gas models

1. Thermal Physics
IBO DP Topic 3.2 Modelling a gas

2. Pressure
What is pressure?
How can a magician lie or stand on a bed of nails?

3. Pressure
Pressure of a gas is due to the particles
bombarding the walls of the container. Each
collision with the wall causes a momentum change
(as there is a change in direction). The force on the
wall by one molecule = average rate of change of
pressure.

4. Pressure
If the temperature increases, the average KE of
the particles increases. The increase in velocity of
the particles leads to a greater rate of collisions
and hence the pressure of the gas increases as
the collisions with the side have increase.
Also the change in momentum is greater,
therefore greater force.

5. When a force is applied to a piston in a cylinder
containing a volume of gas the particles take up
a smaller volume.

6. Smaller area to collide with and hence collisions are
more frequent with the sides leading to an increase
in pressure.
Also, as the piston is being moved in. It gives the
particles colliding with it more velocity.Therefore they
have more KE. Therefore the temperature of the gas
rises because the collisions are perfectly elastic.
There is no loss of KE as a result of the collisions.

7. An Ideal Gas
An ideal gas is a theoretical gas that obeys the
gas laws and thus fit the ideal gas equation
exactly.
How does this fit with your shared knowledge?
TOK

8. Real Gases
Real gases conform to the gas laws under
certain limited conditions.
But they condense to liquids and then solidify if
the temperature is lowered.
Furthermore, there are relatively small forces of
attraction between particles of a real gas. This is
not the case for an ideal gas

9. The Kinetic Theory of Gases
When the moving particle theory is applied to
gases it is generally called the kinetic theory.
The kinetic theory relates the macroscopic
behaviour of an ideal gas to the microscopic
behaviour of its molecules or atoms.

10. The Postulates
1. A gas consists of a large number of identical particles
in continual random motion.
Evidence
1g of hydrogen atoms contain 6.023 x 1023 particles
Smoke particles move in a haphazard path which are
bombarded by gas molecules which in turn have velocities
that are random in magnitude and direction.
Gas fills a volume
Only gravity holds the atmosphere in.

11. The Postulates continued
2. Collisions between particles and the wall of the
container are elastic.
Evidence
A gas left at constant p, V & T do not change
properties
3. The particles in a gas are considered to be point
masses with no rotational Ek. This simplification
allows us to make straight forward calculations.

12. The Postulates continued
4. Volume occupied by the gas particles
themselves is assumed to be negligible
compared with the volume of the container as
you can compress a gas.
5. There are no forces acting between the
particles themselves or between the particles
and the wall of the container except during
collisions. There are large distances between
particles  Fg is negligible.

13. Macroscopic Behaviour
6. Time between collisions is large compared
with the time for collisions. There are no forces
between particles  particles travel in straight
lines and constant v and then collide elastically
and then changes v. Forces act on particles only
during collision  collision time is small.

14. Thermal Properties of Gases
A large volume of gas can be described by bulk
characteristics - pressure (p), volume (V) & temperature
(T). All three terms are inter-related.
Investigations involved the measurement of
• Pressure
• Volume
• Temperature
These experiments used these macroscopic properties
of a gas to formulate a number of gas laws.

15. Units
Temperature is always measured in K
Volume is usually in m3
Pressure can be different units as long as you
are consistent
But 1 atm = 1.01 x 105 Nm-2
= 101.3 kPa
= 760 mmHg

16. Gas Laws
1. Boyle’s Law:
Variation of volume with pressure (const T).
eg Bike pump.

17. Gas Laws
Boyle’s experiment can be
repeated by using a flexible J-tube
containing mercury, and a
barometer to measure atmospheric
pressure.
Air was trapped in one end and
varied the pressure on it, by raising
the other end of the tube as shown
on right:

18. Gas Laws
The gas pressure of the
enclosed air is equal to the
atmospheric pressure plus,the
pressure resulting from, the
height difference of Hg in the
two arms.
Taking a number of pressure
and volume readings the graph
looks like:

19. A more revealing graph is obtained by plotting p vs 1/V
As a straight line is obtained that
passes through the origin, we can say
mathematically:
p  1/V

20. Gas Laws
If the same experiment is
performed at different
temperatures, a curve is
still obtained but at
different positions.

21. The curves are called isotherms as the temp at each
point on the curve is the same.

22. Gas Laws
We have assumed that there is a fixed amount
of gas, maintained at a constant temperature.
We can also write, p = k/V or pV = k
k is a constant of proportionality and is equal to
the slope.

23. Gas Laws
For different points on the graph;
p1V1 = k and p2V2 = k
 p1V1 = p2V2 (for const. temp)

24. Gas Laws
Boyle’s law states at constant temperature, the
volume of a fixed mass of gas, varies inversely with
its pressure.
P  1/V or PV = constant
When the conditions are changed
P1V1 = P2V2

25. The Experiment
Air from
foot pump
Bourdon Pressure
gauge
Volume of
dry air
oil

26. What to do
A column of trapped dry air in a sealed tube by
the oil.
The pressure on this volume of air can be varied
by pumping air in or out of the oil reservoir to
obtain different pressures.
Wait to allow the temperature to return to room
temperature.

27. The Results
P
V
P
1/ V
PV
P

28. Gas Laws
2. Charles’ Law:
Variation of volume with temperature (const p).
To verify this a capillary tube is sealed at one
end, and a small mercury piston is included,
trapping a sample of air.
A thermometer is strapped to the tube and both
are placed in glycerol. The pressure is fixed and
is equal to atmospheric pressure.

29. Gas Laws
The glycerol is heated and the
length of the air column is noted,
at various temperatures.
The length of the column can be
read as the volume in arbitrary
units, as the cross section is
uniform.

30. The results obtained can be graphed as below:

31. Gas Laws
This graph does not go
through the origin.
When extrapolated it
cuts the temp axis at -
273oC

32. Gas Laws
This creates a new scale defined as The
absolute temperature scale, where zero degrees
kelvin (0K) = -273oC.
A 1K rise = 1oC rise.
ToC =T K -273.
A real gas would liquefy before it reaches 0K and
so no gas is ideal.

33. Gas Laws
Any extrapolation should not
continue past the liquefaction
point.
No gas can reach a
temperature below 0K as no
gas could have a negative
volume.
Graphing for two different
pressures:

34. Gas Laws
Using the Kelvin scale we can say mathematically:
V  absolute T
V = kT or k = V/T
where k is the gradient of the graph. For two points,
(V1,T1) and (V2,T2)
V1/T1 = V2/T2

35. Gas Laws
Charles’ law states at constant pressure, the
volume of a fixed mass of gas, varies directly
with its absolute temperature.
V  T or V/T = constant
When the conditions are changed
V1/T1 = V2/T2

36. The Experiment
Tap 1
Tap 2 Tap 3
Water reservoir
Fixed mass
of gas
Mercury in U tube

37. What to do
Fill the mercury column with mercury using the
right hand tube (tap 1 open, tap 2 closed).
With tap 1 open drain some mercury using tap 2,
then close tap 1 and 2. To trap a fixed mass of
gas.
Fill the jacket with water (make sure tap 3 is
closed).

38. Change the temperature of the water by draining
some water from tap 3 and adding hot water.
Equalise the pressure by leveling the columns
using tap 2.
Read the volume from the scale.

39. The Results
V
T K
V
T oCA value for
absolute zero

40. Gas Laws
3. Gay-Lussac’s Law: (Law of pressures)
Variation of pressure with temperature (const V).
A bulb enclosing a fixed volume of gas is placed
in a beaker of water, and attached to a mercury
manometer, as shown below:

41. Gas Laws

42. Gas Laws
The water is warmed the air expands in the bulb
and tube, and causes the Hg, to be forced up the
manometer.
One arm is then raised to increase pressure, and
compress the air, back to the original volume.
The height difference is recorded.

43. Gas Laws

44. As the graph does not pass through the origin it must be
extrapolated.

45. Gas Laws
We can now mathematically say:
p  absolute T
p = kT or k = p/T;
where k is the gradient of the graph.
Using two points on the graph;
p1/T1 = p2/T2

47. Gas Laws
The law of pressures state at a constant volume,
the pressure of a given mass of gas, varies
directly with the Kelvin temperature.
P  T or P/T = constant
When the conditions are changed
P1/T1 = P2/T2

48. The Experiment
Bourdon gauge
Ice
Water
Fixed
Mass of
gas
Heat

49. What to do
Change the temperature of the water by heating
it.
Record the pressure of the gas.

50. The Results
P
T K
P
T oCA value for
absolute zero

51. Absolute Zero and the Kelvin
Scale
Charles’ Law and the Pressure Law suggest that
there is a lowest possible temperature that
substances can go
This is called Absolute Zero
The Kelvin scale starts at this point and
increases at the same scale as the Celsius
Scale

52. Therefore -273oC is equivalent to 0 K
∆1oC is the same as ∆1 K
To change oC to K, add 273
To change K to oC, subtract 273

53. GENERAL GAS EQUATION
All three laws studied above can be combined
into one.
From Boyle’s law:

54. The curves plotted from the above situations are shown:

55. Gas Laws
Going from (1) to (2)
V by raising T at constant pressure.
V1/T1 = V*/T2 (Charles’ Law) 
Going from (2) to (3) temp constant, the piston is
withdrawn causing V & P.
p1V* = p2V2 (Boyle’s Law) 
Multiplying u and v

56. Gas Laws
(V1 x p1V*)/T1 = (V* x p2V2)/T2
(p1V1)/T1 = (p2V2)/T2
pV/T = constant
This is called the ideal gas law and is accurate
except, at low temperatures, and high pressures.

57. Gas Laws
The only gas that approximates the ideal gas
law, over a wide range of conditions, is Helium,
to 4.2K.

58. Gas Laws
p1V1/T1 = C
C depends on the number of moles, n, and;
Universal gas constant, R
(8.315 J mol-1 K-1 if Pressure is in kilopascals(kPa) but
R = 0.0821 L atm K-1 mol-1 if Pressure is in atmospheres(atm)
For one mole of gas, pV = RT
For n moles of gas, pV = nRT

59. Gas Laws
One mole of a substance contains;
6.023 x 1023 particles.
This value is called Avogadro’s number and is
given the symbol NA.
To find the moles present (n) divide the total
number of molecules N, by Avogadro’s number.

60. Gas Laws
n = N/NA
m grams of a substance contains m/mA moles,
where the molar mass is expressed in grams.
n = m/mA

61. The Mole
The mole is the amount of substance
which contains the same number of
elementary particles as there are in 12
grams of carbon-12
Experiments show that this is
6.02 x 1023 particles
A value denoted by NA and called the
Avogadro Constant (units mol-1)

62. Molar Mass
Molar mass is the mass of one mole of the
substance
SI units are kg mol-1

63. Example
Molar mass of Oxygen gas is
32 x10-3 kg mol-1
If I have 20g of Oxygen, how many moles do I
have and how many molecules?
20 x 10-3 kg / 32 x10-3 kg mol-1
 0.625 mol
 0.625 mol x 6.02 x 1023 molecules
 3.7625 x 1023 molecules

64. Use your periodic table to…
Find out how many atoms are in
1. 46g of sodium
2. 64mg of copper
3. 180g of water
Find out how many moles is
1. 10g of Argon
2. 1kg of Iron

65. What is the molar mass of…
Nitrogen
Chlorine
Neon
Water
Sulfuric acid (H2SO4)
Calcium carbonate (CaCO3)

66. Proof of ideal gas law
Consider, a particle of mass m, in a box of side
L, moving with velocity v
As particle collides with wall there is a change
in momentum of
-mv – mv = -2mv
This is only in 1 of the 3 dimensions so we
need to denote this by saying x
Fx = change in p/t = 2mvx/t
(force of molecule on wall)

67. t is the time between collisions so we know that
Distance = 2L = vx * t
therefore t = 2L/ vx
So, Fx = 2mvx/t now becomes
Fx = 2mvx/ (2L/ vx )
Fx = mvx
2/L

68. Fx = mvx
2/L
Now, this is only 1 particle of N so there are a range of N speeds.
In addition the mean speed in 3D can be shown as
vx
2 + vy
2 + vz
2 = v 2 as vx
= vy
= vz so vx
2 = v2 /3
Giving,
F = Nmv2/3L

69. F = Nmv2/3L
Now, pressure, p = F/A
where A = L2
So, p = Nmv2 /3L * L2 = Nmv2/3L3
and L3 = V
Giving, p = Nmv2/3V
Nm/V is density so this equations becomes p = 1/3ᵨ v2

70. Show pV =nRT and p =1/3ᵨ v2
leads to U = 3/2 NkT

71. Combining the Laws
The gas laws can be combined to give a single
equation
For a fixed mass of gas its pressure times its
volume divided by its absolute temperature is a
constant
PV/T = k
So that P1V1/T1 = P2V2/T2

72. The Ideal Gas Equation
PV = nRT
Where n is the number of moles
R is the universal gas constant 8.31 J mol-
1 K-1

73. Real vs Ideal
A real gas can be considered an ideal gas when
at low pressure, moderate temperature and low
density.
When is a model good enough?
TOK