.14  20  100  16
1
315  30
  2.3  0.2219
2  3.14  20  84
= 5 × 10–4 cal–cm–1–sec–1–°C–1
Ex.
A heating wire of 0.0005 metre diameter is embedded
along the axis of a cylinder of 0.12 metre diameter. When
a current is passed through the wire it gives out a power of
3 kilowatt per metre of its length. If the temperature of the
wire be 1500 °C and that of the outer surface of this
cylinder by 20°C, compute the thermal conductivity of the
material of the cylinder.
Sol:
Let the length of the wire be l metre. The power being
generated
= 3 l kilowatt = 3 × 103 l watt
= 3 × 103 l joules/sec (1 watt = joule/sec)
3  103 l
cal / sec.
=
4.18
The radial flow through a cylinder of length l having
internal and external radii r1 and r2 respectively is given by
79
Q=
K.2l  T1  T2 
r 
loge  2 
 r1 
3  103 l
cal / sec.,
Here Q =
4.18
T1 = 1500°C,
T2 = 20°C, r1 = 0.0005 metre
and r2 = 0.12 metre.
Hence
3  103 l K.2  3.14  l 1500  20

4.18
 0.12 
2.3026log10 
 0.0005 
3  103 K.2  3.14  1480

or
4.18
2.3026log10 240
3  103 K.2  3.14  1480

or
4.18
2.3026  2.3802
3  10  2.3026  2.23802
or K =
3
4.18  2  0.14  1480
= 0.423 cal/(metre-sec-°C).
80
Ex.
A thin metal pipe of 1 meter length and 1 cm radius carries
steam at 100°C. This is covered by two layers of lagging.
The thermal conductivity of outer layer, which is 2 cm thick
is 3.6 × 10–4 cal.cm–1.deg–1, sec–1 while that of the inner
layer, which is 1 cm thick is 1.2 × 10–4 cal. cm–1 deg–1 sec–
1
. If the outer surface of the lagging is at 30°C, find (a) the
temperature of the cylindrical interface of the two lagging
materials (b) the mass of steam condensed per second.
Given loge 2 = 0.6931.
Sol:
The situation is shown in fig. (17). The rate of flow of heat
through cylindrical tube is given by
Q=
K.2l  T1  T2 
r 
loge  2 
 r1 
Let T be the temperature of the interface of two lagging
materials. The rate of flow of heat through inner layer is
given by
Q1 =
K12l  T1  T 
r 
loge  2 
 r1 
1.2  10  2  3.14  100 100  T
4
=
loge 2
…(1)
The rate of flow of heat through outer layer
Q2 =
K2 2l  T  T2 
r 
loge  2 
 r1 
81
3.6  10  2  3.14  100  T  30
4
=
 4
loge  
 2
But Q1 = Q2
1.2  10  2  3.14  100 100  T 

4
loge 2
3.6  10  2  3.14  100  T  30
4
=
loge 2
1.2 × (100 – T) = 3.6 × (T – 30)
T = 4.5°C.
Let the mass of steam condensed per second be m gm. The
heat taken away fro the steam per second = m × 540 cal.
This is equal to Q1 or Q2. Thus
1.2  10  2  3.14  100 100  47.5  m  540
4
loge 2
1.2  10  2  3.14  100 52.5
or m =
4
540  0.6931
or m = 0.0106 gm.
82
Ex.
Find the temperature distribution in the space between two
coaxial cylinders of radii R1 and R2 filled with a uniform
heat conducting substance if the temperatures of the
cylinders are constant and are equal to T1 and T2
respectively.
Sol:
In equilibrium state
2rK
dT
  A  constant
dr
or dT = 
A  dr 
 
2K  r 
Integrating his expression we get
A

T =   K loge r  C
2

…(1)
where C = constant of integration.
To obtain the value of C, we apply the condition that when
r = R1, T = T1
and when r = R2, T = T2.
Hence
T1 = 
A
loge R1  C
2K
and T2 = 
…(2)
A
loge R 2  C …(3)
2K
From these equations, we have
T2 – T1 =
R
A
loge 1  C
2K
R2
83
or A =
2K  T2  T1 
R 
loge  1 
 R2 
…(4)
Substituting the values of C from eq. (2) and A from eq.
(4) in eq. (1), we get
T= 
A
A
loge r  T1 
loge R1
2K
2K
= T1 
= T1 
 r 
A
loge  
2K
 R1 
 T2  T1 
 r 
loge  
R 
 R1 
loge  2 
 R1 
Ex.
Solve the above problem for the case of tw0 concentric
spheres of radii R1 and R2 and temperatures T1 and T2.
Sol:
In equilibrium state
4r2K
dT
  A  constant
dr
or dT = 
A dr
4K r2
Integrating the expression, we get
T= 
A
1
 C
4K r
…(1)
Applying boundary conditions,
i.e., T = T1 when r = R1
and T = T2 when r = R2
84
T1 =
A
1

C
4K R1
T2 =
A
1

C
4K R 2
Solving we get
A=
4K  T2  T1 
 1
1

 R
R1 
2
T 
=
A
1
A
1
  T1 

4K r
4K R1
A 1 1 

 T1
4K  r R1 
T = T1 
 T2  T1 
1 1 

 1
1   r R1 
 R  R 
2
1
Ex.
A liquid takes 5 minutes of cool from 80°C to 50°C. How
much time will it take to cool from 60°C to 30°C? The
temperature of surrounding is 20°C.
Sol:
According to Newton's law,
dQ
 k ' T
dt
(Excess of temperature over surroundings)
If T be the excess of temperature over surroundings
ms
dT
 k ' T
dt
85
where m is the mass of a body of specific heat s and it
cools through a small range of temperature dT. Negative
sign is used to indicate that there is a fall of temperature

dT
k'

T  kT
dt
ms
where k = k' m/s
or
dT
  k dt
T
Integrating loge T = – kt
If T1 be the temperature excess over surrounding at time t1
and T2 at time t2, then
loge
T1
 k  t2  t1   kt
T2
…(1)
where t is the time elapsed.
or 2.3026 [log10 T1 – log10 T2] = kt
In first case,
T1 = (80 – 20) = 60°C,
T2 = (50 – 20) = 30°C
and t = 5 min.
 2.306 [log10 60 – log10 30] = 5 k
…(2)
In second case,
T1 = (60 – 20) = 40°C,
T2 = (30 – 20) = 10°C
and t = ?
86
 2.306 [log10 40 – log10 10] = kt …(3)
Dividing equation (3) by equation (2), we get
log10 40  log10 10
t

log10 60  log10 30 5
or t =
5log10 4
log10 2
= 10 minute.
Ex.
A solid copper sphere cools at the rate of 2.8°C per minute,
when its temperature is 127°C. Find the at which another
solid copper sphere of twice the radius lose its temperature
at 27°C, if in both the cases, the room temperature is
maintained at 27°C.
Sol:
We know that the heat lost per second
 dT 
= ms
 dt 
4

 Heat lost by first sphere =  r3  s  2.8
3

By Stefan's law, heat lost is given by

A T 4  T04

4
4
   4r2  400  300 


=
4 3
r   s  2.8
3
2.8
4
4
rs
or   400  300  


3
…(1)
87
For second sphere
4
dT
3
 2r   s
3
dt
2
4
4
=  4  2r  600  300 


2
dT
4
4
 600  300   rs
…(2)

 3
dt
Dividing eq. (2) by eq. (1), we set
2
3
dT 600  300
6  3




4
4
4
4
3 2.8 dt
 400  300  4  3
4
4
4
4
4
4
dT  6  3 
or
 4
 1.4
4
dt
  4  3 
= 9.72 °C/min.
Ex.
A copper ball cools from 62°C to 50°C in 10 minutes and to
42°C in the next 10 minutes. Calculate its temperature at
the end of next 10 minutes and what is the temperature of
surroundings?
Sol:
We know that
dT
  k  T1  T2 
dt
where T2 is the temperature of surrounding.
Rate of cooling during first 10 minutes is
62  50
 k 62  T2  .
10
…(1)
88
Rate of cooling during next 10 minutes is
60  42
 k 50  T2  .
10
…(2)
Dividing equation (1) by equation (2), we get
12 62  T2

8
50  T2
Solving we get,
T2 = 26°C
and k =
1
.
30
Let in the next 10 minutes, the temperature of copper balls
falls to T0, then
42  T0
1

  42  26
10
30
42 – T0 =
16
3
or 126 – 3 T0 = 16
T0 =
Ex.
110
 36.7C.
3
A cubical tank of water of volume 2m3 is kept at a constant
temperature of 65°C by 2 kW heater. At time t = 0, the
heater is switched off. Find the time taken by the tank of
cool down to 50°C, given the temperature of room is
steady at 15°C.
89
Density of water = 103 kg/m3 and specific heat of water s
= 1.0 cal/gm°C. (Do not assume average temperature
during cooling). Take 1 kW = 240 cal/sec.
Sol:
Rate of heat supplied by heater
= rate of heat lost by tank by radiation
(at steady state)
 2kW = C(65 –15)
(Where C is a constant)
or C =
2  240 48

cal / s C
65  15
5
At a temperature T between 65° and 50°C, the rate of heat
lost can be expressed as

dQ
 C  T  15
dt
where 15 = room temperature.
or  ms
or 
dT
 C  T  15
dt
dT
C

 dt
 T  15 ms
Integrating the expression within proper limits, we have

50
65
dT
C

dt
 T  15 ms 
65
C
t
or loge  T  15 50 
ms
90
 65  15 
C
t
or loge 

50

15
ms




C
 50 
or loge   
t
 35 ms
t 
ms
 50 
loge  
 35
C
2  10   1 log
6
=
 48 


5
e
 50 
 
35
Solving we get
t = 7.431 × 104 sec
= 20.64 hr
Ex.
Energy incident on the earth from sun is 1.4 kW m–2. If the
radius of the sun's sphere is 7.5 × 105 km and the distance
of the earth from the sun is 1.5 × 108 km, calculate the
black body temperature of the sun. Stefan's constant for
black body = 5.7 × 10–8 W m–2 K–4.
Sol: The black body radiation of sun
= AT4,
where  = Stefan's constant
= 5.7 × 10–8 m–2 K–4,
A = Surface area of sun
= 4 (7.5 × 108)2 m2
91
T = Black body temperature of sun.
 Black body radiation of sun =  AT4
= (5.7 × 10–8) {4 . (7.5 × 108)2} T4
…(1)
Here the energy distribution at a distance of 1.5 × 1011 m
from the sun is equal to the energy incident on earth per
unit area. The total energy radiated by the sun is given by
4 × (1.5 × 1011)2 × 1.4 × 103 watt,
…(2)
(because energy incident incident on the earth per unit
area is 1.4 × 103 watt m–2).
Now, black body radiation of sun =
sun
Energy
radiated
by
 (5.7 × 10–8) {4 (7.4 × 108)2} T4
= 4 × (1.5 × 1011)2 × 1.5 × 103
1.5  10 
11
4
or T =
2
 1.4  103

5.7  108  7.4  108

2
Solving we get
T = 5.6 × 103 K.
92
Ex.
A body which has a surface area 5.00 cm2 and a
temperature of 727°C radiates 300 joules of energy each
minute. What is its emissivity? Boltzmann constant = 5.67
× 10–8 watt/m2 (°K)4.
Sol:
The total energy radiated by the body of emissivity e,
surface area A at temperature T is given by
Q = EA t = e  (T4 – T04) At,
where T0 = temperature of surrounding and t = time.
When T > T0, then
Q = e  T4 At.
 300 = e × (5.67 × 10–8) (1000)4
(5.00 × 10–4) (60).
Solving we get,
e = 0.18.
93
Ex.
Consider a block of copper of radius 5 cm. Its outer surface
is coated black. How much time is required for the block to
cool down from 1000 K to 300 K? Density of copper = 9 ×
103 kg/m3 and specific heat = 4 kJ/kg.
Sol:
When the copper block is coated black, it behaves a
perfectly black body. The rate of heat energy radiated out
at any instant is given by
dQ
 AT 4
dt
…(1)
where A is the surface area and T is the temperature of
black body at that instant. Here it should be remembered
that T changes continuously with time as black body cools.
The heat given out by a block of mass m and specific heat
c, for unit change of temperature is given by
dQ
 mc
dT

…(2)
dT
A 4

T
dt
mc
…(3)
Negative sign is used to show that temperature falls as
time increases.
From eq. (3),
dt = 
mc dT
A T 4
…(4)
Hence, time required for the block to cool down from T1 to
T2 is given by

t
0
dt  
mc
A

T2
T1
dT
T4
94
Integrating, we get
t=
mc  1
1
 3  3
3  A  T2 T1 
mc  T13  T23 
=


3  A  T13 T23 
If  be the density of copper and r be the radius of sphere,
then
4

m =   r3  
3

and A = 4r2
r  c  T13  T23 
t 


9  T13 T23 
Here r = 5 cm = 5 × 10–2 m,
 = 9 × 103 kg/m3,
c = 4 × 103 J/kg,
T1 = 1000 = 103 K,
T2 = 300 K
and  = 5.67 × 10–8 J/m2 s K4.
Substituting these values and solving we get
t = 127 × 103 sec.
95
Ex.
A solid copper sphere (density  and specific heat c) of
radius r at an initial temperature 200 K is suspended inside
a camber whose walls are at almost 0°K. What is the time
required for the temperature of the sphere to drop to 100
K? (e = emissivity of copper).
Sol:
Let A be the area of copper sphere. Then according to
Stefan's law
E = e A  T4 (e = emissivity) …(1)
The rate of loss of heat is given by
dQ
 dT 
  mc 
 dt 
dt
…92
(negative sign is used to show that temperature decreases
with time)
From eqs. (1) and (2), we get
 dT 
 mc 
 e A  T4

 dt 
or 
dT e A 

 T4
dt
mc
 4
Here m =    r3
 3
and A = 4 r2 (r = radius)

or
dT 3e 4

T
dt
cr
cr  dT 
  dt
3e  T 4 
96
Integrating this expression within proper limits, we have


t
0
dt 
cr
3e

100
200
 dT 
 T 4 
100
cr  1 
t=
9e  T 3 200
=
7 cr
 106 sec.
72 e
Ex.
A hollow sphere of mass m is made of a material having
specific heat capacity s J/kg K. Inner and outer surfaces of
the sphere have are a m2 and A m2 respectively. Sphere is
suspended by a light non-conducting thread from ceiling of
a room. Assuming sphere material to be highly conducting
and its emissivity e, calculate time taken by the sphere to
cool from T1 °C to T2 °C when room temperature is T0 °C.
(Take Stefan's constant = W m–2 K–4).
Sol:
Here heat is radiated only from the outer surface of the
sphere as it is exposed to the room.
Let at some instant of time, the difference between
temperatures of sphere and room be T.
The net rate of radiation from surface of sphere
E = e A     T    4 


4
where   273  T0 


4T 
 1
E = e A  4  1 

 


= 4 e A 3 T
97
(
neglecting higher powers)
Further,
 dT 
E   ms 
 dt 
 dT 
  ms 
 4 e A  3 T

 dt 
dT
4 eA3

dt
or
T
ms
Integrating this expression within proper limits we get

T2  T0
T1  T0
dT
4eA3

T
ms

1
0
dt
Solving, we get
t=
 T1  T0 
ms
log
e 
4eA3
 T2  T0 
98
Ex.
Radiant energy from the sun strikes the earth at a rate of
1.4 × 103 watt/m2. Calculate the temperature of the
surface of sun. Radius of sun = 7 × 108 m, radius of earth'
orbit = 1.5 × 1011 m and Stefan's constant = 5.7 × 10–8 W
m–2 K–4.
Sol:
We know that
S R
T =  
r
2
4
 SR 
or T =  2 
 r 
2
1
4


1
1.4  103  1.5  1011 2  4

T
 5.7  108  7  108 2 




= 5801 K.
99
Ex.
Energy falling on 1.0 m2 area placed at right angles to a
sun beam just outside the earth's atmosphere is 1.35 K
joule in one second. Find sun's surface temperature. Mean
distance of earth from sun is 1.50 × 108 km, mean
diameter of sun = 1.39 × 106 km and Stefan's constant
5.67 × 10–8 watt m–2 K–4.
Sol:
We know that
2
S R
T =   .
r
4
Here
S 1.35 kilo joule / m2  sec

 5.67  108 watt / m2  K2
1.35  103 watt / m2
=
5.67  108 watt / m2  K4
(
watt = joule/sec)
= 2.38 × 1010 K4
R
1.50  108 km

 215.8
and
r
0.695  106 km

 T 4  2.38  1010 K4
 215.8
2
= 1108 × 1012 K4
T = 5.770 × 103 K
or T = 5770 K.
100
Ex.
A cylindrical block of length 0.4 m an area of cross-section
0.04 m2 is placed coaxially on a thin metal disc of mass 0.4
kg and of the same cross-section. The upper face of the
cylinder it maintained at a constant temperature of 400 K
and the initial temperature of the disc is 300 K. If the
thermal conductivity of the material of the cylinder is 10
watt/m-K and the specific heat of the material of the disc in
600 J/kg-K, how long will it take for the temperature of the
disc to increase to 350 K? Assume, for purposes of
calculation, the thermal conductivity of the disc to be very
high and the system to be thermally insulated except for
the upper face of the cylinder.
Sol:
The rate of flow of heat through the block at temperature
400 K and T is
dQ KA  400  T 

dt
L
=
10  0.04   400  T 
0.4
= (400 – T)
This heat is taken by the disc. The rate at which the heat is
taken by disc is given by
ms
dT
dT
dT
 0.4  600 
 240
dt
dt
dt
As the system is thermally insulated
240
dT
  400  T 
dt
 dT 
or dt = 240 
 400  T 
101
Integrating the above expression, we get
t = 240

350
300
 dT 
 400  T 
=  240 log  400  T 300
350
= – 240 [log 50 – log 100]
 100 
= 240 log 
 50 
= 240 loge 2
= 240 × 2.303 log10 2
= 240 × 2.303 × 0.3010
= 166.3 sec.
Ex.
An iron wire of diameter 1 mm and length 10 cm is placed
in an evacuated chamber. Estimate the equilibrium
temperature of the wire if it carries a current of 10 amp.
Assume that all heat transfer is by radiation and that the
surface of the wire radiates according to Stefan's law. Take
the temperature of the chamber walls to be 27°C. Specific
resistance of the wire is 10 × 10–8 ohm meter.
Sol:
If T K be the equilibrium temperature, then the radiation
from the surface of the wire
= s (T4 – 3004) per m2
…(1)
Surface area of wire = 2rl
 radiation of the wire = 2rl  (T4 – 3004)
102
Heat produced per second = i2 R
l 

=  i2 r2 

 


l
 R   r2 ,    specific resistance 


…(2)
From eqs. (1) and (2), we get
l 

2  r l  T 4  3004   i2 r2 

 


Solving we get
i2 
4
T =

300


22 r3 
4
10  10  108 
4

300


2
2  3.14  0.5  109 5.67  108 
2
=
= 178.6 × 108 + 81 × 108
= 401 K
= 128°C
103
Ex.
A wire of length 1.0 m and radius 10–3 m is carrying a
heavy current and is assumed to radiate as a black body.
at equilibrium its temperature is 900 K while that of the
surroundings is 300 K. the resistivity of the material of the
wire at 300 K is 2 × 10–8 m and its temperature
coefficient of resistance is 7.8 × 10–3 per °C. Find the
current in the wire.
(Given:
K–4)
Sol:
Stefan's
constant
=
5.68
×
10–8
W
m–2
According to Stefan's law
E = (T4 – T04)
A = (T4 – T04) × 2  rl
= (5.68 × 10–8) [(900)4 – (300)4] [2 × 10–3 × 1.0]
= 5.68 × 6480 × 2 × 10–3 Watt …(1)
The resistivity of wire at 900 K is given by
900  300 1   T 
= 2  108 1  7.8  103  600
= 2  5.68  108 ohm-metre…(2)
The resistance of the wire at 900 K is given by
R900 = 900 

1
l
 900  2
a
r

= 2  5.68  108 

1.0
 103
=   5.68  102 ohm.

2
…(3)
104
Now P = i2


R900 = i2   5.68  102 watt
…(4)
In equilibrium
P=E


 i2   5.68  102  5.68  6480  2  103
i2  648  2  1296
or I = 36 amp.
Ex.
The peak emission from a black body at a certain
temperature occurs at a wavelength of 9000 Å. On
increasing the temperature, the total radiation emitted is
increased 81 times. At the initial temperature when the
peak radiation from the black body is incident on a metal
surface it does not cause photo-emission from the surface.
After the increase of temperature the peak radiation from
the black body caused photo-emission. To bring these
photoelectrons to rest, a potential equivalent to the
excitation energy between the n = 2 to n = 3 Bohr's levels
of hydrogen atom is required. Find the workfunction of the
metal.
Solution: Suppose the initial absolute temperature of the black
body be T1 which is increased to T2. According to Stefan's
law
E  T14
and 81E  T24
105
4
T 
  2   81
T 
1
or
T2
3
T1
or T2 = 3T1
…(1)
Now applying Wien's displacement law, we have
1T1  2 T2  2  3T1 
or 9000 T1 = 2 × (3T1)
or 2 = 3000 Å
…(2)
This radiation of
emission.
wavelength 3000
Å
causes photo-
Let V0 be cut-off potential. Then
e V0 
hc
W
2
where W = work-function.
W 
hc
 eV0
2
….(3)
Here,


6.62  1034 3  108
hc

2
3000  1010



= 6.62 × 10–19 joule
=
1.62  1019
eV
1.6  1019
106
= 4.14 eV
eV0 = difference between n = 2 and n = 3 Bohr's levels
= E 3 – E2
 1 1  17
= 13.6    
9
9 4


En  
13.6

eV
2

n
= 1.89 eV
Substituting these values in eq. (3), we get
W = 4.14 – 1.89
= 2.25 eV
Ex.
An indirectly heated filament is radiating maximum energy
of wavelength 2.16 × 10–5 cm. Find the net amount of heat
energy lost per second per unit area, the temperature of
surrounding air is 13°C. Given b = 0.288 cm-K,  = 5.77 ×
0–5 erg/s-cm2-K4.
Sol:
If T is the surface temperature of the filament, then
mT  b
or T =
b
0.288

m 2.16  105
= 13333.3 K
The temperature of surrounding air
T0 = 13 + 273 = 286 K
107
The net amount of energy radiated per unit area per
second

E   T 4  T04

4
4
= 5.77  105 13333.3  286 


= 1.824 × 1012 erg/s-cm2 Ans.
Ex.
Two bodies A and B have thermal emissivity of 0.07 and
0.81, respectively. The outer surface areas of the two
bodies are the same. The two bodies emit total radiant
power at the same rate. The wavelength B corresponding
to maximum spectral radiancy in the radiation from B is
shifted from the wavelength corresponding to maximum
spectral radiancy in the radiation from A by 1.00 µm. The
temperature of A is 5802 K. Find temperature of B and
wavelength corresponding to maximum spectral radiancy.
Sol:
If TA and TB are the temperatures of the bodies A and B
respectively, then
 ATA4  BTB4
 
 TB   A 
 B 
1
4
1
 0.01 4
TA = 
 5802
 0.81
= 1934 K
108
By Wien's displacement law, we have
 A TA  B TB
or  A  5802  B  1934
or  A 
B
3
…(i)
It is also given
 B   A  106
…(ii)
Solving equations (i) and (ii), we get
B  1.5  106 m
= 1.5 µm Ans.
Solar spectrum
When light from the sun is seen through a spectrometer,
there observed several dark lines over continuous
spectrum. These dark lines are called Franhoffer liens. By
comparing the wavelengths of these dark lines with those
emitted by elements on the earth, we have indentified
various elements like H, He, Na, N2 etc. in the atmosphere
of the sun.
109
Gas Laws
Matter has three states namely solid, liquid and gas. When external
pressure is applied on the solids and liquids, the change in their volume is
negligible. But when an external pressure is applied on a gas, its volume
changes considerably. Similarly, change in temperature of a gas has also a
great impact on the volume and pressure of the gas. Thus, to state the
condition of a gas, the volume, pressure and temperature of the gas must
be specified. A change in anyone of these quantities (i.e. temperature,
volume and pressure) affects the other quantities.
The relationship between any two physical quantities used to specify the
state of a gas keeping the third physical quantity to be constant is known
as gas law.
Now let us discuss the various gas laws:
(i) Boyle's law: According to this law, the volume (V) of a fixed mass of a
gas is inversely proportional to the pressure (P) of the gas, provided
temperature of the gas is kept constant.
V1/T if T = constant
Cons tan t
Or V 
P
Or PV = constant ...(1)
If P1 and V1 be the initial pressure and initial volume of a gas respectively
at constant temperature and P2 and V2 be the final pressure and final
volume of a gas respectively at the same temperature, then eqn. (1) can
be written as
r1V1 = P2V2 ...(2)
The variation of pressure (P) with volume (V) at constant temperature is
shown in figure (a).
The variation of P with 1/V at constant temperature is shown in figure (b).
110
Boyle's law in terms of molecules per unit volume
Let us suppose there are N molecules in the volume V of a gas. Then the
number of molecules per unit volume of the gas is given by
N
n
V
N
V
...(3)
n
Vn
Using eqn. (3) in eqn. (1), we get
N
P    constant
n
P constant

n
N
Since number of molecules (N) in a given volume of gas are constant,
hence eqn. (4) can be written as
P
 Cons tan t
...(5)
n
If pressure of a gas changes from P1 to P2 at constant temperature and the
corresponding number of molecules per unit volume changes from n1 to n2,
then eqn. (5) can be written as
P1 P2

...(6)
n1 n2
Condition under which gases obey Boyle's law
The variation of the volume (V) of a gas with the pressure (P) at different
constant temperatures are studied. The graphs showing the variation of P
with V at different temperatures (T 1 and T2' T2 > T1) are represented by
the dotted lines (Figure). The solid curves in figure represent the variation
of P with V of the gas at T1 and T2.
It is clear from figure that the experimental curve (dotted) and the
theoretical curve (solid) are almost identical only at high temperature and
111
low pressure. Thus, Boyle's law is obeyed by gases only at high
temperature and low pressure.
(ii) Charle's law: According to this law, the volume (V) of a given mass
of a gas is directly proportional to the temperature of the gas, provided
pressure of the gas remains constant.
Let Vo be the volume of a gas at O°C and Vθ be the volume of the gas at
θ°C. Now change in volume is directly proportional to (i) original volume
i.e. volume at 0°C and (ii) change in temperature.
i.e. Vθ - Vo  VO (θ - 0)
or Vθ - Vo = Voγpθ
... (1)
where γp = constant of proportionality and known as coefficient of
expansion of gas at constant pressure or volume coefficient of the gas.
Eqn. (1) can be written as
Vθ = Vo (1+ γpθ)
V  V0
1
From eqn. (1), p 

(exp erimentally)

V0
273.15
Hence eqn. (2) can be written as



V  V0 1 
...(3)
273.15 

Eqn. (3) shows that the volume of a given mass of a gas increases or
1
decreases by
of its volume at 0°C for each 1°C rise or fall in
273.15
temperature. !
When θ = - 273 .15°C, then from eqn. (3), it is clear that volume of gas at
- 273· 15°C becomes zero. But if θ is less than - 273 '15°C or the gas is
cooled to a temperature less than - 273.15°C, then the volume of gas
becomes negative. However, the volume of a gas cannot be negative. It
means, the lowest temperature which can be achieved is – 273.15°C.A
temperature scale whose zero coincides with – 273.15°C is known as
absolute scale of temperature or kelvin scale of temperature. The
temperature = – 273.15°C is called absolute zero. The temperature O°C
corresponds to 273.15 K on absolute scale and denoted by To That is, To =
112
273.15K. The temperature (273.15 + θ) can be written as T on absolute
scale.
Now eqn. (3) can be written as Vo = Vo 273.15
Substituting 273.15 = To and 273.15 + θ = T, we get
T
V V0
V  V0
or

T0
T
T0
V  T, provided P = constant
Or V/T = constant ...(4)
Thus, Charle's law can be stated as:
The volume of a given mass of a gas is directly proportional to the absolute
temperature of the gas, provided its pressure remains constant.
The variation of volume (V) of a gas with its absolute temperature (T) must
be a straight line (dotted) as shown in figure. However, the experimental
graph between the values of V and T of a gas is not exactly a straight line
as shown by a solid curve. It is clear from the solid curve that the variation
of V is directly proportional to T only at a high temperature. Thus, gases
obey Charle's law only at high temperature.
Gay Lussac's Law (or Pressure Law)
According to this law, the pressure P of a given mass of a gas is directly
proportional to its absolute temperature T, provided the volume V of the
gas remains constant.
Perfect Gas Equation
A gas which fulfils the following conditions is known as a perfect gas or an
ideal gas:
(i)
the molecules of the gas are point masses i.e. the size of the
molecules of the gas is negligible,
113
(ii)
the intermolecular force among the molecules of the gas is zero
and (iii) the gas obeys gas laws (i.e. Boyle's law, Charle's law
and pressure law).
Perfect Gas Equation
The constant is equal to R, known as universal gas constant. Hence eqn.
can be
PV
 R or PV = RT
T
Above eqn. is known as perfect gas equation for one mole of gas.
If a gas has μ moles, then eqn. becomes PV = μRT
SI Unit of R
PV
R
=
T
Nm2xm3
J


 Jmol1K 1
molxK
molxK
Value of universal gas constant (R)
For 1 mol of perfect gas PV = RT
PV
Or R 
...(1)
T
114
Postulates of Kinetic theory of gases
Kinetic theory of gases is based on the following basic assumptions.
1. A gas consists of very large number of molecules. These molecules are
identical, perfectly elastic and hard spheres. They are so small that the
volume of a molecule is negligible as compared to the volume of the gas.
2. The molecules do not have any preferred direction of motion but their
motion is completely random.
3. These molecules travel in straight lines and are in free motion most of
the time. The time interval of the collision between any two molecules is
very small.
4. The collision between molecules and the wall of the container is
perfectly elastic. It means kinetic energy and linear momentum are
conserved in such collision.
5. The path travelled by a molecule between two successive collisions is
called free path and the mean distance travelled by a molecule between
two successive collisions is called mean free path.
6. The motion of molecules is governed by Newton's law of motion.
7. The effect of gravity on the motion of molecules is negligible.
Expression for the Pressure of a Gas
Let us suppose that a gas is enclosed in a cubical box of each side l
(Figure). Let there be n identical molecules, each having mass m. Since the
molecules are of same size and perfectly elastic, so their mutual collisions
result in the interchange of velocities only. Thus, only collisions with the
walls of the container contribute to the pressure by the gas molecules.
Now consider a molecule of mass m, moving with velocity c1. If μ1,1 and
ω1 be the components of the velocity c1 along x-axis, y-axis and z-axis
respectively, then
c12  12  12  12
115
The momentum of the molecule along x-axis = mμ1
When the molecules strike the wall BCHE, it rebounds with the same speed
towards the face ADGF.
 The momentum of the molecule after collision = - mμ1
The change in momentum of the molecule after collision
= – mμ1 – mμ1 = – 2mμ1
The time taken between the successive impacts or collisions
=Distance/Velocity
Dis tance 2l


Velocity
1
The momentum transferred in each collision to the wall of container
= 2mμ1
Time rate of change of momentum due to collision
2
2m1 m 1


2l
l
1
According to Newton's second law of motion, the force due to the impact
of first molecule of the mμ2 gas along x-axis is
m21
F1x 
l
Similarly, forces due to the impact of other molecules of the gas along Xm22 m23
mn2
axis are
respectively.
,
..........
l
l
l
116
Hence, the net force due to the impact of n molecules of the gas along xaxis is
m12 m22 m23
mn2
Fx 


..........
l
l
l
l
m 2
Fx 
1  22  23..........n2
l


The average pressure on the face BCHE due to impacts of the molecules is
Fx Fx m 2
Px 
 2  3 1  22  23..........n2
A
l
l


Similarly, the pressure on the walls of the container along Y and Z direction
are
m
Py  3 12  22  23..........n2
l
m
Pz  3 12  22  23..........n2
l




 Mean pressure exerted by the gas on the walls of the container is
Px  Py  Pz
3
m
 3  12  22  32..........n2 12  22  32..........n2 12  22  32..........n2 

l 
m
 3  12  12  12  22  22  22  .......  n2  n2  n2 

l 
P



 
Using eqn. (1), we get
m
P  3 c12  c22  c32..........cn2
l







But l3 = V, volume of the container
117
m 2
mn (c12  c22  c32..........cn2 )
2
2
2
P
c  c2  c3..........cn 
3V 1
3V
n


(c12  c22  c23..........cn2 )
Now mn = M, mass of the gas and
 C2
n
1
 c12  c22  c23..........cn2  2
or 
 where C is the root mean square velocity.
n


M 2
P 
C
3V
1
...(2)
P  C2
3
M
where   , density of the gas.
V
3P
From (2), C2 

Or
C
3P

...(3)
Relation between pressure and kinetic theory of gas
1
2 1
We know, pressure exerted by a gas is P  C2 = X C2
3
3 2
2
Or P  E
3
1
3
Where E  C2 is the K.E. per unit volume of the gas. Thus E  P
2
2
3
Thus, K.E. per unit volume of the gas is numerically equal to
times the
2
pressure exerted by the gas.
118
Kinetic interpretation of Temperature and Absolute Temperature
From kinetic theory of gases, the pressure exerted by one mole of an ideal
gas is given by
1
1M 2
P  C2 
C
3
3V
1
Or PV  MC2
3
But according to gas equation PV = RT .
1
 MC2  RT
3
or C 
3RT
where M = molecular weight
M
Now M = mNA (where NA is Avogadro's number) 
Multiply both sides by
1
mNAC2  RT
3
3
we get
2
1
3
mNAC2  RT
2
2
1
3 R
T
Or  mC2 
2
2 NA
R
Now
= k, Boltzmann's constant = 1·38X 10-23 J mole-1 K-1
NA
1
3
 mC2  kT
2
2
This is known as the average thermal energy of the atom of the gas.
1
or mC2  T
2
Thus, temperature of a gas is directly proportional to the mean kinetic
energy of per molecule of the gas.

119
Absolute temperature
1
 1

mC2 =0 or C=0. 
m  0
2
 2

Thus, absolute zero temperature is that temperature at which the
molecular motion ceases" (or molecules of the gas come to rest.)
If T=O, then
(i) Boyle's law.
1M 2
1
C or PV  MC2
3V
3
1
Since C2  T, therefore at constant temperature MC2 is also constant.
3
 PV = Constant or V  1/P
If pressure and volume of a given mass of a gas changes from P1, V1 to P2,
V2 then
P1V1 = P2V2 = constant .
..
According to kinetic theory of gases, P 
(ii) Charle's Law.
According to kinetic theory, P 
1M 2
1
1
C or PV  MC2 or PV  mNC2
3V
3
3
N is Avogadro's number.
But mean kinetic energy of a molecule is
1
3
mC2  kT or mC2  3kT
2
2
1
NAx3kT  NAkT
3
NAk
T
Or V 
P
If P is constant, V  T or V/T = constant.
If volume and temperature of a given mass of a gas at constant pressure
changes from V1,T1 to V2,T2 then
V1 V2

 cons tan t
T1 T 2
 PV 
120
(iii) Avogadro's hypothesis.
Consider two samples Aand B of the gases each having volume V and
pressure P.
Let m1 and m2 be the mass of each molecule of gas A and gas B
respectively. N1 and N2 be the number of molecules of gas A and B
respectively.
mNA 2
According to kinetic theory, P 
C ...(i)
3V
1 m1N1 2
 For gas A, P 
C1 where C1 = rms velocity of the molecules of
3 V
gas A.
1 m2N2 2
And for gas B, P 
C 2 ... (ii)
3 V
where C2 = rms velocity of the molecules of gas B.
From (i) and (ii),
1 m1N1 2 1 m2N2 2
C1 
C2
3 V
3 V
Or m1N1C21  m2N2C22 ...(iii)
If the temperature of gas A and gas B is same, then their average
K.E./molecule is also same.
1
1
Thus m1C21  m2C22 or m1C21  m2C22 ... (iv)
2
2
Dividing (iii) by (iv), we have N1 = N2 (Avogadro's hypothesis)
Thus, equal volumes of all gases under similar conditions of temperature
and pressure have the same number of molecules.
(iv) Graham's Law of Diffusion of Gases.
It states that rate of diffusion of a gas is inversely proportional to the
square root of the density of the gas.
121
According to kinetic theory, P 
C
1 2
C
3
3P

Or C 
1
, if P is constant

Since root mean square velocity is proportional to the rate of diffusion (r)
of the gas,
r
1

Thus, denser the gas, the slower is the rate of diffusion.
(v) Regnault's or Gay Lussac's Law.
1M 2
C
3V
Now at constant volume and for a given mass of gas P  C2
But C2  T
 P  T or P/T = constant which is Regnault's or Gay Lussac's law.
According to kinetic theory of gases, P 
(vi) Gas Equation or Equation of state
1M 2
1
According to kinetic theory of gases, P 
C or PV  MC2
3V
3
For a given mass of a gas, PV  C2
But C2  T.  PV  T or PV = RT, where R is universal gas constant.
This is known as gas equation for 1 mole of gas.
For μ moles of a gas I PV = μRT
122
(vii) Dalton's Law of partial pressure.
According to this law, the resultant pressure exerted by a mixture of noninteracting gases is equal to the sum of their individual pressures.
Consider a number of gases in a container which do not interact with each
other. Let ρ1, ρ2, ρ2,……… be their densities and C1,C2,C3,……… be their
rms velocities.
Therefore, pressure exerted by the mixture of the gases is
1
1
1
P  1m1C21  2m2C22  3m3C23.......
3
3
3
1
1
1
Where P1  1C21 ,P2  2C22 ,P3  3C23
3
3
3
P =P1 +P2 +P3 + ...., which is Dalton's law of partial pressure.
Mean, Root Mean Square and Most Probable Speeds
Maxwell and Boltzmann gave a law governing the speed distribution of gas
molecules. According to Maxwell-Boltzmann's speed distribution law, the
fraction of molecules having speeds between  and  + d is given by
3
n()
 M  2 2 M2 2RT
...(1)
f() 
 4 
 e
n
 2RT 
where M = molecular mass of the gas
T = Absolute temperature of the gas
The variation of f() with speed  at a temperature T is shown in figure.
The graph shown by figure indicates that the speed of the gas molecules
varies from 0 to .
123
Mean or average speed of the molecules of a gas is given by

av   f()d
0
3

 M 
  x4 

 2RT 
0
3
 M 
 4 

 2RT 
We know
2
2
e
2 
3
 e
M2

3  ax
 x e dx 
2
3
 M 
 4 

 2RT 
 8RT 


 M 
av
1
2
2RT
2RT
d
d
0
0
av
M2
2
x
1
2
 M 
2

 2RT 
 8RT 


 3.14M 
 2.55RT 


M


1
1
(standard integral)
2a2
1
2
 RT 
 1.6 

 M 
2
1
2
...(2)
Root Mean square speed of the molecules of a gas is given by

rms  c   2f()d
2
2
0

3
 M 
   x4 

 2RT 
0
2
3
 M 
 4 

 2RT 

Now
x e
0
4
 ax2
2 
2
4
 e
2
e
M2
M2
2RT
2RT
d
d
0
 9 
dx  
5 
 64a 
1
2
(standard integral)
124




9
 M  2
2
  3RT
 c  4 

5
M
 2RT    M  
 64 


  2RT  
3
2
rms
12
3RT
 RT 
 1.73 

M
 M 
rms  c 
...(3)
Most probable speed of the molecules of a gas is the speed at which
fraction of molecules f(v) is maximum.
df()
That is
0
d
3
d   M  2 2 M2 2RT 
4
0
e
Or
d   2RT 



3
 M 
4 

 2RT 
2
d  2 M2 2RT 
e
  0
d 
12
2RT
 2RT 
or   
 

M
 M 
This speed is known as most probable speed of molecules and is denoted
by mp
2
12
12
 2RT 
 RT 
mp  

1.41

 M 
 M 


From eqns. (2), (3) and (4), we conclude that
rms > av > mp
125
Specimen Numerical
The molar mass of nitrogen is 14 gmol-1. Calculate (i) average speed
(ii) root mean square speed and (iii) the most probable speed of nitrogen
molecule at 27°C.
Here, M= 14gmol-1=14 x 10-3 kg mol-1; T=27 +273 =300K;
R = 8.31 Jmol-1K-1
12
 RT 
av  1.6 

 M 
12
rms
 RT 
 1.73 

 M 
mp
 RT 
 1.41 

 M 
12
 8.31x300 
 1.6 
3 
 14x10 
12
 1.6x422  675.2ms1
12
 8.31x300 
 1.73 
3 
 14x10 
 1.73x422  730.06ms1
12
 8.31x300 
 1.41 
3 
 14x10 
 1.41x422  595.02ms1
Degrees of freedom
The total number of independent quantities or coordinates which must be
known to completely specify the position or configuration of a dynamical
system is called the degrees of freedom of that system.
Or
The total number of independent ways in which the particles of a system
can acquire energy are called the degrees of freedom of that system.
In general, degrees of freedom (f) In a system is given by f=3N-K ...(1)
where N = number of particles of a system, K = number of common
coordinates of the particles of the system or independent relations.
20 September
126
Illustrations
(i) Monoatomic gas molecule: Monoatomic gas molecule is made of a single
atom. i.e. N = 1, so K = 0  From eqn. (1), f= 3 X 1 - 0 = 3
Thus, the degree of freedom of monoatomic gas molecule is 3.
(ii) Diatomic gas molecule: Diatomic gas molecule is made of two atoms'
i.e. N = 2, so K= 1  From eqn. (1),f = 3 X 2 - 1 = 5
Thus, the degree of freedom of diatomic gas molecules is 5.
(iii) Triatomic gas molecule: Triatomic gas molecules are of two types:
(a) Linear triatomic gas molecule which has two atoms on either side of a
central atom (Figure). In this case, N = 3 and K = 2.
 f=3 x 3 - 2=7
Thus, degree of freedom of linear triatomic gas molecule is 7.
(b) Non-Linear triatomic gas molecule which has three atoms at the
vertices of a triangle (Figure 7). In this case, N = 3 and K = 3.
f=3x3-3=6
Thus, degree of freedom of non-linear triatomic gas molecule is 6
Degree of freedom of a system or a gas molecule in terms of
energy
A system or a gas molecule can have three types of motion. That is
(i) translational motion (ii) rotational motion and (iii) vibrational motion.
Thus, a system can have three types of energies i.e. kinetic energy of
translation, kinetic energy of rotation and kinetic energy of vibration.
Kinetic energy of Translation is the energy possessed by a system or a gas
1
molecule due to its linear motion. It is given by, T = mv2
2
Kinetic energy of Rotation is the energy possessed bya system or a gas
molecule due to its rotational motion. It is given by:
1 2
Tr =
Iω
2
127
where I is moment of inertia of the system or a gas molecule about the
given axis of rotation and ω is its angular velocity.
Kinetic energy of vibration is the energy possessed by a gas molecule due
to its vibration about its equilibrium or mean position. Since Vibratory
system has kinetic and potential energies, so energy of a vibratory
molecule =K.E. + P.E.
Degree of freedom of Monoatomic gas molecule. Monoatomic gas molecule
can move in a straight line as well as rotate about its axis. Thus, total
energy of monoatomic gas molecule is given by
1
1
E = K.E. of translation + K.E. of rotation = mv2 + Iω2 ...(1)
2
2
Since monoatomic gas molecule behaves as a point mass, so its moment of
inertia about its axis of rotation is zero i.e. I = 0. Hence eqn. (1) becomes
1
E = mv2
... (2)
2
Resolve velocity v into three components Vx, Vy and Vz along x-axis, y-axis
and z-axis respectively.
 2  2x  2y  2z
Hence eqn. (2) can be written as E 
1
m(2x  2y  2z )
2
1
m2x  m2y  m2z
2
Thus, monoatomic gas molecule can acquire energy in three ways i.e.,
along x-axis, v-axis and z-axis respectively. Therefore, degree of freedom
of monoatomic gas = 3
Or E 
128
Degree of freedom of a diatomic gas molecule
Diatomic gas molecule is made of two atoms. These atoms are held
together at a fixed distance by an interatomic force (i.e. bond). This force
is represented by a spring and the atoms are attached to its two ends
(Figure). At ordinary temperature, diatomic molecule possesses translatory
and rotational motion.

Total energy of the molecule is given by
1
1
E= m2 + Iω2
...(1)
2
2
Resolve , I and ω into their components. Hence eqn. (1) can be written as
1
1
1
1
1
1
E  m2x  m2y  m2z  Ix2x  Iy2y  Iz2z
2
2
2
2
2
2
Let us suppose the molecule rotates about x-axis. Then, moment of inertia
of the molecule about x-axis is zero i.e. Ix = 0
Hence eqn. (2) becomes
1
1
1
1
1
E  m2x  m2y  m2z  Iy2y  Iz2z ... (3)
2
2
2
2
2
Thus, diatomic gas molecule can acquire energy in five independent ways
as given by eqn. (3).
Therefore, degree of freedom of diatomic gas molecule at ordinary
temperature = 5
Degree of freedom of a solid
In a solid, each atom can only vibrate about its equilibrium position. Atom
in a solid cannot move from one position to another position and cannot
rotate about the given axis of rotation. Therefore, the atom in a solid does
not possess translational energy and rotational energy. It has only
vibrational energy at high temperature. Vibrational energy of an atom is
the sum of kinetic energy and potential energy. Since it can vibrate along
129
three axis, therefore, the atom in a solid has 6 degrees of freedom (i.e., 3
X 2 = 6) at high temperature.
The law of equipartition of Energy
Consider a molecule of gas of mass m, moving with velocity v. The kinetic
energy of the molecule is given by
1
1
m2  m(2x  2y  2z )
2
2
1
1
1
E  m2x  m2y  m2z
2
2
2
E
[ 2  2x  2y  2z ]
But the mean or average kinetic energy of a gas molecule is given by
1
1
1
3
 E  m2x    m2y    m2z  kT
...(2)
2
2
2
2
Since all the three directions i.e. x-axis, y-axis and z-axis are equally
preferred, so the average kinetic energy of the gas molecule along all the
three directions is equal.
Thus, from eqn. (2),
1
1
1
1
 m2x  m2y  m2z  kT
2
2
2
2
A gas molecule moving in space has three degrees of freedom. So, (he
1
energy associated with each gas molecule per degree of freedom is
kT.
2
This fact is known as law of equipartition of energy.
130
Definition of law of equipartition of energy
According to this law, for any system in thermal equilibrium, the total
energy of the system is equally distributed among its various degrees of
freedom and the energy associated with each degree of freedom per
1
molecule is
kT. Here k = 1·38 X 10-23 JK-1 is the Boltzmann constant and
2
T is the absolute temperature of the system.
Total Energy of a System
Consider a system having N molecules. Let f be the degrees of freedom of
each molecule. Therefore, total degrees of freedom of the system = Nf.
According to law of equipartition of energy, energy associated with each
1
degree of freedom of a molecule = kT
2
 Total energy of the system = Number of degrees of freedom
x energy associated with one degree of freedom
1
1
or U = Nf x kT = fRT
...(3)
2
2
Specific Heats of Gases in Terms of Energy
(i) Monoatomic gas: A monoatomic gas molecule consists of a single
atom. For example, helium, argon etc. A mono atomic gas molecule has 3
degrees of freedom.
According to the law of equipartition of energy,
1
Energy associated with each degree of freedom = 3 x kT ...(3) ( kN = R)
2
131
1
3
kT = kT
2
2
3
Total energy associated with 1 mole of gas, U = NAkT
2
where NA is the number of molecules in 1 mole of a gas.
But, NAk = R
3
Hence eqn. (1) becomes U = RT
...(2)
2
dU 3
 R But Cp - C = R
Now C =
dT 2
3
5
 Cp =C +R = R+R= R
2
2
3 Cp 5 2 R 5

  1.67
Now γ = 
2 C 3 2 R 3
 Energy associated with one molecule =3 x
...(1)
(ii) Diatomic gas: A diatomic gas molecule consists two atoms. For
example, hydrogen, nitrogen, oxygen etc. A diatomic gas molecule has 5
degrees of freedom at room temperature.
According to the law of equipartition of energy,
1
Energy associated with one degree of freedom = kT
2
5
 Energy associated with one molecule = kT
2
5
Total energy associated with one mole of gas, U = NAkT
2
Now
Now,
But NA x k = R
5
U=
RT
2
5
7
dU 5
C 
 R and Cp =C +R =
R+R= R
2
2
dT 2
132
7
Cp 2 R 7
 

  1.40
C 5
5
R
2
At very high temperature, a diatomic gas molecule has 7 degrees of
freedom.
7
 Total energy associated with one mole of diatomic gas, U = RT
2
dU
d 7
 7
C 

RT
  2R
dT dT  2

Cp  C  R 
 
7
9
R R  R
2
2
Cp 9
  1.28
C 7
Triatomic gas
(a) If the triatomic gas molecule is non-linear, f = 6.
Hence total energy associated with one mole of gas is
6
6
U = NAkT =
RT
2
2
dU 6
 R  3R and Cp =C +R = 3R + R = 4R
Now C 
dT 2
Cp 4R
 

 1.33
C 3R
(b) If the triatomic gas molecule is linear, f = 7
7
7
U = NAkT =
RT
2
2
7
9
dU 7
 R and Cp =C +R = R + R = R
Now C 
2
2
dT 2
133
 
Cp 9
  1.28
C 7
Examples of triatomic gases: SO2, CO2, H2S etc.
Important Relations: In general, Cv =
and  
f
f

R, Cp =   1  R
2
2

Cp (f / 2  1)R
2

1
C
f / 2R
f
where f is the degree of freedom of molecule.
According to Dulong and Petit, the average molar specific heat at room
temperature is the same for all metals and is nearly equal to 3R or 6 .0 cal
mol-1K-1 (or 24.93 J mol-1 K-1).
The variation of Cv with temperature T for solids is shown in figure.
According to the law of equipartition of energy, the energy of each atom of
1
metal = 6 x
kT ,= 3kT
2
 Energy associated with N atoms of the metal, U = 3NkT
But N x k=R
U=3RT
dU
 3R cal mol-1 K-1 (or 24.93 J mol-1 K-1)
Now C 
dT
The temperature, at which all metals have constant C, is called debye
temperature.
134
Concept of Mean Free Path
The molecules of a gas move with high speeds at a given temperature but
even then a molecule of the gas takes a very long time to go from one
point to another point in the container of the gas. This is due to the fact
that a gas molecule suffers a number of collisions with other gas molecule
surrounding it. As a result, the path followed by a gas molecule in the
container of the gas is zig-zag as shown in figure. During two successive
collisions, a molecule of a gas moves in a straight line with constant
velocity and the distance travelled by a gas molecule between two
successive collisions is known as mean free path. The distance travelled by
a gas molecule between two successive collisions is not constant and hence
the average distance travelled by a molecule during all collisions is to be
calculated. This average distance travelled by a gas molecule is known as
mean free path.
Let γ1, γ2, γ3, γ4,..... γn be the distance travelled by a gas molecule during n
collisions respectively, then the mean free path of a gas molecule is given
by
1  2  3  ..........n

n
Expression for mean free path
Assumptions
(i) The molecules of a gas are considered as hard spheres, each of
diameter „d‟.
(ii) The collisions between gas molecules are perfectly elastic.
(iii) All molecules of a gas except the molecule under consideration are at
rest.
135
(iv) A molecule of the gas under consideration collides with all those
molecules whose centre is at a distance 'd' (diameter of the molecule) from
the centre of the molecule under consideration.
Consider a gas in a container having n molecules per unit volume. Let 'd'
be the diameter of a molecule (A) which is assumed to be in motion, while
all other molecules are at rest. The molecule A collides with other
molecules like Band C whose centres are at a distance ‟d‟ from the centre
of the molecule A as shown in figure. If the molecule moves a distance L,
then this molecule makes collisions with all the molecules lying inside a
cylinder of volume πd2L.
The number of molecules in a cylinder of volume πd2L
= number of molecules per unit volume x volume of cylinder
= n x πd2L = πnd2L
 Number of collisions suffered by the molecule A
= number of molecules in the cylinder of volume πd2L
= πnd2L
Now, mean free path of a molecule is given by
λ = Total distance travelled/Number of collisions suffered
= L/πd2L
1

...(1)
nd2
Eqn. (1) has been derived by assuming that all molecules of the gas except
one molecule under consideration are at rest. But it is not true. In fact, all
the molecules of the gas are in motion. Under such condition, the molecule
under consideration suffers a large number of collisions. Hence, the mean
free path is less than as calculated by eqn. (1). It has been estimated that
the mean free path is given by
1

... (2)
2nd2
N
Since n 
V
136
eqn. (2) becomes  
But PV = NkT or
 
1
2d2N / V
N
P

V kT
kT
... (3)
2d2P
Factors on which Mean Free Path depends
From eqn. (3), it is clear that mean free path of a gas molecule is :
(i) Directly proportional to the absolute temperature (T) of the gas.
(ii) Inversely proportional to the pressure (P) of the gas.
and (iii) inversely proportional to the square of the diameter of the gas
molecule.
Numerical
i)
ii)
Calculate the mean free path of nitrogen molecule at 27°C when
pressure is 1.5 atm. Given, diameter of nitrogen molecule = 1.5Å,
k = 1.38 x 10-23 KJ-1
If the average speed of nitrogen is 675 ms-1, find the time taken
by the molecule between two successive collisions and the
frequency of collisions.
Solution.
(i) Here T = 27 + 273 = 300 K
P = 1 atm = 1·01 x 105Nm-2
d =1·5Å=1·5x10-10m
k = 1·38 X 10-23 JK-1
kT
Using  
, we get
2d2P
1.38x1023 x300

 4.1x107 m
10 2
5
1.41x3.14x(1.5x10 ) x1.01x10
137
(ii) Time interval between two successive collisions, t =
Dis tance 

Speed

4.1x107

 0.006x106  0.006s
675
1
1
 166.6x106 s1
Collision frequency = 
6
t 0.006x10
Brownian Motion
The random or zig-zag motion of small particles suspended in a fluid is
called Brownian motion. The irregular motion of smoke and dust particles
in air is the example of Brownian motion. Brownian motion was discovered
by Brown, an English botanist in 1827.
Reason of Zig-Zag motion
The suspended particles are extremely large in size as compared to the
molecules of the fluid and are being continually bombarded on all sides by
them. If the particles are sufficiently large, equal numbers of molecules
strike the particles on all sides at each instant. For smaller particles, the
number of molecules striking the various sides of the particle at any instant
may not be equal. Hence the particles at each instant suffer an unbalanced
force causing it to move this way or that way. This is the reason of the zigzag motion of the particles.
Factors affecting the Brownian motion
(i) Temperature of the medium. Brownian motion increases with the
increase of the temperature of the medium in which the tiny particles are
138
suspended. On the other hand, it decreases with the decrease of the
temperature of the medium.
(ii) Viscosity of the medium. The Brownian motion decreases with the
increase in the viscosity of the medium in which particles are suspended
and vice-versa.
(iii) Density of the medium. The Brownian motion increases if the density
of the medium decreases and decreases if the density of the medium
increases.
(iv) Size of the suspended particles. The Brownian motion of particles
which are bigger in size is less than the particles which are smaller in size.
According to molecular theory of matter, the quantity of a substance is
measured in terms of the number of molecules rather than in terms of the
mass of the substance. The quantity or amount of substance is a physical
quantity which is determined by the number of molecules constituting the
substance. The unit of measurement of the amount of substance is mole.
The mole is one of the seven basic units of SI.
i) Temperature. Brownian motion increases with the increase in
temperature.
(ii) Viscosity of medium. Brownian motion decreases with the increase in
the viscosity of the medium.
(iii) Density of medium. Brownian motion increases with the decrease in
density of the medium.
(iv) Size of the particle. Brownian motion decreases with the increase in
the size of the particle.
Q.
Calculate the mass of 6·5 X 1024 atoms of gold. The molar mass of
gold is 197 g mol-1
Solution. Molar mass of gold,
M = 197 g mol-1
Avogadro's number,
Mass of 6.5 X 1024 atoms of gold,
139
m = 6.5 X 1024 X 197 g mol-1
6.02 x 1023 mol-1
= 2127.08 g = 2.13 kg
Find (i) the number of moles and (ii) the number of molecules in 2 litres of
an ideal gas at a pressure of 50 Pa and temperature 300K.
Solution. Here, V = 2 litres = 2 X 10-3 m3
P = 50 Pa
T = 300K
R = 8.31 J mol-1 K-1
NA = 6.02 X 1023 mol-1
(i) Using PV = μRT, we get
PV 50x2x103


RT 8.31x300
= 4.01 x 10-5 mol
(ii) Number of molecules,
N = μNA = 4.01 X 10-5 X 6·02 X 1023
= 2.414 X 1019 molecules
The diameter of an oxygen molecule is 3Å. Find the fraction of molecular
volume to the actual volume occupied by oxygen gas at STP.
Solution. Actual volume of 1 mole of oxygen gas at STP = 22.4 litres.
Diameter of oxygen molecule,
D = 3Å = 3 X 10-10 m
 Radius of oxygen molecule,
T = 1.5 X 10-10 m
140
Isothermal and Adiabatic Processes
1. Isothermal Process
When a system undergoes a physical change under, the condition that the
temperature of the system remains constant then such a process is called
'isothermal process'. For such a process to take place, it is necessary that
the system be surrounded by a perfectly conducting material, so that any
heat produced in the process immediately goes out from the system to the
surroundings, or any heat absorbed comes from the surroundings into the
system and the temperature of the system remains constant.
In practice, there is no material which is perfectly conducting. Hence
perfect isothermal process is impossible. But when a process takes place
very slowly so that heat finds sufficient time to flow, then the process is
approximately isothermal. For example, suppose a gas is filled in a 'metal'
cylinder fitted with an air-tight piston. If we compress the gas by putting
small weights, one by one, on the piston, then the pressure of the gas will
increase slowly and heat will be produced. But since the metal is a good
conductor of heat and the pressure of the gas is being increased slowly,
this heat goes out through the walls of the cylinder. Thus the temperature
of the gas will not increase and the change in pressure will be isothermal.
In a perfect gas, an isothermal process obeys Boyle's law, that is, for a
given mass of gas, we have
PV = Constant
The internal energy of a perfect gas depends only upon the temperature.
Hence an isothermal process does not produce any change in the internal
energy of a perfect gas.
2. Adiabatic Process
When a system undergoes a change under the condition that no exchange
of heat takes place between the system and the surroundings, then such a
process is called 'adiabatic process'. For such a process to take place it is
necessary that the system be perfectly insulated from the surroundings.
141
Under this condition, if any heat is produced, it does not go out from the
system and the temperature of the system is increased. If any heat is
absorbed, no heat can come from the surroundings and the temperature of
the system falls.
In practice, there is no material which is perfectly insulating. Hence perfect
adiabatic process is impossible. But when a process takes place very
rapidly so that the heat does not find time to flow in or out then the
process is adiabatic. For example, if large weights be suddenly placed on
the piston of the cylinder, the pressure of the gas will suddenly increase
and its temperature will rise. This is because the work done in increasing
the pressure is converted into heat and this heat is unable to go out of the
cylinder in such a short time. Hence the temperature of the gas rises. It is
for this reason that during pumping a cycle. The barrel of the pump is
heated. Similarly, if the gas is suddenly expanded, then some of its internal
energy is used in doing work against the external pressure and the
temperature of the gas falls. It is for this reason that when a bicycle-tube
bursts suddenly, the air of the tube expands rapidly into the atmosphere
and becomes cooled. An interesting example of adiabatic expansion can be
seen in CO2.If carbon dioxide gas is filled in a tank at a high pressure and a
piece of cloth be held just infront the nozzle of the tank, then on opening
the nozzle the temperature of the emerging gas falls so much that it settles
down on the cloth in the form-of-white solid particles.
Again it must be remembered that the process being very rapid is not the,
theoretical condition of adiabatic process. If we could have a perfectly
insulating material (whose conductivity is zero) then even a slow process
taking place in a vessel of that material will be adiabatic.
In an ideal gas, the adiabatic change takes place according to Poisson's law
(PVY = constant).
Adiabatic relation between Temperature and Volume: Putting P = RT/V in
the above equation, we get
142
RT 
V  cons tan t
V
TV   cons tant
Adiabatic Relation between Temperature and Pressure: Putting V = RT/P in
eq. PVY = constant, we have
Or
P (RT/P)Y = constant
Or
TY/PY-1 = constant
3. Isothermal and Adiabatic Curves
The graph drawn between the pressure P and the volume V cf a given
mass of a gas undergoing an isothermal change is known as 'isothermal
curve'; and that for a gas undergoing an adiabatic change is known as
'adiabatic curve'.
In Fig. are drawn the isothermal curves for a definite mass of a perfect gas
at two constant temperatures T1 and T2. Suppose the initial pressure,
volume and temperature of the gas are P2, V2 and T2 respectively. This
state of the gas is represented by the point A lying on the isothermal curve
T2 If the temperature of the gas is kept constant at T1 and the gas is
143
(isothermally) expanded then its states will be represented by different
points on this curve.
If, however, the gas be expanded from A adiabatically (so that it cannot
take heat from outside), then along with pressure its temperature will also
fall. Suppose the final pressure, volume and temperature become P2, V2
and T2 respectively. This state of the gas is represented by the point B
lying on the isothermal curve T2. Since the gas has expanded adiabatically
from A to B, the curve joining A and B will be the adiabatic curve.
If the gas is heated at constant pressure, then it will expand according to
Charle's law. In this case the pressure-volume (P-V) curve of the gas will
be a straight line called 'isobaric line' parallel to the volume-axis (Fig. 2). In
other words, the slope of isobaric line with the volume axis is zero.
Of all the three curves drawn in the Fig., the slope of the adiabatic curve is
maximum. The reason is that in both the isothermal and the adiabatic
expansions of the gas, the pressure of the gas falls. But, for the same fall
in pressure, the increase in the volume of the gas during adiabatic
expansion is less than the increase during isothermal expansion, because
during adiabatic expansion the temperature of the gas also falls.
For an Ideal Gas, the Slope or the Adiabatic Curve is γ times the slope of
the Isothermal Curve:
For an ideal gas, the equation for the isothermal curve is
P V = constant.
Differentiating:
PdV + VdP = 0
dP
P

dV
V
Therefore, the slope of the isothermal curve at a point (P1 , V1) is
P1
 dP 

 dV 
V1

ISO
……………… (i)
144
Now, the equation for the adiabatic curve of an ideal gas is
P VY = constant,
where γ = Cp/CV. Differentiating it :
dPVY + P γVY- 1 dV = 0
dP
PV  1
P
 
 

dV
V
V
Therefore, the slope of the adiabatic curve at the point (P1, V1) is
P1
 dP 
 
 dV 
V1

ADIA
Dividing eq. (ii) by eq. (i), we get:
 dP / dV ADIA
 dP / dV ISO
That is,
  ……………… (ii)
slope of adiabatic curve

slope of isothermal curve
The value of y is always greater than 1. It is 1.67 for monoatomic gases
and 1.41 for diatomic gases.
Therefore, the slope of adiabatic curve at any point is greater than the
slope of isothermal curve at that point.
145
4. Work done by an Ideal Gas in Isothermal Expansion
When isothermal expansion takes place in an ideal gas, work is done by
the gas. Suppose μ mole ideal gas at a constant absolute temperature T is
expanded from an initial volume Vi to a final volume Vf. Then, the external
work done by the gas is
Vf
 PdV
W
……………(i)
Vi
where P is instantaneous pressure of the gas during infinitesimal expansion
dV. Substituting the value of P from gas equation
PV = μ RT in eq. (l). we get
Vf
Vf
RT
dV
 Vf 
W 
dV  RT 
 RT log e  
V
V
 Vi 
Vi
Vi
From Boyle's law, PiVi = PfVf or
Vf Pi

Vi Pf
 Pi 
W  RT log e  
 Pf 
Pi
 Vf 
W = 2·3026 μ R T log10   = 2·3026 μ R T log10
Pf
 Vi 
5. Work done by an Ideal Gas in Adiabatic Expansion
Suppose, μ mole ideal gas is expanded adiabatically from an initial volume
Vi to a final volume Vf. Then, the external work done by the gas is
Vf
W
 PdV
Vi
where P is the instantaneous pressure of the gas during infinitesimal
expansion dV. In adiabatic expansion, according to Poisson's law, we have
PVγ = K (a constant)
146
W

Vf
Vf
Vi
Vi

 PdV   KV dV 
Vf
K
V1 
Vi
1 
K
1  K
K 
 Vf1  Vi1  



1 
  1  Vi 1 Vf 1 
But PiVi  PfVf  K
K
W 
1 
W
 PiVi PfVf 
      
Vf 
 Vi
K
PiVi  PfVf 
1  
This formula for work does not contain u , Hence it holds irrespective of
the number of moles of the gas.
Now, let Ti and Tf be the absolute temperatures of the gas before and after
expansion. Then
PiVi = μRTi and
Pf Vf = μRTf
W 
R
 Ti  Tf 
 
For 1 mole of the gas, we shall have
W
R
 Ti  Tf 
 
This equation shows that the work done depends only upon the initial and
final temperatures Ti and Tf. If work is done by the gas (adiabatic
expansion), that is, W is positive, then Tf < Ti, that is, the gas is cooled.
If work is done on the gas, the gas is heated up.
147
6. Volume Elasticities of Gases
Gases possess property of volume elasticity. If we change the pressure of a
given mass of a gas, the volume of the gas changes.
Suppose a given mass of an ideal (perfect) gas has a pressure P and
volume V. When the pressure is increased to P + ΔP, the volume decreases
to V - ΔV. Thus, a pressure-increase ΔP results in a
volume-change ΔV. Then
normal stress = ΔP
volume strain = ΔV/ V
 Volume elasticity E =
stress
P
…………(i)

strain V V
Isothermal Modulus of Elasticity
If the pressure and the volume of the gas change in such a way that the
temperature of the gas remains constant, then the modulus of elasticity of
the gas is called 'Isothermal
Modulus of elasticity' and is represented by ET. In this case, eq. (i) can be
written as
ET 
P
VP

V V
V
………………………(ii)
At constant temperature the gas obeys Boyle's law, according to which
PV = constant.
Applying this law to the above change, we have
V= (P+ ΔP) (V – ΔV)
148
= PV – P ΔV + V ΔP – ΔP ΔV
or V ΔP = P ΔV + ΔP ΔV
or
VP
 P  P
V
Substituting this value in eq. (ii), we get
ET = P + ΔP
If ΔP is infinitesimally small (ΔP  0), then in this limit
ET = P
………………(iii)
Thus, the isothermal modulus of elasticity of a perfect gas is equal to its
pressure.
Adiabatic Modulus of Elasticity
If, during the change in volume and pressure of the gas, there is no
exchange of heat between the gas and the surroundings, then the
temperature of the gas is changed. The modulus of elasticity is then called
'adiabatic modulus of elasticity' and is represented by ES. In this case,
eq. (i) can be written as
ES =
P
VP

……………………(iv)
V
V
V
In an adiabatic change the gas obeys Poisson's law. according to which
PVY = constant.
where y is the ratio of the two specific heats Cp and Cv of the gas.
Applying this law to the above change, we have
PVY = (P + ΔP) (V - ΔV)Y = (P + ΔP) VY(1-ΔV/V)γ
149
or P = (P + ΔP) (1 – ΔV/V)Y
Expanding (1 – ΔV/V)Y by Binomial theorem and neglecting second and
higher powers of the small quantity ΔV/V; we have
P = (P + ΔP)(1 - γ ) = P - γ (P ΔV/V) + ΔP - Y (ΔPΔV/V)
or P  
or
PV
PV

V
V
VP
 P  P  (P  P)
V
Substituting this value in equation (iv), we get
ES = γ(P + ΔP)
If ΔP is infinitesimally small (ΔP  0), then in this limit
ES = γP ………………(v)
Thus, the adiabatic modulus of elasticity of a perfect gas is γ times its
pressure. Dividing eq. (v) by eq. (iii), we have
ES P
CP


EP
P
CV
Thus, the ratio of the two elasticities of a perfect gas is the same as the
ratio of its two specific heats.
From the above, we have
ES = γEr.
This shows that the adiabatic elasticity is γ times the isothermal elasticity:
Since γ > 1, so ES> Er.
Theoretically, this is because for the same decrease in pressure, the
volume of the gas increases less in the adiabatic process than in the
isothermal process.
150
Example
If, at 50° C and 75 cm of mercury pressure, a definite mass of a gas is
compressed (i) slowly, (ii) suddenly, then what will be the final pressure
and temperature of the gas in each case if the final volume is one-fourth of
the initial volume? (y = 1.5)
Solution. (i) Let P be the initial pressure and V the volume of the gas. For
slow (isothermal) compression, we have
P V = constant.
Let P' be the pressure when the volume is changed from V to V/4. Then
PV=P'(V/4).
 P' = 4 P = 4 x 75 = 300 cm (mercury).
As the compression is isothermal, the final temperature will remain 50°C.
(ii) For sudden (adiabatic) compression, we have
PVγ = constant
or P Vγ = P'(VI4)γ .
 P'=P X (4)γ =75 X 41.5 = 75 x 4 x (4)1/2
= 75 x 4 x 2 = 600 cm (mercury).
Again, for adiabatic change,
TVy- 1 = constant
TVy-1 = T'(V/4)γ-1
 T' = T X (4)y-1
Here T = 50°C = 50 + 273 == 323 K and γ = 1.5.
 T' = 323 X (4)0.5 = 323 x 2 = 646 K = 373°C.
151
Example
Air is filled in a scooter tyre at a temperature of30°C and at pressure of
3.375 atmospheres. What will be the final temperature of the tyre if it
bursts suddenly? (Assume for air 1= 1.5)
Solution. Air of the tyre is adiabatically expanded,
Initial pressure P = 3·375 atmosphere, initial temperature T = 3O°C + 273
= 303 K, final pressure P' = 1 atmosphere, final temperature T' = ? γ = 1·5
= 3/2.
For adiabatic expansion, we have
T
T '

P   P ' 

T
P
 T'  P'
 
 
 303 
 T' 


 
 
 
 3.375 


 1 
 303 
T'  
 202K  71 C

 1.5 
152
Example
The initial pressure of a perfect gas is P and the volume is V. It is allowed
to expand under the following three conditions such that the final volume
in each case is 2V and (i) pressure P remains constant, (ii) product PV
remains constant, (Hi) product PV'" remains constant.
(a) Draw P-V diagram for these changes and state how the temperature
changes in each expansion does,
(b) In which expansion is the maximum and in which the minimum
external work done?
(c) In which case the internal energy of the gas will remain unchanged?
Solution. (a) The initial pressure of the gas is P and the initial volume is V.
This state is denoted by the point A on the P-V diagram (Fig.)
(i) For the expansion at constant pressure, the final pressure and volume
of the gas will be P and 2V (point B). Hence the line AB represents this
expansion. In this expansion the temperature will rise (V/T = constant).
(ii) Under the condition PV = constant, the expansion will be isothermal
(temperature constant). The final pressure and volume will be P/2 and 2V
153
(point C). Hence the curve AC will represent the isothermal expansion.
(Fig.)
(iii) Under the condition PVγ = constant, the expansion will be adiabatic. In
this case, the slope of the P- V curve will be more. Hence it will be
represented by a curve like AD. In adiabatic expansion, the temperature
will fall.
(b) Work in expansion AB ; WAB = area ABEF .
Work in expansion AC ; WAC = area A CEF .
Work in expansion AD ; WAD = area ADEF.
The area ABEF is maximum and the area ADEF is minimum, hence
maximum work will be done in the expansion (AB) at constant pressure
and minimum in the adiabatic expansion (AD).
(c) During isothermal expansion AC, the temperature remains constant.
Hence the internal energy of the gas remains unchanged in the isothermal
expansion.
154
Example
Isothermal expansion is carried out of 2 moles of an ideal gas at 27'C, so
that its volume is increased to three times of its initial volume. Calculate
the work done and the heat absorbed by the gas. (R = 8·31 J/mol-K, log10
3 = 0·4771).
Solution. The work done by μ mole ideal gas at absolute temperature T in
expanding isothermally from volume Vi to Vf is given by
W = 2·3026μR T 10g10 Vf/Vi
Here μ = 2 mole, T = 27 + 273 = 300 K and Vf/Vi = 3.
:. W = 2·3026 x 2 mol x 8·31 J mol-1 K-1 x 300K x log103
= 2·3026 x 2 x 8·31 x 300 x 0·4771 = 5-48 x 103 J.
Now, W = J Q. Therefore, the heat absorbed by the gas in doing work is
W 5.48x103 J
Q

 1.31x103 cal
1
J
4.18JCal
155
Example
32 g of oxygen gas at temperature 27·C is compressed adiabatically to 1/3
of its initial volume. Calculate the change in internal energy. y = 1·5 for
oxygen.
Solution. Let Ti and Vi be the initial (absolute) temperature and volume of
oxygen gas, and Tf and Vf the values after adiabatic compression. Then
 1
TV
 Tf Vf 1
i i
 1
Tf  T(V
i
i / Vf )
Here Ti = 27 + 273 = 300 K, Vi/Vf = 3 and γ = 1·5.
Tf= 300 K (3)0.5 = 519·6 K.
When μ mol of a gas is compressed adiabatically from temperature T, to Tf,
the work done by the gas is given by
Here μ =

32g
1
32g
Ti-Tf = 300 – 519.6 = -219.6 K and γ =1.5
1molx8.31J / (mol  K)
W
(219.6K)  3650J
(1.5  1)
By first law of thermodynamics, the change in the internal energy of the
gas is
ΔU = Q - W = - W (Q = 0 for adiabatic process)
= - (- 3650 J) = 3650 J.
The internal increases by 3650 J.
= 40 litre x (2)3/2 = 80
litre = 113·14 litre.
156
Example
A piston divides a closed gas cylinder into two parts. Initially the piston is
kept pressed such that one part has a pressure P and volume 5V and other
part has a pressure 8P and volume V.
The piston is now left free. Find the new pressures and volumes for the
adiabatic and isothermal processes. (For this gas γ = 1·5).
Solution. In the beginning, we have
Let, in isothermal process, the new pressure be P‟ and the change in
volume be , Then, by Boyle's law, we have
P x 5 V = P‟ (5 V - ) and 8 P x V = P‟ (V + ) .
Adding and subtracting these equations, we get
P‟ = (13/6) P and  = (35/13) V.
 new volumes are
V1 = 5 V -  = (30/13) V
and V2 = V +  = (48/13) V.
Let, in adiabatic process, the new pressure be PI and change in volume be
. Then, by Poisson's law, we have
P (5 V)Y = P1 (5V - ) γ
And 8P (V)Y = P1 (V + ) γ
(5)  5V   

Dividing

8
 V 

157
5V  
5
5
5
 1  2 
V
83 4
8 
V=(5/3) 
 new volumes are
V1 = 5 V -  = (10/3) V
And V2 = V +  = (8/3) V.
Now, from the first equation of the adiabatic process, we have
'3/2 «' 5 V 27
27
 5V 
P1 = P = P 

P
 1.84P 8" = 1-84P.

8
10V / 3 
Example
The insulated box shown in Fig. has an insulated partition which can slide
without friction along the length of the box. Initially each of the two
chambers of the box has one mole of a monoatomic ideal gas (γ = 5/3) at
a pressure P0, volume V0 and temperature T0. The chamber on the left is
slowly heated by an electric heater so that its gas, pushing the partition,
expands until the final pressure in both the chambers becomes 243 P0/32.
Determine: (i) the final temperature of the gas in each chamber and
(ii) the work done on the gas in the right chamber.
Solution. (i) The gas in the right chamber is adiabatically compressed.
Suppose for the gas in this chamber the final pressure, volume and
temperature are P2, V2 and T2 respectively. Then
P0V0  P2V2 
243
P0V2
32
158
 32 
 V2  

 243 
1

 32 
V0  

 243 
3
5
3
8
2
V0    V0 
V
27 0
3
Again, T0V0   T2V2 
To VOY-I = T2 V2
(5 3)1
V 
 T2   0 
 V2 
23


V0
T0  

 (8 / 27)V2 
2/3
 27 
T0  

 8 
T0 
9
T
4 0
Now, let the final pressure, volume and temperature in the first chamber
be P1, V1 respectively. Then
P0V0 P1V1  243 
8

1


P0   2V0 
V0 

T0
T1
27  T1
 32 

1
207 1
 243   46  1





T0  32   27  T1
16 T1
 T1 
207
T
16 0
(ii) Work done on the gas in the right chamber is
W
1
 P0V0  P2V2 
 1
 243   8

P0V0  
P0  
V0 
 32
  27    15 P V

(5 / 3)  1
8 0 0
159
Thermodynamics
We have discussed in the previous unit that heat is a form of energy. Heat
can be produced by mechanical work or energy. For example, when we rub
our hands, they become warm. It is due to the fact that the work done to
move one hand over the other hand is converted into heat. This heat
warms our hand. Joule studied the relationship between work and heat.
Joule found that the same amount of work done always produces the same
amount of heat. The study of the relationship between mechanical work,
heat and other forms of energy and energy transfer is known as
thermodynamics.
Thus, thermodynamics is the study of heat, temperature and the
conversion of mechanical work into heat and vice-versa.
In thermodynamics, the main focus is on the macroscopic quantities of the
system such as the pressure, volume, temperature, internal energy,
entropy, enthalpy etc. which have the impact on the internal state of the
system. Thus, thermodynamics provides a macroscopic description of the
system.
Thermodynamic System
A collection of an extremely large number of atoms or molecules confined
within certain boundaries such that it has a certain values of pressure (P),
volume (V) and temperature (T) is called a thermodynamic system.
Anything outside the thermodynamic system to which energy or matter is
exchanged is called its surroundings.
Taking into consideration the interaction between a system and its
surroundings, a system may be divided into three classes:
(a) Open system. A system is said to be an open system if it can exchange
both energy and matter with its surroundings (Figure).
160
(b) Closed system. A system is said to be closed system if it can exchange
only energy (not matter) with its surroundings (Figure 1b).
(c) Isolated system. A system is said to be isolated if it can neither
exchange energy nor matter with its surroundings (Figure 1c).
Thermodynamic Variables
A thermodynamic system can be described by specifying its pressure,
volume, temperature, internal energy, entropy, enthalpy and the number
of moles (μ). These parameters or variables are called thermodynamic
variables. Thus, the variables which are required to specify the state of
thermodynamic system are called thermodynamic variables.
Types of thermodynamic variables
(i) Intensive variables: The variables which are independent of the size of
the system are called intensive variables. Examples: Temperature and
pressure, specific heat capacity.
(ii) Extensive variables: The variables which depends on the size or mass of
the system are called extensive variables. Examples: Volume, energy,
entropy, heat capacity and enthalpy.
Thermal Equilibrium
When two bodies having different temperatures are placed in contact, then
the heat energy flows from a body at higher temperature to a body at
lower temperature. The flow of heat continues from one body to another
body till both the bodies attain the same temperature. When both the
bodies in contact have the same temperature and there is no heat flow
between them, then these bodies are in thermal equilibrium.
Thus, two bodies or systems in contact are said to be in thermal
equilibrium if both are at the same temperature.
161
Consider two systems A and B separated by a wall which does not allow
any exchange of heat between them. Such a wall is known as insulating
wall or adiabatic wall. The third system C is separated from the systems A
and B by a conducting or diathermic wall as shown in figure 2.
Since heat can tie exchanged between the systems A and' C, so both A and
C are in thermal equilibrium. Similarly, heat can be exchanged between the
systems Band C, so both B and C are also in thermal equilibrium. In other
words, both the systems A and B are in thermal equilibrium with the third
system C separately. When adiabatic wall between the systems A and B is
removed, no transfer of heat takes place between them. This shows that
the systems A and B are also in thermal equilibrium with each other.
Zeroth law of thermodynamics: Two systems A and B which are separately
in thermal equilibrium with a third system C are also in thermal equilibrium
with each other.
Concept of temperature from Zeroth law of thermodynamics
According to Zeroth law of thermodynamics, if system A is in thermal
equilibrium with a system C, then temperature of system A = temperature
of system C ...(1)
Similarly, if system B is in thermal equilibrium with the system C, then
temperature of system B = temperature of system C ...(2)
162
Now, from eqns. (1) and (2), we have temperature of system A =
temperature of system B
Thus, temperature of a system or a body can be defined as follows:
Temperature of a system is a physical quantity, equality of which is the
only condition for the thermal equilibrium of two systems or bodies in
contact.
Or
Temperature of a system or a body is a physical quantity which determines
whether the system is in thermal equilibrium with another system in its
contact or not.
Heat, Work and Internal Energy
We know heat energy transfers from a system to the surroundings or from
the surroundings to the system whenever there is temperature difference
between the system and the surroundings. When there is no temperature
difference between the system and the surroundings, no heat is
transferred.
Thus, heat is the energy that is transferred between the system and its
surroundings whenever there is temperature difference between the
system and its surroundings.
Sign conventions for heat
(i) When heat energy is transferred to the system from its surroundings,
then heat energy is taken as positive.
(ii) When heat energy is transferred to the surroundings from the system,
then heat energy is taken as negative.
Work Done
Work is said to be done if a body or a system moves through a certain
distance in the direction of the applied force.
163
Consider a cylinder fitted with a frictionless piston of area of cross section
A (Figure 3). Let a gas be contained in the cylinder.
The walls of the cylinder are perfectly conducting one so that the gas in
the cylinder is in thermal equilibrium at temperature T with its surrounding.
If P be the pressure 'of the gas in the cylinder, then the force exerted by
the gas on the piston of the cylinder is given by
F =PA
Let the piston moves through a small distance dx during the expansion of
the gas. Therefore, the work done by the gas is given by
dW = F dx = PA dx
Since A dx = dV, change (increase) in volume …………(1)
dW = PdV
If the volume of the gas increases from initial volume Vi to final volume Vf
then the total work by the system or gas can be calculated by integrating
eqn. (1) between V = Vi to V = Vf
Vf
 dW   PdV
Vi
Vf
W  P  dV
Vi
W  P[V]VVif  P[Vf  V]
i
If the system (or gas) is compressed, its volume decreases. Hence work
done on the system (or gas) to change its colume to dV during
compression is given by
dW = -P dV
-ve sign shows that dV decreases.
164
Hence, total work done to decrease the volume from Vi (initial value) to Vf
(final value) is given by
Vf
 dW    PdV  P[V
f
 V]
i
Vi
W  P[Vf  V]
i
Sign conventions for work done.
(i) Work done on the system (or gas) is taken as negative.
(U) Work done by the system (or gas) is taken as positive.
Internal Energy
Internal energy of a system is defined as the sum of the kinetic energies of
the constituent particles of the system plus the sum of the potential
energies of the constituent of the system.
It is denoted by U.
Thus, U = Σ K.E. of constituent particles of the system + Σ P.E. of the
constituent particles of the system.
Sign conventions for Internal energy
(i) Increase in internal energy is taken as positive.
(ii) Decrease in internal energy is taken as negative.
The first law of thermodynamics is the law of conservation of energy.
Let a system absorbs Q amount of heat energy from the external source.
As a result of this, let W be the work done by the system on its
surrounding and ΔU be the change in the internal energy of the system.
According to the law of conservation of energy,
Q=ΔU+W ...(1)
165
which is the mathematical statement of the first law of thermodynamics.
Thus, first law of thermodynamics may be stated as:
The energy entering the system in the form of heat energy is equal to the
sum of the increase in the internal energy of the system and the energy
leaving the system in the form of work done by the system on its
surrounding.
Since Q and W may be positive, negative or zero, so change in internal
energy ΔU can be positive, negative or zero.
Definition of internal energy from first law of thermodynamics
From the first law of thermodynamics,
ΔU = Q - W ...(2)
Thus, the change in internal energy of a system is defined as the difference
between the hear energy added to the system or taken from the system
and the work done by the system or work done on the system.
Some other processes in Thermodynamics
a) Isochoric process. A thermodynamic process that takes place at
constant volume is called isochoric process. It is also known as isovolumic process. In this cased dV = O.
(b) Isobaric process : A thermodynamic process that takes place at
constant pressure is called isobaric process. In this case, dP = a
(c) Cyclic process: A cyclic process consists of a series of changes which
return the system back to its initial state.
P-V Diagram or Indicator Diagram
A graph representing the variation of pressure with the variation of volume
is called P-V diagram or indicator diagram (Figure 6)
166
Work done by the thermodynamic system is equal to the area under P-V
diagram.
Consider a gas (thermodynamic system) in a cylinder fitted with a
frictionless piston. Let the gas expands from the state A (P1,V1) to the state
B (P2, V2. The variation of pressure and volume of the gas from one state
to another is shown in figure.
Now consider two points a and b on P-V diagram. Points a and b are so
close to each other that the pressure (P) corresponding to these two points
is same. Now draw ad and be perpendicular on the
volume axis. Here dc = dV (change in volume). Now work done by the
system (i.e. gas) against pressure is given by
dW = P dV = ad x dc = area of the strip abcd
Total work done by the system to change its state from A to B is given
by:
W   dW   PdV
= Sum of areas of all strips like abed within ABCD
or W = Area under P-V diagram
167
Applications of First law of Thermodynamics
1. Isothermal Process:
According to the first law of thermodynamics, Q = ΔU +W
In an isothermal process, temperature is constant so change in internal
energy, ΔU = 0
 Q=W
Thus, during isothermal process
Heat added (or removed) = Work done by (or on) the system.
2. Adiabatic process:
According to the first law of thermodynamics Q = ΔU + W
In an adiabatic process, dQ = 0
ΔU= -W
Hence, if work is done by the system during adiabatic process, then the
internal energy of the system decreases by an amount equal to the work
done by it.
If work is done on the system (i.e., W is negative), then ΔU = - (- W) = W.
Thus, the internal energy of the system increases by an amount equal to
the work done on it, during adiabatic process.
3. Isochoric process:
In isochoric process, volume of the system remains constant. It means V =
constant or dV = O.
Therefore, work done, W = PdV = o. Hence, first law of thermodynamics Q
= ΔU + W reduces to ΔU = Q
This shows that (i) if heat is added into the system or heat is absorbed by
the system, then the internal energy of the system increases and (ii) if heat
168
is taken out from the system or the system loses heat energy then the
internal energy of the system decreases.
4. Isobaric process (e.g., boiling water at constant pressure):
When a liquid boils, it changes into vapours and its temperature and
pressure remains constant till whole of the liquid is converted into vapours.
The energy absorbed by the liquid is used in (a) increasing the internal
energy of liquid molecules and (b) in doing external work during expansion
of liquid at constant pressure P.
Let Vi = initial volume of the liquid; Vf = final volume of vapours.
P = constant pressure at which liquid converts into vapours.
 Work done: dW = P (Vf - Vi) ...(i)
Let latent heat of vaporisation = L
mass of liquid = m
 Quantity of heat given to the liquid to convert it into vapours
Q = mL ...(ii)
Putting the values of Eqs. (i) and (ii) in the first law of thermodynamics
i.e.,
Q = ΔU + W, we have
mL = ΔU + P(Vf - Vi) ...(iii)
or ΔU = mL - P (Vf- Vi)
where ΔU is the change in internal energy of the system.
Knowing the values of m, L, P, Vi and Vp we can calculate ΔU.
169
5. Cyclic process:
In a cyclic process, the initial and final states of a system are the same.
Since, internal energy of a system depends upon its initial and final states,
so during cyclic process, there is no change in internal energy of the
system i.e.; ΔU = O.
From the first law of thermodynamics
Q = ΔU + W, get
Q = W for a cyclic process
Thus, heat added to the system or heat absorbed by the system is equal to
the work done by the system during cyclic process. Also heat taken out of
the system or heat lost by the system is equal to the work done on the
system during cyclic process.
A cyclic process is represented by a closed loop on a P-V diagram as shown
in figure.
170
Solution. Since process is cyclic, so change in internal energy of the gas is
zero i.e., ΔU = °
According to first law of thermodynamics,
Q = ΔU+ W, get 0 = W
Since W = area under P-V diagram. Q = area of closed curve ABCD
=
x base x height =
x BC x AC
=
(40 - 10) x (30 - 10) =
x 30 x 20 = 300 J
Comparison of the Work done during Isothermal and Adiabatic Processes
(i) Expansion. Isothermal and adiabatic expansions between the same
values of initial and final volumes i.e. V1and V2 are represented by the
curves M and MA respectively as shown in figure 11.
171
Now, work done during isothermal expansion is given by
WISO 
V2
 PdV = Area under isothermal curve (MI)
V1
= Area MIV2V1
Work done during adiabatic expansion is given by
WADI 
V2
 PdV = Area under adiabatic curve (MA)
V1
= AreaMAV2V1
Since Area MIV2V1 > Area MAV2V1  Wiso > Wadi
Thus, work done during isothermal expansion is more than the work done
during adiabatic expansion between the same values of initial and final
volumes.
172
(ii) Compression. Isothermal and adiabatic compressions between the
same values of initial and final volumes (i.e. V2 and V1) are represented by
the curves MI and MA respectively as shown in figure 12.
Work done during isothermal compression is
Wiso = Area under isothermal curve (MI) = Area MIV1V2
Work done during adiabatic compression is
Wadi = Area under adiabatic curve (MA) = Area MAV1V2
Since, Area MAV1V2 > Area MIV1V2
 Wadi> Wiso
Thus, work done during adiabatic compression is more than the work done
during isothermal compression between the same values of initial and final
volumes.
173
Reversible Process
A process is said to be reversible when the various stages of an operation
to which it is subjected can be traversed back in the opposite direction in
such a way that the substance passes through exactly the same conditions
at every step in the reverse process as in the direct process. Thus, if in a
particular step in the direct process, heat is absorbed by the substance,
then in the same step in the reverse process, the same amount of heat will
be given out by the substance.
Comparison of the Slopes of an isothermal and adiabatic curve
For an isothermal process, PV= constant
Differentiating, we get PdV + VdP = 0 or =
dP
P

dV
V
dP
represents the slope of isothermal curve.
dV
 dP 
P

  

 dV ISO
V
For an adiabatic process, PVγ = constant
Differentiating, we get
PγVγ-1- dV + VγdP = 0
P
 dP 



.........(2)

V
 dV ADI
From (1) and (2). we have
 dP 
 P 

  

 dV ADI
 V ISO
174
the change in volume of the gas is dV = Adx,
:. dW = PdV
For finite change in volume from V1 to V1‟ this equation is then integrated
between V1 to V1‟ to find the network also equal to the area under P-V
graph.
Case 1. When volume is constant
V = constant
WAB = 0
Case 2. When volume is increasing
175
V is increasing
WAB > 0
WAB = Shaded area
Case 3. When volume is decreasing
V is decreasing
WAB < 0
WAB = Shaded area
Case 4. Cyclic process
Wclockwisecycle= + Shaded area
Wanticlockwisecycle= + Shaded area
176
Work Done in Clockwise Cycle
WAB I Positive
WAB
II
Negative
Wcyclic = WAB I Positive + WAB II Negative = area of close path
Ex. For one complete cycle as shown in the P-V diagram here are, (a) ΔEint
for the gas and (b) the net heat transfer Q positive, negative, or zero?
Sol. (a) Zero (b) Negative Wnet = -ve
For the case of closed cycle, ΔEint = 0
and Q = W = area enclosed by the close path.
177
First Law of Thermodynamics
If some quantity of heat is supplied to a system capable of doing external
work, then the quantity of heat absorbed by the system is equal to the
sum of the increase in the internal energy of the system and the external
work done by the system.
δQ = dU + δW
[i] The first law of thermodynamics is essentially a restatement of the law
of conservation of energy, i.e., energy can neither be created nor be
destroyed but may be converted from one form to another.
[ii] In applying the first law of thermodynamics, all the three quantities,
i.e., δQ, dU and δW must be expressed, in the same units, Le., either in
units of work or in units of heat.
[iii] This law is applicable to every process in nature.
[iv] This law is applicable to all the three phases of matter, i.e., solid, liquid
and gas.
[v] dU may be any type of internal energy-translational kinetic energy,
rotational kinetic energy, binding energy etc. It is a characteristic of the
state of a system.
[vi] The first law of thermodynamics introduces the concept of internal
energy.
178
Limitations of First Law of Thermodynamics
(a) It does not explain the direction of heat flow.
(b) It does not explain how much amount of heat given will be converted
into work.
Significance: The first law of thermodynamics tells us that it is impossible
to get work from any machine without giving it an equivalent amount of
energy.
When a system is taken from, state a to state b, in fig. along the path
a → c → b, 60 J of heat flow into the system, and 30 J of work are done:
(i) How much heat flows into the system along the path a → d → b if the
work is 10 J.
(ii) When the system is returned from b to a along the curved path, the
work done by the system is -20 J. Does the system absorb or liberate heat,
and how much?
(iii) If, Ua = 0 and Ud = 22 J, find the heat absorbed in the process a → d
and d → b.
Sol. For the path a, c, b,
dU = dQ - dW = 60 - 30 = 30 J or Ub - Ua = 30 J
(i) Along the path a, d, b, dQ = dU + dW = 30 + 10 = 40 J
179
(ii) Along the curved path b, a,
dQ = (Ua - Ub) + W = (-30) + (-20) = -50 J, heat flows out the system
(iii) Qad = 32 J; Qdb = 8 J
Free expansion
Consider a gas which initially occupies one compartment of a twochambered container as illustrated in fig. A membrance separates the two
chambers, and the one on the right is evacuated. The entire assembly is
insulated from the exterior. Suppose now that the membrane separating
the two chambers spontaneously breaks and the gas expands freely to fill
the entire container. This process is called a free expansion.
The change in the internal energy of the gas can be calculated by applying
the first law of thermodynamics to the free-expansion process.
The process is adiabatic because of the insulation, so Q = O.
No part of the surroundings moves (we consider the rupturing membrance
to be an inert part of the system) so the system does no work on its
surroundings.
For ideal gas :
(δW)ext.= work done against external atmosphere
= P dV = 0 (because P = 0)
(δW)int = work done against internal molecular forces = 0
180
0 = dU + 0
(δQ = dU + δW)
Therefore, the internal energy does not change. QU = OU → const. T →
const.
The initial and final states of this gas have the same internal energy.
Which implies that the internal energy of an ideal gas does not depend on
the volume at all. The free-expansion process has led us to the following
conclusion:
The internal energy U(T) of an ideal gas depends only on the temperature.
Zeroth Law of Thermodynamics
If objects A and B are separately in thermal equilibrium with a third object
C (the thermometer), then objects A and B are in thermal equilibrium with
each other.
Zeroth law of thermodynamics introduces thermodynamic quantity called
temperature.
Two objects (or systems) are said to be in thermal equilibrium if their
temperatures are the same.
In measuring the temperature of a body, it is important that the
thermometer be in the thermal equilibrium with the body whose
temperature is to be measured.
181
Temperature
(a) Temperature is a macroscopic physical quantity related to our sense of
hot and cold.
(b) Temperature is basically a measure of degree of hotness or coldness of
a body.
(c) The natural flow of heat is from higher temperature to lower
temperature, i.e. temperature determines the thermal state of a body
whether it can give or receive heat.
d) Two bodies are said to be in thermal equilibrium if and only if they are
at same temperature.
In is situation heat in the two bodies mayor may not be equal.
(e) Temperature of a body is directly proportional to the kinetic energy of
the random motion of the mo molecules or atoms of the substance.
(f) Temperature is one of seven fundamental quantities with dimensions
(θ). It is a' scalar physical quantity with SI unit kelvin (K).
(g) The highest possible laboratory temperature is about 108 K (in fusion
test reactor) while lowest 108 K (achieve in 1990 through nuclear spin
cooling).
(h) Theory has established that 0 K can never be achieved practically.
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Thermal equilibrium
Heat is the transfer of energy from one object to another object as a result
of a difference in temperature between them.
Thermal equilibrium is a situation in which two objects in thermal contact
cease to exchange energy by the process of heat.
Different types of temperature scales
(a) The Kelvin temperature scale is also known as thermodynamic scale.
The SI unit of temperature is the kelvin and is defined as (1/273.16) of
the temperature of the triple point of water. The triple point of water is
that point on a P-T diagram where the three phases of water, the solid, the
liquid and the gas, can coexist in equilibrium.
(b) In addition to Kelvin temperature scale, there are other temperature
scales also like Celsius, Fahrenheit, Reaumer, Rankine etc. Temperature on
one scale can be converted into other scale by using the following identity:
= constant for all scales
Hence,
t C  0 C
t F  32
tK  273.15






100  0
212  32
373.15  273.15
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