03. KINETIC THEORY OF GASSES AND RADIATION
Q1. What is Equa on of State?
● Equa on of State : For a gas, its state is specified by a
number of physical quan es such as pressure P,
temperature T, volume V, internal energy E, etc. Hence,
the equa on rela ng these quan es is known as the
equa on of state.
● For a fixed mass m of an enclosed gas.
1) Boyle's law : V ∝ 1/P at constant T
2) Charle's law : V ∝ T at constant P
3) Gay-Lussac's law : P ∝ T at constant V
● Combining the three laws into a single rela on yields
ideal gas equa on.
PV ∝ T
● For n moles of gas,
PV ∝ n T
∴P V = n R T
R is propor onality constant, known as 'universal gas
constant.
● Note that
n = M / MO = N / NA
Where,
M- mass of the gas.
MO- molar mass ie mass of 1 mole of gas.
N - number of molecules in the gas.
NA -Avogadro number i.e. number of molecules in one
mole of gas.
∴PV=(N/NA)RT
∴PV=N(R/NA)T
∴PV=NkBT
● Where kB is the Boltznann constant. (kB=R/NA)
Q2. Explain, what is an ideal gas.
●The gas laws are strictly valid for real gases, only if
the pressure of the gas is not too high and the
temperature is not close to the liquefac on
temperature of the gas.
●A gas obeying the equa on of state (PV = nRT) at
all pressures, and temperatures is Called as an
ideal gas.
Q3. Derive the equa on for pressure exerted by a
gas enclosed in a container.
●Fig. shows, N molecules of an ideal gas enclosed
in a cubical box.
● L is length of each side of box
∴volume of box (V) = L³
● The gas molecules are moving con nuously and
randomly in various direc ons.
● Pressure is Adjusted so that, the molecules do
not collide with each other, and collide only with
the walls.
● Fig. Shows a molecule moving towards shaded
wall with velocity v.
● The collision between molecule and wallis
elas c.
● During collision, the X- component of velocity (vX)
get reversed to (-vX)
● Here the y and z components of velocity remains
unchanged.
● Hence the change in momentum of the par cle is
given by
∆ pX = final momentum - ini al momentum
∆ pX = (-mvX) - (mvX)
∆ pX = - 2 mvX
●Thus, the momentum transferred to the wall during
one collision is + 2 m vX
●A er colliding with the shaded wall, the rebounced
molecule travels towards opposite wall and reflected
back, again to collide with the shaded wall.
∴ molecule travels a distance of 2L in between two
collisions.
●Hence me between two successive collisions of
molecule with shaded wallis 2 L / vX
∴ Average force exerted on the shaded wall by molecule
1 is given as
Average force = Average rate of change of momentum
= +2 m v X1 / ( 2L / vX1)
= m vX1 ² / L
where vX1 is the x component of the velocity of
molecule1
●Considering other molecules 2, 3, 4 ... with the
respec ve x components of veloci es vX2, vX3, vX4,... the
total average force on ]the wall from is
2
2
2
= m (vX1 + vX2 + vX3 +... + vXN²) / L
∴The average pressure P is given by
P = (Average force) / (Area of shaded wall)
P = m (vX12 + vX22 + vX32 +... + vXN2) / L × L2
P = m (vX1² + vX2² + vX32 +...+ vXN2 ) / v
But vX2 = (vX12 + vX2² + vX3² +.....+ vXN2) / N
∴ P = m N vX2 / V
Now, v 2= vX² + v Y2+ vZ2
By symmetry,
vX2 = vY2 = vZ2
∴ v 2 = 3vX2 ∴ vX2 = v 2/3
2
∴P=Nmv /3V
Above equa on for pressure exerted by gas on the walls
01 of container is valid for containers of any shape.
Note:
1) We have assumed that there are no intermolecular collisions. The number of molecules in
the container is so large (of the order of 10²³) that
even if molecular collisions are taken into account,
the above expression does not change.
2)If a molecule acquires a velocity with
components different than vX,vY, vZ a er collision,
some other molecule having different ini al
velocity will acquiring the velocity with the
components vX, vY, vZ. As the gas is steady (in
equilibrium).
Q6. Explain why gases have two specific heat
capacity?
● When the temperature of a gas is increased, even
a small rise causes considerable change in volume
and pressure.
● Therefore two specific heats are defined for
gases, namely specific heat at constant volume CV
and specific heat at constant pressure CP.
Q 4. Derive equa on for root Mean Square (rms)
Speed of gas molecules.
● Equa on for pressure exerted by gas molecules
enclosed in a container is given by,
P = N m v² / 3 V
∴ v² = 3 P V / N m
● For ideal gas,
PV = n R T
∴ v² = 3 n R T / N m
● We have, n = N / NA
∴ v² = 3NRT / NAN m
∴ v² = 3RT / NAm
∴ v² = 3RT/M0
where
M0 = NAm = molar mass of the gas.
● Vr.m.s.= √ v²
∴ Vr.m.s.= √ 3 RT/M0
●Note:
At 300 K,
1) For nitrogen gas, the rms speed is 517 m/s.
2)For oxygen gas rms speed is 483 m/s.
Q5. Derive equa on for average kine c energy
per molecule of a gas and show that it is
propor onal to absolute temp. (T) of the gas.
●Equa on for pressure exerted by gas molecules
enclosed in a container is given by P = Nmv²/3V
∴ PV = Nmv²/3 ∴ PV = (2/3) N (1/2) mv²
● The internal energy of an ideal gas is purely
kine c, because the molecules are non
interac ng, and poten al energy is zero.
∴ N (1/2) mv²= (3/2) PV
●Where (1/2) mv² is the average transla onal
kine c energy of a molecule, and N (1/2) mv² is the
average total energy (E) ∴ E = (3/2) PV
●From Ideal gas equa on PV = n RT
∴ E = (3/2) nRT ∴ E = (3/2)(N/N₀) RT
∴ E = (3/2) NkBT
Where, kB = R/N₀ =Boltzmann constant.
∴ E/N = (3/2) kBT
●From above equa on, the average energy per
molecule is propor onal to the absolute
temperature T of the gas.
02
Q7. Derive Mayer's rela on connec ng two
specific heats of gas.
● Consider one mole of an ideal gas enclosed in a
cylinder by light, fric onless air ght piston.
● Let P, V and T be the pressure, volume and
temperature of the gas.
● When gas is heated at constant volume, so that
its temperature rises by dT.
Then all the heat supplied to the gas (dQ₁) is used to
increase the internal energy of the gas (dE).
●In this case volume of the gas is constant, so no
work is done in moving the piston.
∴ dQ₁ = CV dT = dE .................. (1)
where CV is the molar specific heat of the gas at
constant volume.
● When gas is heated at constant pressure, so that
its temperature rises by dT. Volume of the gas
increases by an amount dV.
● In this case heat supplied to the gas (dQ₂) is used
to increse the internal energy of the gas by (dE), as
well as to perform work (dW) during expansion of
gas i.e. moving the piston.
∴ dQ₂ ₌ CP dT ₌ dE + dW ............... (2)
where CP is the molar specific heat of the gas at
constant pressure.
● The work done to move the piston
dW = P dV ..................... (3)
● Using eq'n (1) and eq'n (3) in eq'n (2)
CP dT = CV dT + P dV
∴(CP - CV ) dT = P dV .............. (4)
● For 1 mole of gas, PV = RT
∴P dV = R dT ................... (5)
Since pressure is constant
● Using eq'n (5) in eq'n (4)
(CP - CV ) dT = R dT
∴CP - CV = R
This is Mayer's rela on between CP and CV
● As supplied heat is measured in calories and
workdone is measured in Joules, Mayer's rela on
is modified as
CP - CV = R / J
Where J is mechanical equivalent of heat
● Also CP = MO SP and CV = MO SV
Where MO - molar mass of the gas
SP and SV - respec ve principle specific heats
∴MO (SP - SV) = R / J
∴(SP - SV) = R / MO J
Q.8 Use the law of equipar on of energy and
calculate the specific heat of gases, for
monoatomic, diatomic, and polyatomic gases.
(a) For a monatomic gas enclosed in a container,
held at a constant temperature T and containing NA
atoms, each atom has only 3 transla onal degrees
of freedom (dof).
●Therefore, average energy per atom is
(3/2) kB T
∴ total internal energy per mole will be,
E = (3/2)NAkB T
= (3/2)R T
●Molar specific heat at constant volume is given as
CV = (dE/dT) = (3/2)R
●For one mole of gas we have CP - CV = R
∴ CP = R + (3/2)R ∴ CP = (5/2) R
●we have γ = CP/CV
∴ γ = (5/3)
(b) For a diatomic gas (rigid rotator) enclosed in a
container, held at a constant temperature T and
containing N A atoms, each molecule has 3
transla onal, and 2 rota onal dof.
●Therefore, average energy per atom is
(3/2) kB T + (2/2) kB T = (5/2) kB T
∴ total internal energy per mole will be,
E = (5/2) NA kB T
= (5/2) R T
● Molar specific heat at const. volume is given as
CV = (dE/dT) = (5/2)R
●For one mole of gas we have CP - CV = R
∴ CP = R + (5/2)R
∴ CP = (7/2)R
●we have γ = CP / CV
∴ γ = (7/5)
(c) For a diatomic gas (containing non rigid
vibra ng molecules) enclosed in a container, held
at a constant temperature T and containing NA
atoms, each molecule has 3 transla onal, 2
rota onal and 1 vibra onal dof.
●Therefore, average energy per atom is
(3/2) kB T + (2/2) kB T + (2/2) kB T = (7/2) kB T
∴ total internal energy per mole will be,
E = (7/2) NA kB T
= (7/2)R T
●Molar specific heat at constant volume is given as
CV = (dE/dT)
= (7/2)R
●For one mole of gas we have CP - CV = R
∴ CP = R + (7/2)R ∴ CP = (9/2)R
●we have, γ = CP /CV
∴ γ = (9/7)
(d) For a polyatomic gas (containing non linear
vibra ng molecules) enclosed in a container, held
at a constant temperature T and containing NA
atoms, each molecule has 3 transla onal, 3
rota onal and f vibra onal dof.
●Therefore, average energy per atom is
(3/2) kB T + (3/2) kB T + f kB T
= 3 kB T + f kB T = (3+f) kB T
∴ total internal energy per mole will be,
E = (3+f)NA kB T
= (3+f) R T
●Molar specific heat at constant volume is given as
CV = (dE/dT) = (3+f)R
●For one mole of gas we have CP - CV = R
∴ CP = R + (3+f) R
∴ Cp = (4+f)R
●we have γ = CP /CV
∴ γ = (4+f)/(3+f)
Q.9 Explain the term 'radia on'.
●'radia on' is the fastest mode of transfer of heat,
in the form of electromagne c waves.
●Radia on does not require any material medium
since electromagne c waves travel through
vacuum.
●Due to radia on two bodies can exchange heat
without physical contact.
●The con nuous emission of energy from the
surface of any body is because of thermal energy of
that body.
●This emi ed energy is termed as radiant energy.
●Radia on get absorbed, When falls on a body,
which is not transparent to it, (floor or our hands),
and heats that body.
●Thus, electromagne c radia on emi ed by the
bodies, which are at higher temperature with
respect to the surroundings, is known as thermal
radia on.
Q.10 Define.
1) Coefficient of absorp on or absorp ve power
or absorp vity (a) : "The ra o of amount of heat
absorbed to total quan ty of heat incident is called
the coefficient of absorp on."
a = Q a /Q
2) Coefficient of reflec on or reflectance(r): "The
ra o of amount of radiant energy reflected to the
total energy incident is called the coefficient of
reflec on."
r = Qr/Q
3) Coefficient of transmission or transmi ance (t):
"The ra o of amount of radiant energy transmi ed
to total energy incident is called the coefficient of
transmission."
t = Qt/Q
Note:
Since all the three quan es a, r and tare ra os of
thermal energies, they are dimensionless
quan es.
03
Q.13 Is it possible that the same object can be
athermanous, diathermanous as well as a good
absorber ?
● Values of coefficient of absorp on, reflec on,
and transmission (a, r and t) depend on
1)Material of the object on which heat radia on is
incident.
2)wavelength of the incident heat radia on.
●Hence, it is possible that the same object may be
athermanous, diathermanous or a good absorber
for incident heat radia ons of different
wavelengths.
Q.11 Derive the rela on between a,r and t.
OR
Show that, a + r + t =1
Let
Q - Total amount of radient heat energy incident on
the surface of an object.
Qa - Part of heat radient heat energy absorbed.
Qr - Part of radient heat energy reflected.
Qt - part of radient heat energy transmi ed.
∴ Qa + Qr + Qt = Q
∴ (Qa + Qr + Qt)/Q = Q/Q
∴ (Qa/Q) + (Qr/Q) + (Qt/Q) = 1
∴ a + r + t =1
Where a, r, t are the coefficients of absorp on,
reflec on and transmission, respec vely.
Q.14 Are good absorbers also good emi ers?
OR
Explain why the walls of vacuum bo les or
thermos flasks are silvered.
●Consider two objects, which are opaque to
thermal radia on, having the same temperature
and same surface area.
●The surface of one object is well-polished and the
surface of the other object is painted black.
●The well-polished object reflects most of the
energy falling on it and absorbs li le.
●On the other hand, the black painted object
absorbs most of the radia on falling on it and
reflects li le.
●But the rate of emission of thermal radia on
must be equal to rate of absorp on
for both the objects, so that temperature is
maintained.
●Black painted object absorbs more, hence it must
radiate more to maintain
the temperature.
●Therefore, good absorbers are always good
emi ers and poor absorbers are poor emi ers.
●Since each object must either absorb or reflect
the radia on incident on it, a
poor absorber should be a good reflector and vice
versa.
●Hence, a good reflector is also a poor emi er. This
is the reason for silvering the walls of vacuum
bo les or thermos flasks.
Q.12 Explain the terms.
1) Perfect transmi er : "A substance through
which en re incident radiant energy is transmi ed
is called as perfect transmi er, or completely
transparent object."
● For perfect transmi er r = 0 and a = 0, ∴ t= 1,
2)Diathermanous substance: "A substance
through which incident radiant energy can pass
(transmit) is called as a diathermanous substance."
●For a diathermanous body, t≠ 0.
●A diathermanous body is neither a good absorber
nor a good reflector.
eg. glass, quartz, sodium chloride, hydrogen,
oxygen, dry air etc.
3)Athermanous substances: "The substance
through which incident radiant energy can not
pass (transmit), is called as an athermanous
substance or an opaque body.
● For athermanous substance. t = 0 ∴ a + r = 1,
eg. water, wood, iron, copper, moist air, benzene
etc.
4) Perfect reflector: "The substance which reflects
en re radiant energy incident on it is called as
perfect reflector"
● For a perfect reflector t= 0 and a = 0 ∴ r = 1
● A good reflector is a poor absorber and a poor
transmi er.
5) perfect blackbody : "The body which absorbs
en re radiant energy incident on it is called as ideal
or perfect blackbody"
● For perfect blackbody r = 0 and t= 0 ∴ a = 1
●Surface of perfect blackbody absorbs all the
radient energy incident on it, and does not reflect
any energy, therefore, appears black (unless its
temperature is very high to be self-luminous).
● Lamp black or pla num black absorb nearly 97%
of incident radiant heat, resemble a perfect
blackbody.
Q.15 Explain construc on and working of Ferry's
Blackbody.
OR
What is a perfect blackbody ? How can it be
realized in prac ce?
perfect blackbody : "The body which absorbs
en re radiant energy incident on it is called as ideal
or perfect blackbody"
04
Q.17 Explain Prevost's theory of heat exchange.
According to Prevost's theory of heat exchange,
●All bodies at all temperatures (above the absolute
zero temp. 0 K) emits thermal radia on.
●Rate of emission of thermal radia on depends on
1)Temperature of body.
2)Nature of emi ng surface.
3)Area of body.
construc on :
● It consists of a double walled hollow sphere
having a ny hole or aperture.
● The space between the walls is evacuated.
● Outer surface of the sphere is silvered.
● Inner surface of sphere is coated with lampblack.
● There is a conical projec on on the inner surface
of sphere opposite the aperture.
Working :
● Due to conical projec on, radia on arriving
along the axis of the aperture do not incident
normally on the inner surface.
● Therefore any radia on entering through the
small hole can not reflected back along the same
path and can not escape.
● Radia on suffers mul ple reflec ons inside the
sphere and get completely absorbed.
● Thus, the aperture behaves like a perfect
blackbody.
Note :
● For construc on of perfect black body, Wien
used a double walled hollow cylinder with a
ver cal slit as the aperture.
● This gives greater effec ve area as a perfect
blackbody.
Q.16 Write a short note on cavity radiator.
● Cavity radiator consists of a block of material
with internal cavity.
● The inner and outer surfaces are connected by a
small hole.
● The radia on falling on the block enters through
the hole, and cannot escape back.
● Hence, the cavity acts as a blackbody.
● When the block is heated to high temperature,
thermal radia on is emi ed, called as 'cavity
radia ons'
● Cavity radia ons are similar to radia on emi ed
by a blackbody.
● Its nature depends only on the temperature of
the cavity walls and not on the shape and size of
the cavity or the material of the cavity walls.
●At the same me, all bodies absorb radia on coming
from the surroundings, so that
1)If rate of emission is more than rate of absorp on,
temperature of body decreases.
e.g. cup of hot tea.
2)If rate of absorp on is more than rate of emission,
temperature of body increases.
e.g. plate containing ice kept on a table,
3)If rate of absorp on is equal to rate of emission, the
temperature of the body remains constant and the
body is said to be in thermal equilibrium with its
surroundings.
e.g. any body at room temperature.
Note :
1)Ho er bodies emits heat radia on at higher rate than
the cooler bodies.
2)Light coloured bodies reflect most of the incident
radia on where as dark coloured bodies absorb most of
the incident radia on.
05
Q.18 Explain the rise in temperature of a body,
when it absorb incident radia on.
● When a body, absorb incident radia on
● Being energy, radia on increases kine c energy
of the atoms of body oscilla ng about their mean
posi ons.
● We know that, average transla onal kine c
energy determines temperature of the body.
● Therefore absorbed radia on causes a rise in
temperature of the body.
Note :
● A body at room temperature emits thermal
radia on having wavelengths longer than
wavelengths of visible light. So we do not see
them.
● A hot body of temperature around 800 ºC, emits
thermal radia on having shorter wavelengths, in
the visible range and body appears red.
● At around 3000 ºC, body looks white hot. e.g.
The filament of a tungsten lamp.
Q.19 What is emissive power of a body?
●Amount of heat radia ons emi ed by a body
depends on
1) The absolute temperature of the body (T)
2) The nature of the body (the material and nature
of surface)
3) Surface area of the body (A)
4) Time dura on for which body emits radia on (t)
● For given body at a given temperature the
amount of heat radiated (Q ) is directly
propor onal to the surface area (A) and me
dura on (t).
∴Q∝At
∴Q=RAt
∴ R = Q/ A t
● Where R is constant called as emissive power of
given body at given temperature.
● Emissive power(R) : " The quan ty of heat
radia on emi ed per unit me per unit area of a
body is called as emissive power (OR radiant
power) of that body at given temperature."
● SI unit - J/ m² s or W/m²
● Dimensions [L⁰M¹T-³]
●At a given temperature, a perfect blackbody has
maximum emissive power.
●Now Kirchhoff's law of thermal radia on can also
be stated as follows
"for a body in thermal equilibrium, the emissivity is
equal to its absorp vity."
(e = a)
●Thus, if a body has high emissive power, it also
has high absorp ve power and vice versa.
∴ Good emi ers are good absorbers.
Q.22 State Kirchhoff's law of thermal radia on
and give it's theore cal proof.
● Kirchhoff's law of thermal radia on : "For a
body in thermal equilibrium, the emissivity is
equal to its absorp vity."
(a = e)
● Consider an ordinary body A and a perfect
blackbody B having iden cal shapes, placed in an
enclosure.
● Both bodies are In thermal equilibrium with
enclosure i.e. temperature of A, B and enclosure is
same.
● Let
R - Emissive power of A.
RB - Emissive power of B
a - Coefficient of absorp on of A.
● If Q is quan ty of radiant heat incident on each
body per unit me per unit area.
1) Quan ty of radiant heat absorbed by the body
A, per unit area per unit me is (aQ)
2) Quan ty of radiant heat absorbed by the body B,
per unit area per unit me is (Q)
● Since emissive power is the quan ty of radient
heat emi ed from unit area in unit me.
1) Quan ty of radiant heat emi ed by the body A,
per unit area per unit me is (R)
2) Quan ty of radiant heat emi ed by the body B,
per unit area per unit me is (RB)
● For a body In thermal equilibrium, radiant heat
emi ed in unit me, is equal to radiant heat
absorbed in in unit me.
So
1) For body O
a Q = R ------(1)
2) For body B
Q = RB ------(2)
●Dividing Eq.(1) by Eq.(2), we get
a =R/RB
∴a=e
(As e = R/RB)
Hence, Kirchhoff's law is theore cally proved.
Q.20 State and explain Coefficient of Emission or
Emissivity.
● Coefficient of emission or emissivity : "The ra o
of the emissive power R of the surface to the
emissive power RB of a perfect black surface, at the
same temperature is called as Coefficient of
emission or emissivity (e)"
e = R/RB
● For a perfect blackbody e = 1
● For a perfect reflector e = 0 .
● For an ordinary body,
0 < e < 1 depending on the nature of the surface
(colour and composi on).
e.g. emissivity of copper is 0.3
● Emissivity is larger for rough surfaces and smaller
for smooth and polished surfaces.
● Emissivity also varies with temperature and
wavelength of radia on to some extent.
Q.21 State Kirchhoff's Law of Heat Radia on and
show that Good emi ers are good absorbers.
●Kirchhoff's law of thermal radia on: At a given
temperature, the ra o of emissive power to
coefficient of absorp on of a body is equal to the
emissive power of a perfect blackbody at the same
temperature."
∴ (R/a) = RB
∴ (R/RB) = a
∴ e=a
06
Q.23 Explain spectral distribu on of blackbody
radia on.
Q.24 State and explain Wien's Displacement Law.
● Wien's displacement law : "The wavelength
(λmax), for which emissive power of a blackbody is
maximum, is inversely propor onal to the absolute
temperature of the blackbody (T)."
∴ λmax ∝ 1 / T
∴ λmax = b / T
where b is called the Wien's constant and its value
is 2.897 × 10-3 m K.
Note :
● Wien's displacement law is useful to determine
temperatures of distant stars, Sun, moon etc. as
follows
●λmax indicates the dominant wavelength at which
blackbody radiates, so it corresponds to the
dominant colour and temperature of the radia ng
body.
●e.g.
1) White dwarfs are hot stars with surface
temperature ~ 10000 K
2) Red giants are cooler corresponding to
surface temperature ~ 3000 K.
●Lummer and Pringsheim studied the energy
distribu on of blackbody radia on as a func on of
wavelength.
●They kept a black body at different constant
temperatures and measured the radiant power
per unit area (RB) for different wavelengths(λ).
●It gives spectral distribu on of blackbody
radia on. Different curves in fig. shows that,
1) At a given temperature, the energy is not
uniformly distributed as a func on of wavelength.
Q.25 State and explain Stefan-Boltzmann Law of
2) At a any given temperature, ini ally the radiant
Radia on.
power(RB) increases with increase of wavelength,
● Stefan-Boltzmann Law of Radia on : “The
reaches it's maximum value and then decreases.
energy radiated by a perfect blackbody per unit
(λmax) is the emi ed wavelength for which radiant
area per unit me is directly propor onal to the
power(RB) is maximum.
fourth power of its absolute temperature”.
3) Value of λ max decreases with increase in
RB ∝ T4
temperature.
4
∴ RB = σ T
4) Area under each curve represents total radiant
where σ is Stefan's constant.
power(RB) for all wavelengths.
σ = 5.67 × 10−8 J / m2 s K4 or W / m2 K4
5) At higher temperatures, area under the curve
dimensions: [L0M1T-3K-4].
increases.
● Note :
6) At room temperature (300 K) radiant power is
1) Emissive power of a perfect blackbody depends
maximum for waves having wavelength about 5 ×
-6
only
on its temperature
10 m.
and not on any other characteris cs such as colour,
For other waves radiant power is very small.
materials,
nature of surface etc.
So all the radiant energy have wavelength longer
2)For perfect blackbody we have
than wavelength of red light (i.e.infrared
RB=Q/At
radia ons).
Where
Note :
Q - Amount of radiant energy emi ed in me t
For theore cal explana on of spectral distribu on
A - Surface area.
of black body radia on.
∴ Q / A t = σ T4
1) Wien gave an expression from laws of
∴ Q = A t σ T4
thermodynamics. which can explain experimental
observa ons only for short wavelengths.
3)For ordinary (non black) body, eqn. of emissivity
is,
2) Lord Rayleigh and Sir James Jeans gave a formula
from the equipar on of energy, which can
e = R/RB
explain experimental observa ons only for long
∴ R = e RB
4
wavelength regions.
∴R=eσT
3) Therefore Planck propose a new model (which is
As, R = Q /A t
combina on of the two models) and gives an
Q /A t = e σ T4
empirical formula to explain the behaviour of
∴Q = A t e σ T4
blackbody.
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Q.26 Using Stefan-Boltzmann Law of Radia on
derive the equa on for net loss of radient energy
per unit area per unit me for a perfect black
body.
●Consider a perfect blackbody kept in a
surrounding at a lower temperature.
●Let
T - Absolute temperature of body.
T0 - Absolute temperature of surrounding.
●then
1)Energy radiated by body per unit area per unit
me
4
=σT
2) Energy absorbed from surroundings per unit
area per unit me = σ T04
● Therefore net loss of energy by perfect
blackbody per unit area per unit me = σT4 - σ T04
= σ(T4 - T04)
Note :
1) For an ordinary body, net loss of energy per unit
area per unit me = e σ (T4 - T04)
e is emissivity of ordinary body.
2) If the body is at a temperature lower than the
4
4
surrounding i.e., T< T0 , then e σ (T0 - T ) will be the
net gain in thermal energy of the body per unit
area per unit me.
Q.29 If surfaces of all bodies are con nuously
emi ng radiant energy, why do they not cool
down to 0 K?
●According to Prevost's theory of heat exchange,
all the bodies emit as well as absorb the heat
radia on con nuously.
● E m i s s i o n o f h e at ra d i a o n d e c re a s e s
temperature of body, at the same me absorp on
of heat radia on from surroundings increases
temperature of body.
●Therefore any body do not cool down to 0 K.
Q.30 The wavelength corresponding to maximum
intensity for the Sun (λmax) is in the blue-green
region of visible spectrum. Why does the Sun then
appear yellow to us?
●In Sunlight the wavelength corresponding to
maximum intensity, (λmax) is in the blue to green
region of visible spectrum.
●When sunlight travels through Earth's
atmosphere blue to green colours get sca ered
due to their short wavelength.
●So only longer wavelengths of sunlight can reach
to our eyes.
●As a result, sun appears yellow to us.
Q.27 Why are the bo oms of cooking utensils
blackened and tops polished?
● Bo oms of cooking utensils are blackened to
absorb maximum heat from the flame, as dark
coloured rough surface absorbs most of incident
radia ons.
● Tops of cooking utensils are polished to reduce
heat loss due to radia on, as emissive power of
light coloured polished surface is small.
Q.28 A car is le in sunlight with all its windows
closed on a hot day. A er some me it is observed
that the inside of the car is warmer than outside
air. Why?
●Glass is diathermanous for radia ons having
small wavelengths.
●So high frequency radia ons coming from hot
sun get transmi ed inside car through glass
windows.
●It increases the inside temperature of car.
●But glass is athermanous for radia ons having
long wavelengths.
● Therefore low frequency heat radia ons from
inside the car can not escape through glass
windows.
● That's why A er some me it is observed that
the inside of the car is warmer than outside air.
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