The document discusses Avogadro's constant and its relationship to moles. It defines a mole as the amount of a substance containing 6.02x1023 particles, which may be atoms, molecules, or ions. It then discusses how molar quantities allow expressing amounts of substances in moles rather than grams. For example, one can refer to the molar volume or molar mass of a gas. The document also discusses the kinetic theory of gases and how it relates the pressure and temperature of a gas to the motion of its molecules.
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kinetic theory of gases ppt by Mr. B.Sesha Sai
If you want this slides you can contact me.
It contains about kinetic theory of gases.
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Wk 6 p3 wk 7-p8_1.2-1.3 & 10.1-10.3_ideal gases
1. PHYSICAL QUANTITIES AND
UNITS
h) show an understanding that the Avogadro constant is the number
of atoms in 0.012 kg of carbon-12
(i) use molar quantities where one mole of any substance is the
amount containing a number of particles equal to the Avogadro
constant
2. Avogadro’s Constant
• The Avogadro constant, NA, is the number of atoms in 0.012 kg of Carbon-
12 or the number of particles in one mole of any substance.
• The substance may be in a solid, liquid or gaseous state, and its basic
fundamental unit may either be atomic, molecular or ionic in form.
• NA equals to 6.02 x 10²³ atoms per mole.
• A molar quantity is the amount per mole unit (6.02 x 10²³ particles) e.g.:
– molar volume of a gas at s.t.p. is 22.4 dm³.
– Number of particles per mole 6.02 x 1023 atoms/mol
– Molar mass in g/mol
• E.g the number of atoms in 6 g of Carbon-12 is 6/12 x 6.02 x 10²³ or 3.01 x
10²³.
2
3. MOLAR QUANTITIES
• Avogadro’s constant = no. of particles
mole
• Molar mass = mass
mole
• Molar gas volume = volume
mole
4. 4
Ideal gases
• 10. Ideal Gases
• Content
• 10.1 Equation of state
• 10.2 Kinetic theory of gases
• 10.3 Pressure of a gas
• 10.4 Kinetic energy of a molecule
• Learning Outcomes
• (a) recall and solve problems using the equation of state for an ideal gas expressed as
pV = nRT. (n = number of moles)
• (b) infer from a Brownian motion experiment the evidence for the movement of
molecules.
• (c) state the basic assumptions of the kinetic theory of gases.
• (d) explain how molecular movement causes the pressure exerted by a gas and hence
deduce the relationship, p = ⅓ Nm/V < c2> . (N = number of molecules)
[A simple model considering one-dimensional collisions and then extending to three dimensions
using
1
3
< 𝑐2
> = < cx
2
>
cx is sufficient]
• (e) compare pV = ⅓ Nm< c2> with pV = NkT and hence deduce that the average
translational kinetic energy of a molecule is proportional to T.
5. 5
Microscopic model of a gas
• Physics describes and explains the behaviour of various
systems
• In some cases it is impossible to describe what
happens to each component of a system e.g properties
of a gas in terms of the motion of each of its molecules
• In such cases we describe the average conditions in
the gas rather than describing the behaviour of each
molecule i.e macroscopic instead of microscopic
• Kinetic theory of an ideal gas is one such example
where we make simple assumptions to derive at a law
relating kinetic energy to temperature
6. 6
What is the evidence that gas molecules
are moving around all the time?
• 1 cm3 of atmospheric air contains approx 3 x 1025 molecules
• 1827 biologist Robert Brown observed tiny pollen grains suspended
in water with a microscope with a illuminating beam
• Although the water was completely still, the grains were always
moving in a jerky, haphazard manner : we call this Brownian
motion
• Hence the assumption that the water molecules are in rapid,
random motion under the bombardment from all sides of the water
molecules
• We can see the same movement of tiny soot particles in smoke
• A century later, Mr. Einstein did a theoretical analysis of Brownian
motion and estimated the diameter of a typical atom in the order
of 10-10 m
• It has been found that the average spacing of atoms of liquids is up
to 2 times that of solids or still in the order of 10-10 m. For gases it is
of the order of 10-9 m
7. 7
The gas laws
• Experiments in the 17th and 18th centuries
showed that the volume, pressure and
temperature of a given sample of gas are all
related
8. 8
Boyle’s Law
Robert Boyle 1627-91
Boyle’s Law states that the pressure of a given
mass of gas at constant temperature is
inversely proportional to its volume.
p 1/V or pV = constant
If p1,V1 are the initial pressure and volume of
the gas, and p2,V2 are the final values after a
change of pressure and volume carried out at
constant temperature, then
p1V1 = p2V2
9. 9
Charles Law
Jacques Charles 1746-1823
Charles’Law states that the volume of a given mass of any
gas at constant pressure is directly proportional to its
thermodynamic (absolute) temperature T
V T or V/T = constant
If V1,T1 are the initial volume and temperature of the gas,
and V2,T2 are the final values after a change of volume and
temperature carried out at constant pressure, then
V1/T1 = V2/T2
T K = (273 + ) degrees C
10. 10
Pressure Law
Joseph Gay-Lussac 1778-1850
Pressure Law or Gay-Lussac’s Law states that the
pressure of a given mass of gas at constant volume
is directly proportional to its absolute
temperature.
p T or p/T = constant
If p1,T1 are the initial pressure and temperature of
the gas, and p2,T2 are the final values after a change
of pressure and temperature carried out at constant
volume, then
p1/T1 = p2/T2
11. 11
Ideal Gas Equation
In practice, real gases obey the gas laws only at moderate
pressures and at temperatures well above the temperature at
which the gas would liquefy.
An ideal gas is one which obeys the 3 gas laws and is given
by:
pV T or pV/T = constant
i.e. p1V1/T1 = p2V2/T2
All the above laws relate to a fixed mass of gas
It is also found by experiments that the volume of a gas is
proportional to its mass giving a combined equation of
pV mT or pV = AmT
where m is the mass of the gas and A is a constant of
proportionality
As A is different for different gases we express the fixed
mass of gas in terms of the number of moles of gas present
12. 12
Avogadro’s Hypothesis
Avogadro’s hypothesis states that equal
volume of all gases, at the same
temperature and pressure, contain equal
number of molecules.
As such, gases like O2, N2, CH4, CO2,
CO, H2 etc all contain the equal number
of molecules at the same temperature
and pressure.
13. 13
Mole (abbreviated as mol)
A mole is the number of elementary
units (atoms or molecules) of any
substance which is equal to the amount
of atoms in 12 g of carbon-12.
The number of molecules per mole for
all substances is the same and is called
the Avogadro’s constant or number NA
NA is 6.02 x 1023/mol
14. 14
Relative atomic mass & atomic mass
unit
The relative atomic mass Ar is the ratio of the mass of an atom
to one-twelfth of the mass of an atom of carbon-12
The relative atomic mass is numerically equal to the mass in
grams of a mole of atoms
The relative molecular mass Mr is the ratio of the mass of a
molecule to one-twelfth of the mass of an atom of carbon-12,
and is numerically equal to the mass in grams of a mole of
molecules
The atomic mass unit u is one-twelfth of the mass of an atom
of carbon-12, and has the value 1.66 x 10-27 kg (also called the
unified atomic mass constant)
Simple way to find the mass of 1 mole of an element is to take
its nucleon number expressed in grams
e.g nucleon number of argon 40Ar is 40. Therefore, 1 mole
of argon then has a mass of 40 g
15. 15
Gas laws in terms of moles & molar gas constant
• Hence in terms of moles n,
pV nT
and putting in a new constant of proportionality R,
pV = nRT
• R is called the molar gas constant (or sometimes universal
gas constant as it has the same value for all gases)
• R has a value of 8.3 J K-1mol-1
• The above equation is expressed in the form pVm = RT
where Vm is the volume occupied by 1 mole of the gas
• Another version is pV = NkT where N is the number of
molecules in the gas and k is a constant called the
Boltzmann constant which has a value of 1.38 x 10-23 J K-1
• The molar gas constant R and the Boltzmann constant k are
connected through the Avogadro constant NA
nR = kNA
16. 16
Experiment has confirmed that one mole
of all gases at s.t.p. ( 0 degrees C and
760mm Hg or 1.01 x 105 Pa ) occupies 22.4
litres (1 litre is 1000 cm3).
From pV/T = k , substituting the above
values, the constant for one mole R, called
the molar gas constant, is therefore the
same for all gases and has a value of 8.31
J/K/mol
At s.t.p.:
pressure = 1.01 x 105Pa
temperature = 273.15 K
molar gas volume = 22.4dm3
17. 17
Equation of state
• The precise relation between volume,
pressure, temperature and the mass of the
given gas in a sample is called the equation of
state of the gas
• An ideal gas is one which obeys the equation
of state:
pV = nRT
• For approximate calculations the ideal gas
equation can be used with real gases
18. 18
Exercise
• Find the volume occupied by 1 mole of air at
stp (273K and 1.01 x 105 Pa), taking R as 8.3 J
K-1 mol-1 for air.
• Solution:
pV = nRT, so V = nRT/p
= 2.24 x 10-2 m3
19. 19
Exercise
• Find the number of molecules per cubic metre of air at stp.
Solution:
from above example, 1 mole of air at stp is 2.24 x 10-2
m3
but 1 mole of air contains NA molecules where NA is the
Avogadro constant.
thus the number of molecules per cubic metre of air is
6.02 x 1023/2.24 x 10-2 = 2.69 x 1025 m-3
20. 20
Exercise
• A syringe contains 25 x 10-6 m3 of helium gas at a
temperature of 20° C and a pressure of 5.0 x 104 Pa.
The temperature is increased to 400° C and the
pressure on the syringe is increased to 2.4 x 105 Pa.
Find the new volume of gas in the syringe.
Solution:
Using p1V1/T1 = p2V2/T2 with T1 = 293 K & T2 =
673 K
V2 = 12 x 10-6 m3
21. 21
Exercise
Oxygen gas contained in a cylinder of volume 1 x 10-2 m3 has a
temperature of 300 K and a pressure 2.5 x 105 Pa. Calculate the
mass of the oxygen used when the oxygen pressure has fallen to
1.3 x 105 Pa.
22. 22
Exercise
Oxygen gas contained in a cylinder of volume 1 x 10-2 m3 has a
temperature of 300 K and a pressure 2.5 x 105 Pa. Calculate the
mass of the oxygen used when the oxygen pressure has fallen to
1.3 x 105 Pa.
Solution
Let the initial and final number of moles of oxygen be n1 and n2
respectively.
n1 = p1V/(RT), n2 = p2V/(RT)
no. of moles used = n1 – n2 = (p1 – p2)V/(RT)
= [(2.5 – 1.3) x 105 x 10-2]/(8.31 x 300)
= 0.48 moles
= 0.48 x 32 x 10-3 kg = 0.015 kg.
23. 23
The kinetic theory
• Robert Boyle in 17th century developed an explanation
of how a gas exerts pressure
• Daniel Bernoulli in 18th century explained it in greater
detail
• The basic idea was that gases consist of atoms or
molecules moving about at great speed (later visualised
by Robert Brown) and that the gas exerts a pressure on
the walls of its container because of the continued
impacts of the molecules with the walls
It’s value is the total rate of the momentum change of
the molecules per unit area of the wall.
• The assumptions of the kinetic theory of an ideal gas
are:
All molecules behave as identical, hard, perfectly elastic spheres
The volume of the molecules is negligible compared with the volume of the containing vessel
There are no forces of attraction or repulsion between molecules
There are many molecules, all moving randomly so that average behaviour can be considered
24. Derivation of pressure of gas
1. Derive the equation:
p V= 1/3Nm<c2>
p = 1/3ρ<c2>
2. Define the following:
a) the mean speed <c>
b) the mean speed squared <c>2
c) the mean square speed <c2>
d) the rms speed.
c
25. 25
Exercise
• The speed of 7 molecules in a gas are numerically
equal to 2,4,6,8,10,12 and 14 units.
Find the numerical values of:
a) the mean speed <c>
b) the mean speed squared <c>2
c) the mean square speed <c2>
d) the rms speed.
a) <c> = 8 units
b) <c>2 = 64 units2
c) <c2> = 80 units2
d) rms speed = 8.94 units
26. 26
Exercise
• Calculate the rms speed of hydrogen gas at
s.t.p. where its density is 0.09 kg m-3.
• Solution
At s.t.p., p = 1.013 x 105 Pa
27. 27
Mean Energy of Molecules
The kinetic energy of a molecule moving at an
instant with a speed v is ½ mv2
The average kinetic energy <Ek> of translation
of the random motion of the molecule of a
gas is therefore ½ m<c2>.
Show that:
i. 3/2kT = 1/2m<c2>
ii. <c> = √(3kT/m)
28. 28
Conclusion
• The r.m.s. speed or velocity of the molecules of a gas
T.
• The r.m.s. speed of the molecules of different gases at
the same temperature 1/M, so gases of higher
molecular mass have smaller r.m.s. speeds
• Hence the rms speed is proportional to the square root
of the thermodynamic temperature of the gas, and
inversely proportional to the square root of the mass of
the molecule.
• Thus at a given temperature, less massive molecules
move faster, on average, than more massive molecules
i.e the higher the temperature, the faster the
molecules move.
• The molecules of air at normal temperatures and
pressures have an average velocity of the order of 480
m/s
29. 29
Exercise
• The r.m.s. speed of hydrogen gas at s.t.p. is 1840 m/s.
Calculate its new r.m.s. speed at 1000 C and the same
pressure. What is the r.m.s. speed of oxygen at s.t.p.?
• Let cr = r.m.s. speed of hydrogen at 1000 C
cs = r.m.s. speed of oxygen at s.t.p.
H2 is 2, O2 is 32
cr = 2150 m/s
cs = 460 m/s
30. 30
Exercise
• Find the total kinetic energy of the molecules
in one mole of an ideal gas at standard
temperature.
Solution:
Ek = 3400 J mol-1
31. 31
Exercise
• Find the rms speed of the molecules in
nitrogen gas at 27° C. The mass of a nitrogen
molecule is 4.6 x 10-26 kg.
• Solution:
crms
= 520 ms-1