Cambridge A2 Physics

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