Thermal Physics
IBDP - Lecture Notes
Dr. Belal Al Qassem
Ibn Rushd National Academy
Topic 3: Thermal Physics
Syllabus Outline
3.1 – Thermal concepts
•
Molecular theory of solids, liquids and gases
•
Temperature and absolute temperature
•
Internal energy
•
Specific heat capacity
•
Phase change
•
Specific latent heat
3.2 – Modelling a gas
•
Pressure
•
Equation of state for an ideal gas
•
Kinetic model of an ideal gas
•
Mole, molar mass and the Avogadro constant
•
Differences between real and ideal gases
Recommended Teaching Hours: 11 Hours
References:
1. Kirk, Tim. Physics: for the IB Diploma. Oxford University Press, 2014.
2. Homer, David and Bowen-Jones, Michael. Physics: course companion. Oxford
University Press, 2014.
3. Tsokos, K. A. Physics: for the IB Diploma. Cambridge University Press, 2014.
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Dr. Belal Al Qassem
Thermal Physics
Thermal Concepts
Solid
•
•
•
•
Solids have a fixed volume and fixed shape.
The molecules in a solid are held close together by
forces (the intermolecular bonds).
The bonds are not absolutely rigid, so the molecules
vibrate/oscillate around a mean (average) position.
The higher the temperature, the greater the
vibrations.
The molecules in a
solid are held close
together by the
intermolecular
bonds.
Model for Solid Material
The springs represent the
bonds.
Liquid
•
•
•
Liquids have a fixed volume but its shape can
change (it takes the shape of the container).
In liquids the forces between the particles are
weaker.
The particles are able to move around the
volume of the liquid and the liquid will take the
shape of the container in which it is placed.
However, the inter-particle forces between the
particles in a liquid are sufficiently strong that
the particles cannot move far from each other.
Model for Liquid Material
•
Gas
•
In gases the inter-particle forces are very weak so
as to be almost negligible. The only time
significant forces exist between the particles is
during collisions.
Model for gas
Molecules are free to move
randomly.
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Dr. Belal Al Qassem
Temperature
• Temperature usually described as the “degree of hotness” of an object.
• Temperature is a scalar quantity and is measured in units of degrees
Celsius (°C) or kelvin (K) using a thermometer.
• The absolute Temperature scale (Kelvin) is defined as:
𝑇(𝐾 ) = 𝑇( 𝑜𝐶) + 273
Notes:
1.
∆𝑻(𝑲) = ∆𝑻( 𝒐𝑪)
2. The lowest possible temperature on the absolute scale is zero
kelvin, 0 K, on the Celsius scale this is equivalent to -273 oC.
• Temperature is a measure of the average kinetic energy of the
molecules in a substance.
Heat and Temperature
Thermal energy naturally flows from hot to cold. Eventually, the two objects
would be expected to reach the same temperature. When this happens, they
are said to be in thermal equilibrium.
• Heat is energy that is transferred from one body to another as a result
of a difference in temperature.
• The unit of heat energy is Joules (J).
• It is a scalar quantity.
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Dr. Belal Al Qassem
Internal Energy
• Internal energy is the total random kinetic energy of the particles of a
substance, plus the total inter-particle potential energy of the particles.
Internal Energy = Total random KE + Total inter-particle PE
• The internal energy of a system can changes as a result of heat added or
taken out and as a result of work performed.
Heat capacity (C)
The energy required to increase the temperature of a body by
one kelvin.
𝑪=
𝑸
∆𝑻
or
𝑸 = 𝑪∆𝑻
Where
C is the heat capacity (in J K-1)
Q is the heat energy (in J)
T is the change in temperature (in K or oC)
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Dr. Belal Al Qassem
Specific Heat capacity (c)
The heat energy needed to raise the temperature of a unit
mass (1 kg) of the body by one kelvin (or degree)
𝒄=
𝑸
𝒎∆𝑻
or
𝑸 = 𝒎𝒄∆𝑻
Where
c is the specific heat capacity (in Jkg-1K-1)
Q is the heat energy (in J)
m is the mass of the substance (in kg)
T is the change in temperature (in K or oC)
Relationship between Heat Capacity and Specific Heat Capacity
Simply
𝐶 = 𝑚𝑐
or
𝑐=
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𝐶
𝑚
Dr. Belal Al Qassem
Measuring Specific Heat Capacity
1. Electrical Method
The experiment would be set up as follows (Figure 1 (a) for solid
material and Figure 1 (b) for liquid).
The electric heater will produce heat energy and we will assume that
The heat energy (Q) = The electrical heat energy
Or
𝑄 = 𝑃. 𝑡 = 𝑉𝐼𝑡
So
𝒄=
𝑽𝑰𝒕
𝒎(𝑻𝟐 − 𝑻𝟏 )
Sources of Errors
• Loss of thermal energy from the apparatus.
• It will take some time for the energy to be shared uniformly through the
substance.
• The container for the substance and the heater will also be warmed up.
In this case, we can take into consideration the heat energy absorbed by
the container.
𝑄 = 𝑉𝐼𝑡 = 𝑚𝑠 𝑐𝑠 (𝑇2 − 𝑇1 ) + 𝑚𝑐 𝑐𝑐 (𝑇2 − 𝑇1 )
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Dr. Belal Al Qassem
1. Method of Mixtures
The experiment would be set up as follows:
•
•
•
•
•
measure the masses of the liquids mA and mB
measure the two starting temperatures TA and TB
mix the two liquids together.
record the maximum temperature of the mixture Tmax
If no energy is lost from the system then,
energy lost by hot substance cooling down = energy gained by cold
substance heating up
𝑚𝐴 𝑐𝐴 (𝑇𝐴 − 𝑇𝑚𝑎𝑥 ) = 𝑚𝐵 𝑐𝐵 (𝑇𝑚𝑎𝑥 − 𝑇𝐵 )
Sources of Errors
• The main source of experimental error is the loss of thermal
energy from the apparatus; particularly while the liquids are being
transferred.
• The changes of temperature of the container also need to be taken
into consideration for a more accurate result.
In minimize the error, we can take into consideration the heat
energy absorbed by the container.
𝑚𝐴 𝑐𝐴 (𝑇𝐴 − 𝑇𝑚𝑎𝑥 ) = 𝑚𝐵 𝑐𝐵 (𝑇𝑚𝑎𝑥 − 𝑇𝐵 ) + 𝑚𝐶 𝑐𝐶 (𝑇𝑚𝑎𝑥 − 𝑇𝐶 )
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Dr. Belal Al Qassem
Change of Phase
When heat is provided to a body or removed from it, the body may not
necessarily change its temperature. The body may change phase instead.
Changes of phase happen at constant temperature and include:
• melting – when a solid changes to a liquid (heat must be provided to the
solid)
• freezing – when a liquid changes into a solid (heat must be taken out of
the liquid)
• vaporisation (or boiling) – when a liquid changes into vapour (by giving
heat to the liquid)
• condensation – when a vapour changes into a liquid (by taking heat out
of the vapour).
Specific Latent Heat (L)
The amount of energy required to change the phase of a unit mass at
constant temperature.
or
𝑄
𝐿=
𝑚
𝑄 = 𝐿𝑚
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1kg
water at
100oC
1kg Ice
at
0 oC
Heat
Heat
1kg
steam at
100oC
1kg
water at
0 oC
Dr. Belal Al Qassem
Notes:
1. we use the term fusion if the change is from solid to liquid, and we use
the term vaporization for a change from liquid to gas.
2. During the change of phase, the kinetic energy of the molecules stays the
same (so the temperature is constant).
3. The amount of heat given to the molecules will increase the potential
energy of the molecules by breaking the intermolecular bonds.
4. The specific latent heat of vaporization is greater than specific latent heat
of fusion.
𝑳(𝒗𝒂𝒑𝒐𝒓𝒊𝒔𝒂𝒕𝒊𝒐𝒏) > 𝑳(𝒇𝒖𝒔𝒊𝒐𝒏)
Why?
This is because the increase in separation of the molecules is much larger
when going from the liquid to the vapour phase than when going from the
solid to the liquid phase. More work is needed to achieve the greater
separation, and so more energy is required.
1kg Ice
at -20oC
1kg Ice
at 0oC
Heat
𝑄 = 𝑚𝑐∆𝑇
Heat
𝑄 = 𝐿𝑚
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1kg ice
at 0oC
1kg
water at
0 oC
1kg
water at
0 oC
1kg
water at
100oC
Heat
𝑄 = 𝑚𝑐∆𝑇
Heat
𝑄 = 𝐿𝑚
1kg
water at
100oC
1kg
steam at
100oC
Dr. Belal Al Qassem
Example
Calculate the heat energy needed to convert 5 kg of ice at -20 oC into steam at
100 oC.
•
•
•
•
The specific heat capacity of ice is 2100 J/kg.oC.
The specific heat capacity of water is 4200 J/kg.oC.
The specific latent heat of fusion of ice is 3.3 × 105 J kg–1
The specific latent heat of vaporization of water is 2.3 × 106 Jkg–1
Solution
5kg Ice
at -20oC
5kg Ice
at 0oC
5kg
water at
0 oC
5kg
water at
100oC
Heat
𝑄1 = 𝑚𝑐∆𝑇
Heat
𝑄2 = 𝐿𝑚
Heat
𝑄3 = 𝑚𝑐∆𝑇
Heat
𝑄4 = 𝐿𝑚
5kg ice at
0 oC
5kg water
at 0oC
𝑄1 = 𝑚𝑐∆𝑇 = 5 × 2100 × 20 = 210000 𝐽
𝑄2 = 𝐿𝑚 = 5 × 3.3 × 105 = 16.5 × 105 𝐽
5kg water
at 100oC
𝑄3 = 𝑚𝑐∆𝑇 = 5 × 4200 × 100 = 2100000𝐽
5kg steam
at 100oC
𝑄4 = 𝐿𝑚 = 5 × 2.3 × 106 = 11.5 × 106 𝐽
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝑄1 + 𝑄2 + 𝑄3 + 𝑄4
= 1.55 × 107 𝐽
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Dr. Belal Al Qassem
Measuring the Specific latent heat
1. A method for measuring the specific latent heat of vaporization of water
a.
b.
c.
d.
e.
Measure the mass of water.
Connect the circuit as shown in the diagram.
Once the water starts boiling, turn on the stopwatch to measure the time.
After 5 min, turn off the circuit and measure the final mass of water.
Use the following equation to measure the specific latent heat of fusion:
𝑽. 𝑰. 𝒕 𝒑. 𝒕
𝑳=
=
∆𝒎
∆𝒎
2. A method for measuring the specific latent heat of fusion of water
𝑻𝒉𝒆 𝒉𝒆𝒂𝒕 𝒍𝒐𝒔𝒕 𝒃𝒚 𝒉𝒐𝒕 𝒘𝒂𝒕𝒆𝒓
= 𝑻𝒉𝒆 𝒉𝒆𝒂𝒕 𝒖𝒔𝒆𝒅 𝒕𝒐 𝒎𝒆𝒍𝒕 𝒕𝒉𝒆 𝒊𝒄𝒆 + 𝒕𝒉𝒆 𝒉𝒆𝒂𝒕 𝒖𝒔𝒆𝒅 𝒕𝒐 𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒕𝒉𝒆 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒎𝒊𝒙𝒕𝒖𝒓𝒆.
𝒎𝒘𝒂𝒕𝒆𝒓 𝒄𝒘𝒂𝒕𝒆𝒓 (𝑻 − 𝑻𝒎𝒊𝒙 ) = 𝒎𝒊𝒄𝒆 𝑳𝒇𝒖𝒔𝒊𝒐𝒏 + 𝒎𝒊𝒄𝒆 𝒄𝒘𝒂𝒕𝒆𝒓 (𝑻𝒎𝒊𝒙 − 𝟎)
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Dr. Belal Al Qassem
Practice Problems (P1)
1.
2.
The Kelvin temperature of an ideal gas is a measure of the
A.
average speed of the molecules.
B.
average momentum of the molecules.
C.
average kinetic energy of the molecules.
D.
average potential energy of the molecules.
Two ideal gases X and Y, are contained in a cylinder at constant temperature. The
mass of the atoms of X is m and of Y is 4m.
Which one of the following is the correct value of the ratio
average kinetic energy of the atoms of Y
?
average kinetic energy of the atoms of X
3.
4.
A.
1
B.
2
C.
4
D.
16
A substance changes from solid to liquid at its normal melting temperature. What
change, if any, occurs in the average kinetic energy and the average potential energy of
its molecules?
Average kinetic energy
Average potential energy
A.
constant
constant
B.
increases
constant
C.
increases
decreases
D.
constant
increases
The specific latent heat of vaporization of a substance is greater than its specific latent
heat of fusion because
A.
boiling takes place at a higher temperature than melting.
B.
thermal energy is required to raise the temperature from the melting point to the
boiling point.
C.
the volume of the substance decreases on freezing but increases when boiling.
D.
the increase in potential energy of the molecules is greater on boiling than on
melting.
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Dr. Belal Al Qassem
5.
6.
Which of the following is the internal energy of a system?
A.
The total thermal energy gained by the system during melting and boiling.
B.
The sum of the potential and the kinetic energies of the particles of the system.
C.
The total external work done on the system during melting and boiling.
D.
The change in the potential energy of the system that occurs during melting and
boiling.
A hot liquid X has specific heat capacity SH. It is mixed with an equal mass of a cold
liquid Y of specific heat capacity SC.
The best estimate of the ratio
temperature fall of liquid X
temperature rise of liquid Y
is
7.
A.
SC
SH
B.
SH
SC
C.
2
SC
SH
D.
2
SH
SC
Heat lost by hot = heat gained by cold
𝑚𝑆𝐻 . ∆𝑇𝑋 = 𝑚. 𝑆𝐶 . ∆𝑇𝑌
∆𝑇𝑋 𝑆𝐶
=
∆𝑇𝑌 𝑆𝐻
A copper block and a steel block each have the same mass. The copper block is at a
higher temperature than the steel block.
The blocks are placed in thermal contact and they then reach thermal equlibrium. There
is no energy exchange with the surroundings.
How do the magnitude of the change in temperature ∆T and the magnitude of the change
in internal energy ∆U of the two blocks compare?
∆T
∆U
A.
same
same
B.
same
different
C.
different
same
D.
different
different
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Dr. Belal Al Qassem
8.
A container holds 20 g of neon (mass number 20) and also 8 g of helium (mass number
4).
What is the ratio
number of atoms of neon
?
number of atoms of helium
A.
0.4
1 mole of Ne → the mass of it = 20 g → # of atoms = 6 x 1023
B.
0.5
1 mole of He → the mass of it = 4 g → # of atoms = 6 x 1023
C.
2.0
D.
2.5
6 x 1023
1
=
23
2 × 6 × 10
2
9. A temperature of 23 K is equivalent to a temperature of
A.
°C.
B.
°C.
C.
°C.
D.
°C.
10. The specific latent heat is the energy required to change the phase of
A. one kilogram of a substance.
B. a substance at constant temperature.
C. a liquid at constant temperature.
D. one kilogram of a substance at constant temperature.
11. The internal energy of any substance is made up of the
A. total random kinetic and potential energy of its molecules.
B. total potential energy of its molecules.
C. total random kinetic energy of its molecules.
D. total vibrational energy of its molecules.
12. A sealed container contains water at 5 °C and ice at 0 °C. This system is thermally isolated from
its surroundings. What happens to the total internal energy of the system?
A.
It remains the same.
B.
It decreases.
C.
It increases until the ice melts and then remains the same.
D.
It increases.
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Dr. Belal Al Qassem
13. A 1.0 kW heater supplies energy to a liquid of mass 0.50 kg. The temperature of the liquid
changes by 80 K in a time of 200 s. The specific heat capacity of the liquid is 4.0 kJ kg–1 K–1.
What is the average power lost by the liquid?
A. 0
B. 200 W
C. 800 W
D. 1600 W
𝑝 = 1000𝑊
𝑄 = 𝑚𝑐∆𝑇 = 0.5 × 80 × 4000 = 𝟏𝟔𝟎𝟎𝟎𝟎𝑱
𝑚 = 0.5𝑘𝑔
𝐵𝑢𝑡
∆𝑇 = 80𝐾
𝑡 = 200 𝑠
𝑐 = 4000 𝐽𝑘𝑔−1 𝐾 −1
𝑄ℎ𝑒𝑎𝑡𝑒𝑟 = 𝑝. 𝑡 = 1000 × 200 = 𝟐𝟎𝟎𝟎𝟎𝟎𝑱
𝑄𝑙𝑜𝑠𝑡 = 200000 − 160000 = 𝟒𝟎𝟎𝟎𝟎𝑱
𝑄𝑙𝑜𝑠𝑡 40000
𝑝𝑙𝑜𝑠𝑡 =
=
= 200 𝑊
𝑡
200
14. The graph shows the variation with time t of the temperature T of two samples, X and Y. X and Y
have the same mass and are initially in the solid phase. Thermal energy is being provided to X
and Y at the same constant rate.
What is the correct comparison of the specific latent heats LX and LY and specific heat capacities in
the liquid phase cX and cY of X and Y?

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Dr. Belal Al Qassem
15. A mass m of ice at a temperature of –5 °C is changed into water at a temperature of 50 °C.
Specific heat capacity of ice = ci
Specific heat capacity of water = cw
Specific latent heat of fusion of ice = L
Which expression gives the energy needed for this change to occur?
A. 55 m cw + m L
B. 55 m ci + 5 m L
C. 5 m ci + 50 m cw + m L
D. 5 m ci + 50 m cw + 5 m L
16. A container with 0.60kg of a liquid substance is placed on a heater at time t=0. The specific
latent heat of vaporization of the substance is 200kJkg–1. The graph shows the variation of the
temperature T of the substance with time t.
𝑄 = 𝐿𝑚 = 0.6 × 200 = 120𝑘𝐽
𝑃=
𝑄 120000
=
= 3000 𝑊
𝑡
40
What is the power of the heater?
A. 1200 W
B. 3000 W
C. 4800 W
D. 13 300 W
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Dr. Belal Al Qassem
17. Two objects are in thermal contact and are at different temperatures. What is/are determined
by the temperatures of the two objects?
I.
The direction of thermal energy transfer between the objects
II.
The quantity of internal energy stored by each object
III.
The process by which energy is transferred between the objects
A.
I only
B.
II only
C.
I and II only
D.
I, II and III
18. Molecules leave a boiling liquid to form a vapour. The vapour and the liquid have the same
temperature.
What is the change of the average potential energy and the change of the average random kinetic
energy of these molecules when they move from the liquid to the vapour?

19. Equal masses of water at 80°C and paraffin at 20°C are mixed in a container of negligible
thermal capacity. The specific heat capacity of water is twice that of paraffin. What is the final
temperature of the mixture?
𝑚2𝑐 (80 − 𝑇) = 𝑚𝑐(𝑇 − 20)
A. 30°C
B. 40°C
2(80 − 𝑇) = 𝑇 − 20
160 − 2𝑇 = 𝑇 − 20
C. 50°C
D. 60°C
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Dr. Belal Al Qassem
20. A block of iron of mass 10 kg and temperature 10°C is brought into contact with a block of iron
of mass 20 kg and temperature 70°C. No energy transfer takes place except between the two
blocks. What will be the final temperature of both blocks?
A. 30°C
B. 40°C
C. 50°C
D. 60°C
21. A sample of solid copper is heated beyond its melting point. The graph shows the variation of
temperature with time.
During which stage(s) is/are there an increase in the internal energy of the copper?
A. P, Q and R
B. Q only
C. P and R only
D. Q and R only
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Dr. Belal Al Qassem
Practice Problems (P2)
1.
This question is about an experiment to measure the temperature of a flame.
(a)
Define heat (thermal) capacity.
.....................................................................................................................................
.....................................................................................................................................
(1)
A piece of metal is held in the flame of a Bunsen burner for several minutes. The metal is
then quickly transferred to a known mass of water contained in a calorimeter.
flame
water
calorimeter
container
Bunsen burner
lagging (insulation)
The water into which the metal has been placed is stirred until it reaches a steady
temperature.
(b)
Explain why
(i)
the metal is transferred as quickly as possible from the flame to the water;
..........................................................................................................................
.
(1)
(ii)
the water is stirred.
..........................................................................................................................
(1)
The following data are available:
heat capacity of metal
= 82.7 J K–1
heat capacity of the water in the calorimeter = 5.46 × 102 J K–1
heat capacity of the calorimeter
= 54.6 J K–1
initial temperature of the water
= 288 K
final temperature of the water
= 353 K
(c) Assuming negligible energy losses in the processes involved, use the data to
calculate the temperature T of the Bunsen flame.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(4)
(Total 7 marks)
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Dr. Belal Al Qassem
2.
This question is about thermal physics.
(a)
Explain why, when a liquid evaporates, the liquid cools unless thermal energy is
supplied to it.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(3)
(b)
State two factors that cause an increase in the rate of evaporation of a liquid.
1. .................................................................................................................................
2. .................................................................................................................................
(2)
(c)
Some data for ice and for water are given below.
Specific heat capacity of ice
Specific heat capacity of water
Specific latent heat of fusion of ice
= 2.1 × 103 J kg–1 K–1
= 4.2 × 103 J kg–1 K–1
= 3.3 × 105 J kg–1
A mass of 350 g of water at a temperature of 25°C is placed in a refrigerator that
extracts thermal energy from the water at a rate of 86 W.
Calculate the time taken for the water to become ice at –5.0°C.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(4)
(Total 9 marks)
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Dr. Belal Al Qassem
3.
This question is about specific heat capacity and specific latent heat.
(a)
Define specific heat capacity.
.....................................................................................................................................
.....................................................................................................................................
(1)
(b)
Explain briefly why the specific heat capacity of different substances such as
aluminum and water are not equal in value.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(2)
A quantity of water at temperature θ is placed in a pan and heated at a constant rate until
some of the water has turned into steam. The boiling point of the water is 100°C.
(c)
(i)
Using the axes below, draw a sketch-graph to show the variation with time t
of the temperature θ of the water. (Note: this is a sketch-graph; you do not
need to add any values to the axes.)
(1)
100°C
°C
0
time at which
heating starts
22 | P a g e
t
time at which
water starts to boil
Dr. Belal Al Qassem
(ii)
Describe in terms of energy changes, the molecular behavior of water and
steam during the heating process.
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
(5)
Thermal energy is supplied to the water in the pan for 10 minutes at a constant rate of
400 W. The thermal capacity of the pan is negligible.
(d)
(i)
Deduce that the total energy supplied in 10 minutes is 2.4 × 105 J.
..........................................................................................................................
.
(1)
(ii)
Using the data below, estimate the mass of water turned into steam as a
result of this heating process.
initial mass of water
= 0.30 kg
initial temperature of the water θ
= 20°C
specific heat capacity of water
= 4.2 × 103 J kg–1 K–1
specific latent heat of vaporization of water = 2.3 × 106 Jkg–1
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
..........................................................................................................................
(3)
(iii)
Suggest one reason why this mass is an estimate.
..........................................................................................................................
..........................................................................................................................
(1)
(Total 14 marks)
23 | P a g e
Dr. Belal Al Qassem
Gas Laws
There are 4 factors to describe the gas:
•
•
•
•
Pressure (P)
Temperature (T)
Volume (V)
Amount of the gas
Boyle’s Law:
For a fixed mas of a gas, the volume of the gas is inversely proportional to its
pressure provided that the temperature is constant.
𝟏
𝑷∝( )
𝑽
𝑷 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 ×
𝟏
𝑽
𝑷𝑽 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
Or
𝑷𝟏 𝑽𝟏 = 𝑷𝟐 𝑽𝟐
Example
Pressure
Volume
1
20
1 X 20 = 20
24 | P a g e
2
10
2 X 10 = 20
4
5
4 X 5 = 20
5
4
10
2
20
20
1
20
20
Dr. Belal Al Qassem
25
20
20
Pressure
Pressure
25
15
10
5
15
10
5
0
0
0
5
10
15
20
0.00
25
0.50
1.00
1.50
1/Volume
Volume
25
20
PV
15
10
5
0
0
5
10
15
20
25
Pressure
PV = Constant
25 | P a g e
Dr. Belal Al Qassem
Pressure
Pressure is defined as Force per unit Area
𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 =
𝑷=
𝑭𝒐𝒓𝒄𝒆
𝑨𝒓𝒆𝒂
𝑭
𝑨
The unit of Pressure is N/m2 = Pascal (Pa).
1 N/m2 = Pascal (Pa).
There are other units for Pressure such as:
1 atm = 1.013  105 Pa  1  105 Pa  101  kPa
1 atm = 76 cm.Hg = 760 mm.Hg
𝑷=
But
𝑭
𝑨
A
𝒎𝒈
𝑷=
𝑨
𝜌=
𝑚
𝑉
h
𝑚 = 𝜌𝑉 = 𝜌ℎ𝐴
𝑷=
𝜌ℎ𝐴𝒈
𝑨
𝑷 = 𝝆𝒈𝒉
Pressure depends on density and Depth.
26 | P a g e
Liquid of
density 
Dr. Belal Al Qassem
Atmospheric Pressure
𝑷 = 𝝆𝒈𝒉
𝑷 = 𝟏𝟑. 𝟔 × 𝟏𝟎𝟑 × 𝟗. 𝟖 × 𝟎. 𝟕𝟔 = 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑷𝒂
If we used water instead of Mercury
𝑷 = 𝟏 × 𝟏𝟎𝟑 × 𝟗. 𝟖 × 𝒉 = 𝟏. 𝟎𝟏𝟑 × 𝟏𝟎𝟓 𝑷𝒂
𝒉 ≈ 𝟏𝟎 𝒎
A
h
h
A
The weight of the water in the Cylindar is Greater.
F = mg
So if we used the equation P=F/A
It seems that the pressure at the bottom of the cylindar is greater !!!!!!!!!
BUT
If we used the equation
𝑃 = 𝜌𝑔ℎ
Then both will have the SAME Pressure at the bottom.
WHO CAN SOLVE THIS PARADOX.
27 | P a g e
Dr. Belal Al Qassem
Why Pressure inversely proportional to Volume?
𝑷=
𝑭
𝑨
𝑭𝒐𝒓𝒄𝒆 = 𝑹𝒂𝒕𝒆 𝒐𝒇 𝑪𝒉𝒂𝒏𝒈𝒆 𝒐𝒇 𝑴𝒐𝒎𝒆𝒏𝒕𝒖𝒎
• Pressure depends on the number of collisions per second between the
molecules and the wall of the container.
•
•
•
Increasing the Volume will decrease the rate of collisions (pressure).
As a result, will decrease the pressure of the gas.
Since the temperature is constant, then the speed of the molecules stays the same.
28 | P a g e
Dr. Belal Al Qassem
Charles’ Law
The volume of a fixed mass of a gas is directly proportional to its absolute temperature
(Temperature in Kelvin) if the pressure is constant.
𝑽∝𝑻
𝑽 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 × 𝑻
𝑽
= 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑻
𝑽𝟏 𝑽𝟐
=
𝑻𝟏 𝑻𝟐
V
V
0
-273
T/K
T/oC
T (K) = T(oC) + 273
So
0 K = -273 oC
Using Charles’ Law, explain why -273 oC (Zero Kelvin) is the lowest possible temperature?
Since at -273 oC, the volume is zero, and below that temperature the volume becomes
negative which is impossible.
29 | P a g e
Dr. Belal Al Qassem
Why Volume is directly proportional to Temperature at constant pressure? (explanation)
A higher temperature means faster moving molecules.
•
•
•
•
Faster moving molecules hit the walls with a greater microscopic force.
If the volume of the gas increases, then the rate at which these collisions take place
on a unit area of the wall must go down.
The average force on a unit area of the wall can thus be the same.
Thus the pressure remains the same.
30 | P a g e
Dr. Belal Al Qassem
The Pressure Law
The pressure of a fixed mass of a gas is directly proportional to its absolute temperature
(Temperature in Kelvin) if the volume is constant.
𝑷∝𝑻
𝑷 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 × 𝑻
𝑷
= 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑻
𝑷𝟏 𝑷𝟐
=
𝑻𝟏 𝑻𝟐
P
P
0
T/K
-273
T/oC
Using the Pressure Law, explain why -273 oC (Zero Kelvin) is the lowest possible
temperature?
Since at -273 oC, the pressure is zero, and below that temperature the pressure becomes
negative which is impossible.
31 | P a g e
Dr. Belal Al Qassem
Why Pressure is directly proportional to Temperature? (explanation)
If the temperature of a gas goes up, the molecules have more average kinetic energy – they
are moving faster on average.
•
•
•
•
•
Fast moving molecules will have a greater change of momentum when they hit the
walls of the container.
Thus the microscopic force from each molecule will be greater.
The molecules are moving faster so they hit the walls more often.
For both these reasons, the total force on the wall goes up.
Thus the pressure goes up.
32 | P a g e
Dr. Belal Al Qassem
33 | P a g e
Dr. Belal Al Qassem
Ideal Gas Law (The equation of state)
𝑷𝑽 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑽
= 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑻
𝑷𝑽
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑻
𝑷
= 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑻
𝑷𝟏 𝑽𝟏 = 𝑷𝟐 𝑽𝟐
𝑽𝟏 𝑽𝟐
=
𝑻𝟏 𝑻𝟐
𝑷𝟏 𝑽𝟏 𝑷𝟐 𝑽𝟐
=
𝑻𝟏
𝑻𝟐
𝑷𝟏 𝑷𝟐
=
𝑻𝟏 𝑻𝟐
𝑷𝑽
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 = 𝒏𝑹
𝑻
n = The number of moles.
R = The molar gas constant = 8.31 J mol-1 K-1.
𝑷𝑽 = 𝒏𝑹𝑻
34 | P a g e
Dr. Belal Al Qassem
Definitions
Ideal gas:
An ideal gas is one that follows the gas laws (PV = nRT) for all
values of P, V and T.
Mole:
The mole is the basic SI unit for ‘amount of substance’. One mole
of any substance is equal to the amount of that substance that
contains the same number of particles as 0.012 kg of carbon–12
(12C). When writing the unit it is (slightly) shortened to the mol.
• The mass of 1 atom of C-12 = 12 u (atomic mass unit)
• The mass of 1 mole of C-12 = 12 g = 0.012 kg
• There are 6  1023 atoms in 1 mole.
The unified atomic mass unit
One unified mass unit is defined as exactly one twelfth the
mass of a carbon-12 atom.
i.e.
𝟏
𝟏𝒖 =
× 𝒎𝒂𝒔𝒔 𝒐𝒇 𝟏𝟐𝟔𝑪 𝒂𝒕𝒐𝒎
𝟏𝟐
Note:
1 u = 1.661 × 10−27 kg
Avogadro
This is the number of atoms in 0.012 kg, of carbon–12 (12C). It is
constant, NA 6.02 × 1023.
Molar mass
The mass of one mole of a substance is called the molar mass. A
simple rule applies. If an element has a certain mass number, A,
then the molar mass will be A grams.
𝑵
𝒏=
𝑵𝑨
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 =
1 𝑚𝑜𝑙
𝑛 𝑚𝑜𝑙𝑒
35 | P a g e
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒂𝒕𝒐𝒎𝒔
𝑨𝒗𝒐𝒈𝒂𝒅𝒓𝒐 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
→ 6 × 1023 𝑎𝑡𝑜𝑚𝑠
→
𝑁 𝑎𝑡𝑜𝑚𝑠
Dr. Belal Al Qassem
Example
a) What volume will be occupied by 8 g of helium (mass number 4) at room
temperature (20 °C) and atmospheric pressure (1.0 × 105 Pa).
ANS:
𝑇 = 20𝑜 𝐶 = 20 + 273 = 𝟐𝟗𝟑 𝑲,
𝑃 = 1 × 105 𝑃𝑎,
𝑉 =?
1 𝑚𝑜𝑙 𝑜𝑓 𝐻𝑒 → 4𝑔
? 𝑚𝑜𝑙 𝑜𝑓 𝐻𝑒 → 8𝑔
? (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠) =
8
= 2 𝑚𝑜𝑙.
4
𝑷𝑽 = 𝒏𝑹𝑻
1.0 × 105 × V = 2 × 8.3 × 293
V = 0.049 m3
b) How many atoms are there in 8 g of helium (mass number 4)?
ANS:
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 =
2=
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒂𝒕𝒐𝒎𝒔
𝑨𝒗𝒐𝒈𝒂𝒅𝒓𝒐 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑁
6.02 × 1023
N = 12 × 1023 = 1.2 × 1024
36 | P a g e
Dr. Belal Al Qassem
Gas Laws – Practice Questions
1.
A container holds 20 g of neon (mass number 20) and also 8 g of helium (mass number
4).
What is the ratio
2.
A.
0.4
B.
0.5
C.
2.0
D.
2.5
number of atoms of neon
?
number of atoms of helium
The equation of state of an ideal gas is
pV = nRT.
In this equation, the constant n is the number of
3.
A.
atoms in the gas.
B.
molecules in the gas.
C.
particles in the gas.
D.
moles of the gas.
A fixed mass of an ideal gas is heated at constant volume. Which one of the following
graphs best shows the variation with Celsius temperature t with pressure p of the gas?
A.
B.
p

0
p
0
C.
0
t /°C
0
t /°C
D.
p
0
37 | P a g e
0
t /°C
p
0
t /°C
0
Dr. Belal Al Qassem
4.
The graph below shows the variation with absolute temperature T of the pressure p of
one mole of an ideal gas having a volume V. R is the molar gas constant.
p
𝑝𝑉 = 𝑛𝑅𝑇
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
𝑝 𝑛𝑅
=
𝑇
𝑉
T
Which of the following is the best interpretation of the intercept on the temperature axis
and the gradient of the graph?
Intercept on temperature axis / K
Gradient of graph
– 273
R
V
0
R
V
0
V
R
– 273
V
R
A.
B.
C.
D.
5.
When a gas in a cylinder is compressed at constant temperature by a piston, the
pressure of the gas increases. Consider the following three statements.
I.
The rate at which the molecules collide with the piston increases.
II.
The average speed of the molecules increases.
III.
The molecules collide with each other more often.
Which statement(s) correctly explain the increase in pressure?
A.
I only
B.
II only
C.
I and II only
D.
I and III only
38 | P a g e
Dr. Belal Al Qassem
6.
7.
The equation of state for an ideal gas, pV = nRT, describes the behaviour of real gases
A.
only at low pressures and large volumes.
B.
only at high temperatures.
C.
only at large volumes and large pressures.
D.
at all pressures and volumes.
Two identical boxes X and Y each contain an ideal gas.
Box X
Box Y
n moles
2n moles
temperature T
temperature
pressure PX
pressure PY
T
3
In box X there are n moles of the gas at temperature T and pressure PX. In box Y there
T
are 2n moles of the gas at temperature
and pressure PY.
3
The ratio
A.
8.
2
.
3
B.
3
.
2
C.
2.
D.
3.
PX
PY
𝑝𝑉 = 𝑛𝑅𝑇
is
For Box X;
𝑝𝑥 =
𝑛𝑅𝑇
𝑉
𝑝𝑦 =
2𝑛𝑅𝑇
3𝑉
For Box Y;
𝑛𝑅𝑇
𝑝𝑥
1 3
= 𝑉 = =
2𝑛𝑅𝑇
2 2
𝑝𝑦
3𝑉
3
The temperature of an ideal gas is reduced. Which one of the following statements is
true?
A.
The molecules collide with the walls of the container less frequently.
B.
The molecules collide with each other more frequently.
C.
The time of contact between the molecules and the wall is reduced.
D.
The time of contact between molecules is increased.
39 | P a g e
Dr. Belal Al Qassem
9.
10.
The Kelvin temperature of an ideal gas is a measure of the
A.
average speed of the molecules.
B.
average momentum of the molecules.
C.
average kinetic energy of the molecules.
D.
average potential energy of the molecules.
Two ideal gases X and Y, are contained in a cylinder at constant temperature. The mass
of the atoms of X is m and of Y is 4m.
Which one of the following is the correct value of the ratio
average kinetic energy of the atoms of Y
?
average kinetic energy of the atoms of X
11.
A.
1
Since the Temperature is constant
B.
2
Then they have SAME average KE
C.
4
Ratio ….. = 1
D.
16
An ideal gas is kept in a container of fixed volume at a temperature of 30C and a
pressure of 6.0 atm. The gas is heated at constant volume to a temperature of 330C.
pressure 6.0 atm
new pressure
temperature 30 C
temperature 330 C
gas
gas
𝑷 𝟏 𝑷𝟐
=
𝑻𝟏 𝑻𝟐
The new pressure of the gas is about
A.
0.60 atm.
B.
3.0 atm.
C.
12 atm.
D.
40 | P a g e
66 atm.
𝟔 × 𝟏𝟎𝟓
𝑷𝟐 × 𝟏𝟎𝟓
=
𝟑𝟎 + 𝟐𝟕𝟑 𝟑𝟑𝟎 + 𝟐𝟕𝟑
𝟔
𝟑𝟎𝟑
=
𝑷𝟐
𝟔𝟎𝟑
→ 𝒑𝟐 = 𝟏𝟐
Dr. Belal Al Qassem
12.
Container X below has volume V and holds n moles of an ideal gas at kelvin temperature
T. Container Y has volume 2V and holds 3n moles of an ideal gas also at kelvin
temperature T.
The pressure of the gas in X is PX and in Y is PY.
The ratio
A.
13.
PX
is
PY
2
.
3
B.
3
.
2
C.
5.
D.
6.
𝑝𝑉 = 𝑛𝑅𝑇
For Box X;
𝑝𝑥 =
𝑛𝑅𝑇
𝑉
𝑝𝑦 =
3𝑛𝑅𝑇
2𝑉
For Box Y;
𝑛𝑅𝑇
𝑝𝑥
1 2
= 𝑉 = =
3 3
𝑝𝑦 3𝑛𝑅𝑇
2𝑉
2
When the volume of a fixed mass of an ideal gas is reduced at constant temperature, the
pressure of the gas increases.
This pressure increase occurs because the atoms of the gas
14.
A.
collide more frequently with each other.
B.
collide more frequently with the walls of the containing vessel.
C.
are spending more time in contact with the walls of the containing vessel.
D.
are moving with a higher mean speed.
A fixed quantity of an ideal gas is compressed at constant temperature. The best
explanation for the increase in pressure is that the molecules
A.
are moving faster.
B.
are colliding more frequently with the container walls.
C.
exert greater forces on each other.
D.
are colliding more frequently with each other.
41 | P a g e
Dr. Belal Al Qassem
16.
Gas leaks slowly out of a cylinder of constant volume. The temperature of the gas in the
cylinder does not change. Which of the following is constant for the gas molecules in the
cylinder?
A.
The number striking unit area of surface in unit time
B.
The number of the collisions between molecules per unit time
n : decreases (leaks)
C.
The number per unit volume
p: decreases, so less number of
collisions.
D.
The average speed
𝑷𝑽 = 𝒏𝑹𝑻
T: Constant, so Same Speed.
17.
Which of the following is not an assumption on which the kinetic model of an ideal gas
is based?
A.
All molecules behave as if they are perfectly elastic spheres.
B.
The mean-square speed of the molecules is proportional to the kelvin temperature.
C.
Unless in contact, the forces between molecules are negligible.
D.
The molecules are in continuous random motion.
42 | P a g e
Dr. Belal Al Qassem
Kinetic Model of an Ideal Gas
The kinetic theory of gases is a statistical treatment of the movement of gas
molecules in which macroscopic properties such as pressure are interpreted by
considering molecular movement. The key assumptions of the kinetic theory
are:
• A gas consists of a large number of identical tiny particles called
molecules; they are in constant random motion.
• The number of molecules is large enough for statistical averages to be
made.
• Each molecule has negligible volume when compared with the volume of
the gas as a whole.
• At any instant as many molecules are moving in one direction as in any
other direction.
• The molecules undergo perfectly elastic collisions between themselves
and also with the walls of their containing vessel; during collisions each
momentum of each molecule is reversed.
• There are no intermolecular forces between the molecules between
collisions (energy is entirely kinetic).
• The duration of a collision is negligible compared with the time between
collisions.
• Each molecule produces a force on the wall of the container.
• The forces of individual molecules will average out to produce a uniform
pressure throughout the gas (ignoring the effect of gravity).
43 | P a g e
Dr. Belal Al Qassem
y
m
L
cx
-cx
m
x
L
z
L
The initial momentum of the molecule = mcx
The final momentum of the molecule = -mcx
The change in momentum of the molecule = -mcx-mcx = -2mcx
𝑭=
∆𝒑 −𝟐𝒎𝒄𝒙
=
𝒕
𝒕
Newton’s 2nd Law
From Newton’s 3rd Law, the force by the wall of the container on the molecule
equals the force by the molecule on the wall of the container but in the opposite
direction, i.e.,
∆𝒑 𝟐𝒎𝒄𝒙
𝑭=
=
Newton’s 3rd Law
𝒕
𝒕
t=?
𝒕=
𝟐𝑳
𝒄𝒙
→
𝟐𝒎𝒄𝒙 𝒎𝒄𝒙 𝟐
𝑭=
=
𝒕
𝑳
44 | P a g e
The force on the
wall of the
container from
ONE molecule.
Dr. Belal Al Qassem
For N molecules
The total force exerted on the shaded face of the container:
NOTE:
𝒎𝒄𝒙𝟏 𝟐 𝒎𝒄𝒙𝟐 𝟐 𝒎𝒄𝒙𝟑 𝟐
𝒎𝒄𝒙𝑵 𝟐
𝑭=
+
+
+ ⋯+
𝑳
𝑳
𝑳
𝑳
𝒎
𝑭 = (𝒄𝒙𝟏 𝟐 + 𝒄𝒙𝟐 𝟐 + ⋯ + 𝒄𝒙𝑵 𝟐 )
𝑳
All molecules
have the same
mass but not
the same
speed.
Where
cx1 : the x component of the velocity of the 1st molecule.
cx2 : the x component of the velocity of the 2nd molecule.
NOTE:
̅̅̅̅𝟐 ≠ 𝑐̅ 2
𝒄
𝒙
𝑥
The mean value of the square of the velocities
(𝒄𝒙𝟏 𝟐 + 𝒄𝒙𝟐 𝟐 + ⋯ + 𝒄𝒙𝑵 𝟐 )
̅̅̅̅
𝟐
𝒄𝒙 =
𝑵
̅̅̅̅𝟐 = (𝒄 𝟐 + 𝒄 𝟐 + ⋯ + 𝒄 𝟐 )
𝑵. 𝒄
𝒙
𝒙𝟏
𝒙𝟐
𝒙𝑵
So we can write the force F as follows:
𝑭𝒙 =
𝑵𝒎 ̅̅̅̅𝟐
. 𝒄𝒙
𝑳
In 3D, we can use Pythagoras theorem, and write
𝟐
̅̅̅̅
̅̅̅̅𝟐
̅̅̅̅𝟐
𝒄̅𝟐 = ̅̅̅̅
𝒄𝒙 𝟐 + 𝒄
𝒚 + 𝒄𝒛 = 𝟑. 𝒄𝒙
𝟐
𝒄̅̅̅̅
𝒙 =
𝒄̅𝟐
𝑐 2 = 𝑐𝑥 2 + 𝑐𝑦 2 + 𝑐𝑧 2
cy
c
𝟑
𝑵𝒎 𝒄̅𝟐
𝑭𝒙 =
.
𝑳 𝟑
cz
cx
The pressure on the wall of the container:
𝑭𝒙 𝑭𝒙 𝑵𝒎 𝒄̅𝟐
𝒑𝒙 =
= 𝟐= 𝟑 .
𝑨
𝑳
𝑳 𝟑
45 | P a g e
Dr. Belal Al Qassem
𝒑𝒙 =
𝑵𝒎 𝟐
. 𝒄̅
𝟑𝑽
In Fluids (liquids and gases), the pressure at a point acts equally in all directions:
𝒑=
𝑵𝒎 𝟐
. 𝒄̅
𝟑𝑽
But
𝑻𝒉𝒆 𝒕𝒐𝒕𝒂𝒍 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 = 𝑵𝒎
and
𝑫𝒆𝒏𝒔𝒊𝒕𝒚 =
𝝆=
So we can write
𝒑=
𝒎𝒂𝒔𝒔
𝒗𝒐𝒍𝒖𝒎𝒆
𝑵𝒎
𝑽
𝟏
𝝆. 𝒄̅𝟐
𝟑
It links a mAcroscopic quantity (such as pressure) with a mIcroscopic quantity
(such as the speed of the molecule).
Molecular Interpretation of Temperature
𝑷=
𝟏 𝑵𝒎 ̅̅̅𝟐 𝟏 ̅̅̅𝟐
. 𝒄 = 𝝆. 𝒄
𝟑 𝑽
𝟑
𝑷𝑽 =
𝟏
̅̅̅𝟐
𝑵𝒎. 𝒄
𝟑
Multiply each side by 3/2
𝟑
𝟑 𝟏
̅̅̅𝟐
𝑷𝑽 = × 𝑵𝒎. 𝒄
𝟐
𝟐 𝟑
𝟑
𝟏
̅̅̅𝟐
𝑷𝑽 = 𝑵𝒎. 𝒄
𝟐
𝟐
Compare it with pV = nRT
46 | P a g e
Dr. Belal Al Qassem
𝟑
𝟏 ̅̅̅𝟐
𝒏𝑹𝑻 = 𝑵. 𝒎. 𝒄
𝟐
𝟐
𝟑𝒏
𝟏 ̅̅̅𝟐
𝑹𝑻 = 𝒎. 𝒄
𝟐𝑵
𝟐
𝒏=
𝑵
𝑵𝑨
𝒏
𝟏
=
𝑵 𝑵𝑨
𝟑 𝑹
𝟏 ̅̅̅𝟐
𝑻 = 𝒎. 𝒄
𝟐 𝑵𝑨
𝟐
BUT
𝒌𝑩 =
𝑹
= 𝟏. 𝟑𝟖 × 𝟏𝟎−𝟐𝟑 𝑱𝑲−𝟏 ,
𝑵𝑨
𝑩𝒐𝒍𝒕𝒛𝒎𝒂𝒏𝒏 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝟑
𝟏 ̅̅̅𝟐
𝒌𝑩 𝑻 = 𝒎. 𝒄
= ̅̅̅̅
𝑬𝒌
𝟐
𝟐
It links the mIcroscopic quantity (Kinetic energy of the molecules) with the
mAcroscopic quantity (the Temperature).
Ideal Gases and Real Gases
An ideal gas is a one that follows the gas laws for all values of p, V and T and thus ideal gases
cannot be liquefied.
Real gases, however, can approximate to ideal behaviour providing that the intermolecular
forces are small enough to be ignored. For this to apply, the pressure/density of the gas must
be low and the temperature must be moderate.
47 | P a g e
Dr. Belal Al Qassem