1. CHAPTER 5.0
STATES OF MATTER
Gas
Liquid
Solid

2. OBJECTIVE... particle, density and compressibility.
At the end of the lesson, student should be able to :
(a) Explain the general properties of gas in terms of arrangement of
particle, density and compressibility.
(b) Explain qualitatively the basic assumptions of the kinetic
molecular theory of gases for an ideal gas.
(c) Define gas laws
(i) Boyle’s Law
(ii) Charles’s Law
(iii) Avogadro’s Law
(d) Sketch and interpret the graphs of Boyle’s and Charles’s laws
(e) Perform calculations involving gas laws.
(f) derive ideal gas equation based on the gas laws

3. (g) perform calculations using the ideal gas equation
(h) determine the molar mass and density of a gas using ideal gas
equation
(i) Define:
Partial pressure
Dalton's law
(j) perform calculation using Dalton's law
(k) compare the ideal and non-ideal behaviours of gases in terms of intermolecular forces and molecular volume
(l) explain the conditions at which real gases approach
the ideal behaviour.
(m) explain qualitatively van der Waals equation and relate the
values of a and b to intermolecular forces and molecular volume of a gas

4. General Properties of Gas
Particles of gas are far apart and fill the available space.
Gases assume the volume and shape of their containers.
can be compressed – due to the particles being
so small and are relatively far apart from one another.
Gases have relatively low densities.

5. Kinetic – Molecular Theory of Gases
Describes the behavior of an ideal gas
Ideal gas : gases which obey the ideal gas equation (PV = nRT)

6. The theory is based on the following assumptions:
1) Gas molecules are very tiny that their size are negligible
compared to the volume of the container.
(having mass but no volume)
2) Gas molecules move in straight lines and are at a constant motion unless they collide.
3) Molecular collisions are elastic – no energy is lost during
collisions.
4) Attractive and repulsive forces between gas particles are negligible.
5) The average kinetic energy of the particles is proportional to the absolute temperature.

7. The Gas Laws Boyle’s Law :
The volume of fixed amount of gas at constant temperature is inversely proportional to the gas pressure
V  (no of mole and temperature are constant)
PV = k
Where:
k = constant
V = volume
P = pressure
T = temperature
n = number of moles

9.  at different pressure and volume :
Where
P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume
P1 V1 = P2 V2

10. pressure is inversely
proportional to volume
pressure is directly
proportional to

11. Graph of PV versus P PV P
PV = constant

12. Example 1
A sample of chlorine gas occupies a volume of 2 L at a pressure
of 1 atm. Calculate the pressure of the gas if the volume is
increased to 5 L at constant temperature.
0.4 atm

13. Example 2
The pressure of a sample of hydrogen gas in a 50.0 mL container is 765 mmHg. The sample is then transferred into another container and the measured pressure is 825 mmHg. What is the volume of the second container?
46.36 mL

14. Charles’s Law :
The volume of a fixed amount of gas at constant pressure is directly proportional to the absolute temperature of the gas (in Kelvin).
V  T (no of mole and pressure constant)
Where :
k = constant
T = absolute temperature (K)
V = volume

16. At different volume and temperature:
Where V1 = initial volume
T1 = initial temperature
V2 = final volume
T2 = final temperature
T = absolute temperature
in Kelvin (K)
T(K) = T°C

17. Graph of volume versus temperature :
T(K)
T(0C)

18. Example 1 A sample of carbon monoxide gas occupies 3.2 L
At 125 °C. The sample is then cooled at constant
pressure until it contracts to 1.54 L. Calculate the
final temperature in degree Celsius. °C

19. Example 2 A sample of gas trapped in a capillary tube by a
plug of mercury at 22 oC has a volume of 4.5 mL.
Calculate the volume of the gas when the capillary
tube is heated to 60 oC.
5.08 mL

20. The Combination of Boyle’s and Charles’s Law
Boyle’s law :
Charles’s law : V  T =

21. Example 1 A sample of methane gas occupies 25.5 L at
K and kPa. Find its volume at STP.
Ans : L

23. Example 2
2 moles of chlorine gas kept in a cylinder with piston occupies a volume of 49 L. When another 3 moles of chlorine gas is pumped into the cylinder at constant temperature and pressure the piston moves upwards to accommodate the gas. Calculate the final volume of the gas.
Ans : 73.5 L

24. C) Avogadro’s Law
At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of the gas present
V  n (P and T are constant)
V = k n
where :
n = number of moles
k= constant

26. Combination of Boyle's law, Charles's law and Avogadro's law :
Where :
R = gas constant
T = Temperature(K)
n = number of moles
V = volume
P = Pressure
PV = nRT
Ideal gas equation

27. or Value of R depend on the unit of pressure and
volume used in the equation.
unit of pressure
unit of volume
value of R
unit of R
atm
L or dm3
L atm mol1 K1
Nm2 or Pa m3
8.314
Nm mol1 K1
J mol1 K1

28. Example 1
A steel gas tank has a volume of 275 L and is filled with kg of O2. Calculate the pressure of O2 if the temperature is 29 oC.
Ans : 1.36 atm

29. Example 2
A sample of chlorine gas is kept in a 5.0 L container at 228 torr and 27 °C. How many moles of gas are present in the sample?
Ans : 0.06 mol

30. MOLAR MASS and DENSITY CALCULATION
Molar mass and density of a gas can be calculated
by rearranging the Ideal Gas Equation:
Mr = molar mass of a gas

32. Example 1
Calculate the density of ammonia (NH3) in grams per litre (g/L) at 752 mmHg and 55 C.
Ans : gL-1

33. Example 2
A chemist has synthesized a greenish-yellow compound of chlorine and oxygen and finds that its density is 7.71 g L-1 at 36 °C and 2.88 atm. Calculate the molar mass of the compound.
Ans : 67.9 gmol-1

34. Example 3
Ans : 8.0 g

35. Dalton’s Law of Partial Pressure
The total pressure of mixture of non reacting gases is the sum of the partial pressures exerted by each of the gas in the mixture
(Partial pressure is the pressure of individual gas component in a mixture).
- For a mixture of 3 gases, A,B and C :
PT = PA + PB + PC

38. Example 1
A gaseous mixture of 7.00 g N2 and 3.21 g CH4 is placed in a 12.0 L cylinder at 25 oC.
a) What is the partial pressure of each gas?
b) What is the total pressure in the cylinder?
Ans :a) atm , 0.41 atm
b) 0.92atm

39. Example 2
A mixture of gases contains 4.53 moles of neon, 0.82 moles of argon and 2.25 moles of xenon. Calculate the partial pressure of the gases if the total pressure is 2.15 atm at a certain temperature.
Ans : P Ne = 1.28 atm,, P Ar = atm, P Xe = 0.63 atm,

40. Example 3 A sample of gas at 5.88 atm contains 1.2 g CH4, 0.4 g H2 and
0.1 g He. Calculate :
The partial pressure of CH4, H2 and He in the mixture.
What is the partial pressure of CH4 and H2 if He is removed?
Ans : a) P CH4= 1.74 atm,P H2 = 3.92 atm, P He= 0.49 atm
b) P CH4= 1.6 atm,P H2 = 4.28 atm

41. One of the applications of Dalton’s Law is to calculate the pressure of a gas collected over water ( for gases that not soluble in water).
The gas collected is actually a mixture of the gas and water vapour.
gas + water vapour
gas
Vapour pressure of water, Pwater = 23.8 torr

42. Example 1 Consider the reaction below : 2KClO3 2KCl + 3O2
A sample of 5.45 L of oxygen is collected over water at a total pressure of torr at 25 °C. How many grams of oxygen have been collected?
(at 25°C, Pwater = 23.8 torr)
Ans : g

43. Example 2
Excess amount of hydrochloric acid is added to 2.5 g of pure zinc. The gas produced is collected over water in a gas cylinder at 28 oC and kNm-2. Calculate :
a) the number of mole of gas produced in the reaction.
b) the volume of gas collected in the cylinder.
Ans : a) mol
b) 0.95 L

44. Ideal gas
any gas that obeys the ideal gas equation and has the properties as outlined by the Kinetic Molecular Theory

45. Deviation from Ideal Behavior
real gas (non-ideal gas) : gases which do not obey ideal gas properties
Real gases do not behave ideally because:
i) gas molecules do have its own volume and they occupy
some space.
ii) gas molecules do have intermolecular forces acting
between them

48. Reason :
At low pressure
At low pressure, the volume of a container is very large
Thus the molecules will be more far apart from one another. Hence the intermolecular forces can be neglected.
At low pressure, the volume of the container is extremely large compared to the size of the molecules, thus the volume can be neglected

51. Van der Waal’s Equation
Since real gas does not exhibit ideal gas behavior
at high pressure and low temperature :
the ideal gas equation (PV=nRT) needs to be
adjusted
adjusting the equation, two parameters need to
be reconsidered :
attractive forces between the gas molecules
volume of the gas molecules

52. a) Attractive Forces Between Molecules
Attractive forces which act between the gas molecules will :
make the molecules move slower
give less impact to the wall
pressure exerted by the real gas is less compared to the ideal gas
since Preal < Pideal
the term pressure need to be corrected by adding coefficient

56. 5.2 Liquids

57. At the end of this topic, students should be able to :
a) Explain the properties of liquids : shape, volume, surface tension, viscosity, compressibility and diffusion.
b)Explain vaporization process and condensation process based on kinetic molecular theory and intermolecular forces.
c) Define vapour pressure and boiling point
d) Relate :
- intermolecular forces to vapour pressure
- vapour pressure to boiling point

58. The properties of Liquids
Volume and Shape
has a definite volume but not a definite shape
the particles are arranged closely but not rigidly
held together by a strong intermolecular forces but not strong enough to hold the particles firmly in place
particles are able to move freely
thus, a liquid flows to fit the shape of its container and is confined to a certain volume

59. 2. Surface tension
Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area
Surface tension is caused by the attractive forces and the direction of the force acting between
molecules
Molecules within a liquid experienced intermolecular attractive forces in all directions by their neighbouring molecules
however, molecules at the surface are pulled inwards and sideways from the neighbouring molecules

61. these intermolecular attractive forces will pull the molecules into the liquid
Thus, cause the surface to stretch and tighten
the stronger the intermolecular attractive forces, the higher the surface tension

62. 3. Viscosity
is a measure of the resistance of a liquid to flow
the greater the viscosity, the more slowly
the liquid flows.
Viscosity may be explained in terms of the intermolecular forces present in a liquid

63. The factors affecting the viscosity :
i) The strength of intermolecular forces
Liquids that have strong intermolecular forces have higher viscosity
ii) The size of the molecules
Liquids with larger size (higher molar mass) is more viscous because it has stronger
intermolecular forces

64. iii) The temperature of the liquid
At higher temperature, molecules have higher
kinetic energy,
the molecules move faster
molecules can overcome the intermolecular
attractive forces more easily
resistance to flow decrease
Viscosity decrease

65. 4. Compressibility
in liquid, the particles are packed closely together
thus, there is very little empty space between the molecules
 liquids are much more difficult to
compress than gas

66. 5. Diffusion
Diffusion is the movement of liquid from one fluid through another.
Diffusion can occur in a liquid because molecules are not tightly packed and can move randomly around one another.

67. Vaporisation Process
a process in which molecules escape from liquid into gaseous state through a surface
molecules in a liquid moves freely, collide with each other and posses different magnitudes of kinetic energy
when the kinetic energy is sufficient enough to overcome the intermolecular forces acting on them, the molecules will escape as vapour.

68. Factors affecting the rate of vaporization :
i) Surface area
~ the surface area , the chances for the molecules to escape from the surface 
 surface area , evaporation rate 

69. ii) Temperature
~ temperature , the kinetic energy 
 more molecules have enough energy to overcome the attractive intermolecular forces and escape from the surface of the liquid
~ thus evaporation rate 

70. Condensation Process
a process whereby gaseous molecules turn to liquid
some of the vapour molecules may lose their kinetic energy during the collision
they do not have enough energy to remain as vapour molecules
they reached the surface of the liquid and trapped by the attractive forces
if they cannot overcome the attractive forces, these vapour molecules return as liquid molecules
the process is known as condensation

79. 5.3 Solid

81. Properties of solid
Particles are closely arranged and regularly in order
Rigid arrangement- particles can vibrate, rotate about fixed position and cannot move freely without disrupting the whole structure.
Has definite shape and volume.
Has strong forces between the particles.
Has high densities.
Incompressible.
Diffusion within the solid is extremely slow.

91. 5.4 Phase Diagram

92. Learning Outcomes :
At the end of this lesson students should be able to:
(a) Define phase .
(b) Sketch the phase diagram of H2O and CO2.
(c) Define triple point and critical point.
(d) Explain the anomalous behaviour of H2O.
(e) Describe the changes in phase with respect to
i. temperature (at constant pressure)
ii. pressure (at constant temperature).

93. Phase is a homogeneous part of a system in contact with other parts of the system but separated from them by a well-defined boundary
Phase Diagram is a diagram that describes the stable phases and phase changes of a substance as a function of temperature and pressure
used to predict the phase that exist under a certain temperature and pressure

94. (i) Phase Diagram of H2O B C T A A
8/16/11

95. - The diagram has three main regions : Solid, liquid and gas
Regions of the diagram
- The diagram has three main regions :
Solid, liquid and gas
- Each region corresponds to one phase of the substance.
- A particular phase is stable for any combination of pressure and temperature.
- Any point along a line shows the pressure and temperature at which the two phases exist in equilibrium.
8/16/11

96. - Known as the triple point
Important Points
1. Point T :
- Known as the triple point
- Triple point is the point at which the vapour, liquid and solid states of a substance are in equilibrium.
- Triple point for water is 0.01°C and 0.06 atm.
2. Point C :
- Known as the critical point
- Critical point is the point on a phase diagram at which
the vapour cannot be condensed to a liquid.
- The liquid – gas line ends at the critical point.
- Above the critical point, the liquid cannot be
distinguished from its vapour form.
8/16/11

97. (ii) Phase Diagram of CO2B C T A
8/16/11

98. Triple point for CO2 is at 5.2 atm pressure and
At point T, the triple point; solid, liquid and vapour are in equilibrium
CO2(s) ⇌ CO2(l) ⇌ CO2(g)
Triple point for CO2 is at 5.2 atm pressure and
temperature 57 OC.
for most compounds, the TB curve slant to the right because solid is denser than liquid
8/16/11

99. ANOMALOUS BEHAVIOR OF H2O
The phase diagram for water is not typical.
The melting temperature line , TB , slopes
slant to the left (negative slope)
i.e. the melting temperature decreases with pressure.
This is because ice is less dense than water, the solid occupies more space than the liquid An increase in pressure, favours the formation of liquid H2O.( the phase that occupy less space)
Thus, the higher the pressure, the lower the
temperature at which solid water melts.

100. Exercise:
a) Describe the phase change when carbon dioxide undergoes isobaric heating at 5.2 atm pressure

101. Exercises:
b) Describe the phase change when pressure is applied to water isothermally at 0.01 C