2. Can you recall?
1. What are different states of matter?
2. How do you distinguish between solid, liquid and gaseous
states?
3. What are gas laws?
4. What is absolute zero temperature?
5. What is Avogadro number? What is a mole?
6. How do you get ideal gas equation from the gas laws?
7. How is ideal gas different from real gases?
8. What is elastic collision of particles?
9. What is Dalton's law of partial pressures?
3. SOLID LIQUID GAS
MOLECULES ATOMS
Matter is made up of tiny particles called molecules.
All types of matter are made up of extremely small particles.
STATES OF MATTER
6. Gas lawโs
โข Boyleโs Law: V โ
๐
๐ท
at constant T
โข Charles Law: V ๐ถ T at constant P
โข Avogadroโs Law: V ๐ถ ๐ง
โข Gay-Lussacโs Law: P ๐ถ T at constant V
Combining these all lawโs, we get
V ๐ถ n T
๐
๐ท
โด ๐ท๐ฝ ๐ถ ๐๐ป
PV = nRT
Where, R is proportionality constant OR Universal Gas Constant
R = 8.314 J ๐๐๐โ๐
๐โ๐
8. ATOM
Weight of one atom of an element =
๐จ๐๐๐๐๐ ๐พ๐๐๐๐๐
๐.๐๐๐ ๐ฟ ๐๐๐๐
Weight of one molecule of a compound =
๐ด๐๐๐๐๐๐๐๐ ๐พ๐๐๐๐๐
๐.๐๐๐ ๐ฟ ๐๐๐๐
For Gas,
n =
๐ต
๐ต๐จ
Where, n โ mole
N โ Number of molecules in a gas
๐ต๐จ โ Avogadro's number OR Number of molecule in 1 mole of gas
9. IDEAL GAS
Equation, PV = nRT
PV =
๐ต
๐ต๐จ
RT โฆโฆโฆโฆ. [n =
๐ต
๐ต๐จ
]
Here, R = ๐ต๐จ๐๐ฉ
Where, ๐๐ฉ is the Boltzmann Constant
So, PV =
๐ต
๐ต๐จ
๐ต๐จ๐๐ฉ๐ป
PV = N๐๐ฉ๐ป
Where, N is the Number of Particles
NOTE: A gas obeying the equation of state i.e., PV = nRT at all
pressure, and temperature is an ideal gas.
10. Behaviour of gas
โข Gas contained in container is characterized by its
pressure, volume and temperature.
โข Particle of gas are always in constant motion.
โข It is very hard to evaluate the motion of single
particle of gas.
โข We find average of physical quantities and then
relate to the macroscopic view of gas.
11. Fig.(a): A gas with
molecules dispersed in
the container: A stop
action photograph.
Fig.(b): A typical
molecule in a gas
executing random
motion.
12. IDEAL gaS
โข A gas obeying ideal gas equation is nothing but an ideal
gas.
โข Intermolecular interaction is different or absent.
REAL GAS
โข Compose of atom/molecule which do interact with
each other.
โข If there is atom/molecule are so apart then we can say
that the real gases are ideal.
โข This can be achieved by lowering pressure and
increasing temperature.
13. MEAN FREE PATH
โข Represented by โ๐โ
โข Defined as average distance travelled by a molecule with constant
velocity between two successive collision.
โข Mean free path is inversely vary with density.
โข Mean free path also inversely proportional to size of the particle of
gases.
Conclusion: ๐ =
๐
๐ ๐๐
๐ =
๐
๐ ๐๐๐ ๐
๐ =
๐
๐ ๐(
๐ต
๐ฝ
)๐ ๐
โฆโฆโฆ.[๐ =
๐ต
๐ฝ
]
14. EAXMPLE
โข Obtain the mean free path of nitrogen molecule at 0 ยฐC and 1.0 atm pressure.
The molecular diameter of nitrogen is 324 pm (assume that the gas is ideal).
Solution: Given T = 0 ยฐC = 273 K, P = 1.0 atm = 1.01ร๐๐๐
Pa and
d = 324 pm = 324 ร ๐๐โ๐๐
m.
For ideal gas PV = N๐๐ฉT, โด
๐ต
๐ฝ
=
๐ท
๐๐ฉ๐ป
Using Eq., mean free path
๐ =
๐
๐ ๐(
๐ต
๐ฝ
)๐ ๐
=
๐๐ฉ๐ป
๐๐ ๐ ๐๐ท
=
๐.๐๐ ๐ฟ ๐๐โ๐๐ ๐ฑ
๐ฒ
(๐๐๐ ๐ฒ)
๐๐ ๐๐๐ ๐ฟ ๐๐โ๐๐ ๐
๐
(๐.๐๐ ๐ฟ ๐๐๐ ๐ท๐)
= 0.8 X ๐๐โ๐
๐
Note that this is about 247 times molecular diameter.
15. Pressure of ideal gas
A cubical box of side L.
It contains n moles of
an ideal gas.
Volume of that cube is, V = ๐ณ๐
,
Where L โ ๐ณ๐๐๐๐๐ ๐๐ ๐๐๐ ๐
We have to find relation between pressure P of gas
with molecular speed
Here, change in momentum in โcomponent
โ ๐ท๐ = ๐ญ๐๐๐๐ ๐๐๐๐๐๐๐๐ โ ๐ฐ๐๐๐๐๐๐ ๐ด๐๐๐๐๐๐๐
= - m๐๐ โ ๐๐๐
= - 2 m๐๐
So, total momentum applied on the wall is 2 m๐๐
The total distance travelled by particle of gas during
collision of molecule and wall is 2L
โ๐๐
m
๐๐
16. โ๐ =
๐๐ณ
๐๐
โฆโฆ[Speed =
๐ซ๐๐๐๐๐๐๐
๐ป๐๐๐
]
For 1-molecule, Average force = Average rate of change of momentum
=
๐ด๐๐๐๐๐๐๐
๐ป๐๐๐
=
๐๐๐๐๐
โ๐
=
๐๐๐๐๐
๐๐ณ
๐๐๐
=
๐๐๐๐๐
๐๐ณ
x ๐๐๐
Average Force (๐ญ๐) =
๐๐๐๐
๐
๐๐ณ
Similarly, for particle 2,3,4, โฆโฆwith respect to x-component velocities are
๐๐๐
, ๐๐๐
, ๐๐๐
, โฆ . .
Average Force (๐ญ) =
๐(๐๐๐
๐ +๐๐๐
๐ + ๐๐๐
๐ + โฆโฆ..)
๐ณ
20. INTERPRETATIONOF TEMPERATURE IN KTG
We know that, P =
๐
๐
๐ต
๐ฝ
๐ ๐๐
P V =
๐
๐
๐ต ๐ ๐๐
=
๐
๐
๐ต (
๐ ๐๐
๐
) โฆโฆโฆโฆโฆ(i)
In ideal gas there is no interaction molecule so P.E. = 0.
Here, Total Energy (E) = K.E. + P.E.
= N (
๐ ๐๐
๐
) + 0 = N (
๐ ๐๐
๐
) โฆโฆโฆโฆโฆ.(ii)
Put in equation (i)
P V =
๐
๐
E โฆโฆโฆ.(iii)
Using Ideal Gas, P V = N ๐๐ฉ ๐ป =
๐
๐
E โฆโฆโฆ.(iv)
N ๐๐ฉ ๐ป =
๐
๐
E
E =
๐
๐
N ๐๐ฉ ๐ป
๐ฌ
๐ต
=
๐
๐
๐๐ฉ ๐ป
21. Example
At 300 K, what is the rms speed of Helium atom? [mass of He atom
is 4u, 1u = 1.66 ร ๐๐โ๐๐
kg; ๐๐ฉ= 1.38 ร ๐๐โ๐๐
J/K]
Solution: Given T = 300 K, m = 4 ร 1.66 ร ๐๐โ๐๐
kg
Average K. E =
๐
๐
๐ ๐๐ =
๐
๐
๐๐ฉ ๐ป
๐๐ =
๐๐๐ฉ ๐ป
๐
=
๐ ๐ฟ ๐.๐๐ ๐ฟ๐๐โ๐๐ ๐ฟ ๐๐๐
๐ ๐ฟ1.66 ร ๐๐โ๐๐
= 187.05 X ๐๐๐
๐๐๐๐ = ๐๐ = 13.68 X ๐๐๐
= 1368 m/s
22. ๐ณ๐จ๐พ ๐ถ๐ญ ๐ฌ๐ธ๐ผ๐ฐ๐ท๐จ๐น๐ป๐ฐ๐ป๐ฐ๐ถ๐ต
Energy of single molecule is
K.E =
๐
๐
๐๐๐
๐
+
๐
๐
๐๐๐
๐
+
๐
๐
๐๐๐
๐
For gas at temperature T,
The average K.E per molecule is <K.E.>
<K.E> = <
๐
๐
๐๐๐
๐
> + <
๐
๐
๐๐๐
๐
> + <
๐
๐
๐๐๐
๐>
But the mean distance is
๐
๐
๐๐ฉ ๐ป
๐
๐
๐๐ฉ ๐ป is the molecular energy in single x or y or z component.
23. DEGREEOF FREEDOM
The molecule is free to move in whole space.
Molecule can move in 3-D space.
โDegree of freedom of a system are defined as the total number of co-ordinate
required to describe the position and configuration of the system completely.โ
DIATOMIC MOLECULE
Molecules having 3-translational degrees (x, y, z)
Diatomic molecule rotate around axis
Suppose, a molecule moving along axis z with ๐ฐ๐ with ๐๐
K.E =
๐
๐
๐ฐ๐ ๐๐
๐
For axis y , K.E =
๐
๐
๐ฐ๐ ๐๐
๐
So, Total energy = E (Translational) + E (Rotational)
= [
๐
๐
๐๐๐
๐
+
๐
๐
๐๐๐
๐
+
๐
๐
๐๐๐
๐] + [
๐
๐
๐ฐ๐ ๐๐
๐
+
๐
๐
๐ฐ๐ ๐๐
๐
]
Fig.: The two independent
axes z and y of rotation of
a diatomic molecule lying
along the x-axis.
24. From equipartition of energy:
Each and every molecule separated with
๐
๐
๐๐ฉ ๐ป molecule.
For, Translational motion โ 3 (dof) โ [ 3 X
๐
๐
๐๐ฉ ๐ป ]
Rotational motion โ 2 (dof) โ [ 2 X
๐
๐
๐๐ฉ ๐ป ]
Vibrational motion โ 2 (dof) โ [ 2 X
๐
๐
๐๐ฉ ๐ป ]
โ [
๐
๐
m ๐๐ +
๐
๐
k ๐๐ ]
โข The energy of molecule [
๐
๐
๐๐ฉ ๐ป ] is valid for high temperature and not
valid for extremely low temperature where quantum effects become
important.
25. SPECIFIC HEAT CAPACITY
If temperature of gas increase
(Cause considerable change in volume and pressure)
๐ช๐ฝ โ Specific heat at constant volume
๐ช๐ท โ Specific heat at constant pressure
MAYERโS RELATON
Consider, 1 mole of gas (Ideal) enclosed in cylinder
Let P, V, T be the pressure, volume and temperature
dT = Change in temperature
(Volume is constant โ ๐ต๐ ๐๐๐๐ ๐๐ ๐ ๐๐๐)
d๐ธ๐ โ Increase in the internal energy
dE โ ๐ฐ๐๐๐๐๐๐๐ ๐๐๐๐๐๐
โด ๐ ๐ธ๐ = ๐ ๐ฌ = ๐ช๐ฝ๐ ๐ป โฆ โฆ โฆ โฆ ๐
Where, ๐ช๐ฝ โ ๐ด๐๐๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐
26. If gas is heated with same given temperature at constant pressure
then,
dV โ Change (increase) in volume
Here, work is done to pull backward the piston
dw = p dv
โด ๐ ๐ธ๐ = dE + dw = ๐ช๐ท๐ ๐ป โฆโฆโฆ(ii)
Where, ๐ช๐ท โ Molar specific heat at constant pressure
From equation (i)
โด ๐ช๐ฝ๐ ๐ป + ๐ ๐ = ๐ช๐ท๐ ๐ป
๐ช๐ฝ๐ ๐ป + ๐ท ๐ ๐ฝ = ๐ช๐ท๐ ๐ป
๐ท ๐ ๐ฝ = ๐ช๐ท๐ ๐ป - ๐ช๐ฝ๐ ๐ป
๐ท ๐ ๐ฝ = (๐ช๐ท - ๐ช๐ฝ ) ๐ ๐ป โฆโฆโฆโฆโฆโฆ (iii)
27. For, 1 mole of gas, PV = RT โฆโฆโฆโฆ(n = 1)
But, when pressure is constant
P dV = R dT
(๐ช๐ท - ๐ช๐ฝ ) ๐ ๐ป = R dT
๐ช๐ท - ๐ช๐ฝ = ๐น โฆโฆโฆ(Mayerโs relation)
Now, ๐ช๐ท - ๐ช๐ฝ =
๐น
๐ฑ
Where, J โ mechanical equivalent of heat
Here, ๐ช๐ท = ๐ด๐ ๐บ๐ท
๐ช๐ฝ = ๐ด๐ ๐บ๐ฝ
๐ด๐ = Molar mass of gas
๐บ๐ฝ, ๐บ๐ท = Respective principle of heat
๐ด๐ ๐บ๐ท - ๐ด๐ ๐บ๐ฝ =
๐น
๐ฑ
(๐บ๐ท - ๐บ๐ฝ) =
๐น
๐ด๐๐ฑ
30. For non rigid vibrating molecule
E = Translational Energy + Rotational Energy + Vibrational Energy
E = [3 X
๐
๐
๐ต๐จ๐๐ฉ ๐ป] + [ 2 X
๐
๐
๐ต๐จ๐๐ฉ ๐ป] + [ 2 X
๐
๐
๐ต๐จ๐๐ฉ ๐ป]
=
๐
๐
๐ต๐จ๐๐ฉ ๐ป
๐ช๐ฝ =
๐ ๐ฌ
๐ ๐
=
๐
๐
๐๐ฉ๐ป๐ต๐จ
๐ป
โฆโฆโฆ.(๐๐ฉ๐ต๐จ = ๐น)
๐ช๐ฝ =
๐
๐
๐๐ฉ๐ต๐จ =
๐
๐
๐น
๐ช๐ท = ๐น + ๐ช๐ฝ
= R +
๐
๐
๐น =
๐
๐
๐น
โด ๐ธ =
๐ช๐ท
๐ช๐ฝ
=
๐
๐
๐น
๐
๐
๐น
=
๐
๐
31. b) POLYaTOMICGASES
E = Translational Energy + Rotational Energy + (f x Vibrational Energy)
Where, f โ Number of vibrational dof
E = [3 X
๐
๐
๐ต๐จ๐๐ฉ ๐ป] + [ 3 X
๐
๐
๐ต๐จ๐๐ฉ ๐ป] + [ f X 2 X
๐
๐
๐ต๐จ๐๐ฉ ๐ป]
=
๐
๐
๐ต๐จ๐๐ฉ ๐ป (3+3+2f)
= (3 + f) ๐ต๐จ๐๐ฉ ๐ป
๐ช๐ฝ =
๐ ๐ฌ
๐ ๐
=
(๐+๐)๐๐ฉ๐ป๐ต๐จ
๐ป
โฆโฆโฆ.(๐๐ฉ๐ต๐จ = ๐น)
๐ช๐ฝ = (3+f)๐๐ฉ๐ต๐จ = (3+f) ๐น
๐ช๐ท = ๐น + ๐ช๐ฝ
= R + (3+f) ๐น = ๐ + ๐ ๐น
โด ๐ธ =
๐ช๐ท
๐ช๐ฝ
=
(๐+๐)
(๐+๐)
32. HEAT
Heat is the form of energy which can be formed from one
object to another object.
MODES OF CONDUCTION
1)Conduction
2)Convection
3)Radiation
โข Does not required material medium for transfer
โข Emission of heat โ radiant energy โ Electromagnetic wave
โข Can travel through both material and vacuum medium.
โข Fastest way of heat transfer.
Requires material medium for transfer.
33. INTERACTION OF THERMAL RADIATIONANDMATTER
When any thermal radiation falls on object then some part is
reflected, some part is absorbed and some part transmitted.
Consider,
Q โ Total amount of heat energy incident
๐ธ๐ โ ๐จ๐๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐๐๐๐๐๐
๐ธ๐ โ ๐จ๐๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐๐๐๐๐๐๐
๐ธ๐ โ ๐จ๐๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐
Total energy (Q) = ๐ธ๐ + ๐ธ๐ + ๐ธ๐
โด ๐ = ๐ + ๐ + ๐๐
Where, a =
๐ธ๐
๐ธ
, r =
๐ธ๐
๐ธ
, ๐๐ =
๐ธ๐
๐ธ
Coefficient of absorption, reflection, transmission
34. COEFFICIENT OF HEAT READIATION
โข Coefficient of absorption/absorptive power/absorptivity
โThe ratio of amount of heat absorbed to total amount of heat incident.โ
a =
๐ธ๐
๐ธ
โข Coefficient of reflection/reflectance
โThe ratio of amount of heat reflected to total amount of heat incident.โ
r =
๐ธ๐
๐ธ
โข Coefficient of transmission or transmittance
โThe ratio of amount of heat is transmitted to total amount of heat
incident.โ
๐๐ =
๐ธ๐
๐ธ
35. Cases
a)Perfect transmitter: r = 0, a = 0, ๐๐ โ 1
Object is said to be completely transparent.
b) Di-athermanous substance: ๐๐ โ 0
โSubstance through which heat radiation can pass is called di-
athermanousโ
Neither a good absorber nor good reflector
E.G., Quartz, Sodium Chloride, Hydrogen, Oxygen
c) Athermanous substance: ๐๐ = 0, a + r = 1
โSubstance which are largely opaque to thermal radiation i.e. do
not transmit heat.โ
36. d) Good reflector/poor absorber/poor transmitter: ๐๐ = 0, a = 0, r = 1
e) Perfect blackbody: ๐๐ = 0, r = 0 and a = 1
โAll the incident energy is absorbed by the object such an object is called
a perfect blackbody.โ
NOTE: All the a (absorption), r
(reflection) and t (transmission) depends
upon wavelength of incident radiation.
โA body which absorb entire radiant
energy incident on it, is called an ideal or
perfect blackbody.โ
Platinum black absorb nearly 97 % of
incident radiant heat on it.
Good absorber are good emitter and
poor absorber are poor emitter.
Fig.: Ferryโs blackbody.
37. EMISSION OF HEAT RADIATION
Pierre prevost published a theory of radiation known
as theory of exchange of heat.
At temperature 0 K or above it.
โข Body radiate thermal energy and at same time
they absorb radiation received from surrounding.
โข Emission per unit time depends upon surface of
emission, its area and size.
โข Hotter body radiate at higher rate than the cooler
body
38. AMOUNT OF HEAT RADIATED BY THE BODY DEPENDS ON โ
โข The absolute temperature of body
โข The nature of body (Polished or not, size, colour)
โข Surface area of body (A)
โข Time duration
Amount of heat radiated ๐ถ Surface area of body and time duration
Q ๐ถ At
Q = Rat
Where, R โ Emissive power/Radiant power
R =
๐ธ
๐จ๐
SI Unit โ J ๐โ๐
๐โ๐
Dimension - [๐ณ๐
๐ด๐
๐ปโ๐
]
39. EMISSIVITY
โThe coefficient of emission or emissivity (e) of a given surface is the ratio of the emissive
power R of the surface to emissive power ๐น๐ฉ of perfect black body at same temperature.โ
e =
๐น
๐น๐ฉ
For perfect blackbody, e = 1, Perfect reflector, e = 0
KIRCHOFFโS LAW OF HEATRADIATION
โAt a given temperature, the ratio of emissive power to coefficient of absorption of a body
is equal to the emissive power of a perfect blackbody at the same temperature for all
wavelength.โ
๐น
๐
= ๐น๐ฉ
โด
๐น
๐น๐ฉ
= ๐
โด ๐ = ๐
At thermal equilibrium, emissivity is equal to absorptivity.
40. WEINโS DISPLACEMENTLAW
The wavelength for which emissive power of a blackbody is
maximum is inversely proportional to the absolute temperature
of the blackbody.
๐๐๐๐ ๐ถ
๐
๐ป
๐๐๐๐ = ๐ ๐ฟ
๐
๐ป
Where, b โ Weinโs constant
b = 2.897 X ๐๐โ๐
mk
Law is useful to determine distant length.
41. SREFAN-BOLTZMANN LAWOF RADIATION
โThe rate of emission of radiant energy per unit area or the power
radiated per unit area of a perfect blackbody is directly proportional to
fourth power of its temperature.โ
R ๐ถ ๐ป๐
R = ๐ ๐ป๐
Where, ๐ โ ๐บ๐๐๐๐๐โฒ
๐ ๐๐๐๐๐๐๐๐
๐ = ๐. ๐๐ ๐ฟ ๐๐โ๐
Unit โ J ๐โ๐
๐โ๐
๐โ๐
Dimension โ [๐ณ๐
๐ด๐
๐บโ๐
๐ฒโ๐
]
We know that, R =
๐ธ
๐จ๐
So, R = ๐ ๐ป๐
โ
๐ธ
๐จ๐
= ๐ ๐ป๐
โฆโฆโฆโฆ(For black body)
For ordinary body, R = ๐ ๐ ๐ป๐
42. Let, T โ ๐จ๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐ ๐ ๐ค๐๐๐ฉ ๐๐ญ ๐ฅ๐จ๐ฐ๐๐ซ
๐๐๐ฌ๐จ๐ฅ๐ฎ๐ญ๐ ๐ญ๐๐ฆ๐ฉ๐๐ซ๐๐ญ๐ฎ๐ซ๐ ๐ป๐ถ
Energy radiated per unit time = ๐ ๐ป๐
Energy radiated per unit time in surrounding = ๐ ๐ป๐
๐
Net loss of energy by perfect blackbody per unit time = ๐ ๐ป๐ - ๐ ๐ป๐
๐
= ๐ ( ๐ป๐
- ๐ป๐
๐
)
For ordinary body = ๐๐ ( ๐ป๐
- ๐ป๐
๐
)