PHYSICAL CHEMISTRY - Gases Is the branch of chemistry which deals with measurable quantities. PARTS OF PHYSICAL CHEMISTRY The physical chemistry is studied under the following sub topics. These are 1. States of matter 2. Chemical equilibrium 3. Thermochemistry/ Energetics 4. Electrochemistry 5. Chemical kinetics

1. PHYSICAL CHEMISTRY - Gases
Is the branch of chemistry which deals with measurable quantities.
PARTS OF PHYSICAL CHEMISTRY
The physical chemistry is studied under the following sub topics. These are
1. States of matter
2. Chemical equilibrium
3. Thermochemistry/ Energetics
4. Electrochemistry
5. Chemical kinetics
1. STATE OF MATTER
This is the form in which matter do exist. There are three states of matter. These are
a) Gaseous state
b) Liquid state
c) Solid state
A. GASEOUS STATE
This is the form in which matter do exist in motion.
PROPERTIES OF GASEOUS
Properties of gases are governed by kinetic theory of gases ( ).
THE KINETIC THEORY OF GASES
1. Kinetic theory of gases include the following points or postulates.
2. Gases contain large number of molecules which are in continuous random motion.
3. Molecules of gases are far away such that the force of attraction between individual molecules
is negligible kinetic energy of the molecules is directly proportion to absolute temperature i.e the
increase of temperature will causes the increases in kinetic energy of the molecules.
4. The values of individual molecules of gases is negligible compared to the volume of the
container.
5. Pressure of the gas inside the container is due to the collision between molecules of the gas
and the wall of the container.
6. The collision of molecules of gases are perfectly elastic that is kinetic energy is conserved (Kt
before is equal to K.E after collision).
CLASSIFICATION OF GASES
Gases are categorized into two groups
These are
i) Ideal gases
ii) Real gases

2. i. IDEAL GASES
Ideal gases are gases which obey all assumption of kinetic theory and also they do obey the ideal
gas equation
i.e PV = nRT.
Where by
P = Pressure of the gases
V = Volume of the gases
n = Number of moles of gases
R = Universal of gas constant
T = Temperature
ii. REAL GASES
Real gases are the gases which do not obey all the assumptions of kinetic theory of gases. Such
assumptions are:
i. The volume of individual molecules of the gases is negligible compared to the volume of the
container.
ii. Molecules of gases are far away such that the force of attraction between individual molecules
is negligible.
From kinetic theory of gases, scientist comes up with the formula / equation that is governed by
the theory the equation is
PV = Nm
Where by
P = Pressure of the gas
V = Volume of the gas
N = Number of the molecules of the gas
m = Mass of molecules of the gas
= Means square speed of the molecules
= Is given by
DEDUCTION OF GASES LAWS FROM KINETIC THEORY OF GASES
This is the process by which gas laws are obtained or derived from kinetic theory of gases.
Such laws to be deduced are:
i. Charles’s law.
ii. Boyle’s laws.
iii. Graham's law of diffusion/ effusion.
iv. Dalton’s law of partial pressure.
v. Avogadro’s law.
1. DEDUCTION OF CHARLES’ LAW
It state that “ At constant pressure the volume of a fixed mass of gas is directly proportion to the

3. absolute temperature .
i.e The ratio of volume of temperature is constant.
Thus,
From kinetic theory of gases
PV = Nm
For any gas to obey charle’s law
V = k
From kinetic theory of gases
= T
V = KT
=
QUESTION
a) Define the following terms
i) Real gases.
ii) Ideal gases.
b) State Charles’s law.
c) Deduce Charles law from kinetic theory of gases.
2. DEDUCTION OF BOYLE'S LAW FROM KINETIC THEORY OF GASES
It state that “ At constant thermodynamic volume of a fixed mass of gas is inversely proportional
to pressure.

4. From kinetic of gases
PV = Nm
From gas to obey Boyle’s law
PV = K
Hence Boyle’s law deduced
3. DEDUCTION OF GRAHAM’S LAW FROM KINETIC THEORY OF GASES
Graham’s rate of diffusion or effusion: It states that,
"The rate of diffusion or effusion of gaseous material is inversely proportional to the square root
of its density"
If rate is represented by r and density by ρ
For two gases which are G1 and G2
For Gas1
Comparison
For the ratio of their rates

5. In terms of their density
If the volume is constant
In terms of their masses
4. DEDUCTION OF GRAHAM’S LAW
From Graham’s law
From kinetic theory of gases
Pv = Nm
3PV = Nm
But Nm = m
=
=
For any gases to obey graham’s law
3P = k
constant = 1

6. 3P = 1
Now,
=
=1/
But,
= r2
2 =
Hence grahams law deduced
DALTON’S LAW OF PARTIAL PRESSURE
It state that
“ The total pressure of gases which are not reacting is equal to the sum of pressure of individual
gases in a container"
Consider the two gases which are;
Gas A and Gas B
Each gas will create its own pressure.
Therefore the pressure of gases will be,
PA for gas A
PB for gas B
From Dalton’s law of partial pressure,
PT = PA + PB
Where by
PT is the total pressure of all gases in the container.
DEDUCTION OF DALTON’S LAW OF PARTIAL PRESSURE FROM KINETIC
THEORY OF GASES
From Dalton’s law for gas A and B
PT = PA + PB
From kinetic theory of gases
PV = Nm
Since the volume is common.
For gas A
PAV = NAmA A
Since the volume is common.

7. For gas B
PBV = NBmB B
If Pt is the total pressure
V is the common volume
Nt is the total number of molecules
M is the total mass of molecules
is the sum of of A and B
PtV= Nt m
Nt =
But Nt = NA + NB
Since K. E is conserved ( It is the same before and after )
= MA = MB
PT = PA + PB, Hence Dalton law deduced
QUESTION
Deduced the Avogadro’s law from kinetic theory of gases
THE IDEAL GAS EQUATION
This is the equation which is obeyed by all ideal gases.
Formulation of ideal gas equation
The ideal gas equation is formed from the combination of two gas law . These are
i) Charles’s law
ii) Boyle’s law
i) Charle’s law
V T

8. ii) Boyle’s law
V
Combining
V
V =
V = kT
k represent universal gas constant which is denoted by R.
PV = RT
This is exactly for one mole of a gas.
For n moles of the gas
PV = nRT
Where
P is the pressure of the gas
V is the volume of the gas
n is the number of the moles of the gas
R is the universal constant
T is the absolute temperature
Units for R
R can be written as;
i) R = 0.0821 atm moles
ii) R = 8.314 -1 mol -1
Some questions need just general gas equation.
Recall
DEVIATION OF REAL GASES FROM IDEAL BEHAVIOUR
From kinetic theory of gases it has been observed that gases which do obey all assumption are
termed as ideal gases.
For an ideal gas PV is exactly = 1
PV = nRT
For one mole of a gas.
PV = RT

9. The ratio of PV to RT is exactly 1.0
= 1.0
Real gases do deviate from this behaviour at high pressure and low pressure.Therefore the
variation of pressure cause the fluctuation on the value of and this bring about the deviation
of real gases from ideal behaviour. This deviation is graphically represent when the graph
of v/s Pressure is plotted.
The graph of v/s P (Amagat curve)
TERMS
 Critical temperature
It is the temperature above which the gas cannot be liquified without further cooling.
 Critical pressure
It is the pressure at which the gas start to liquefy.
APPLICATION OF IDEAL GAS EQUATION
Ideal gas equation is applied in the following aspects;
1. Determination of moral mass of the gas.
From ideal gas equation

10. PV = nRT
But, n is the number of moles
n =
PV = RT ×
=
Mr =
Example 1
a) Define the following terms
i) Critical pressure
ii) Ideal gas
iii) Real gas
iv) Critical temperature
b) “Some gases are ideal while other are not
c) 1.27 of sample of oxide of nitrogen believed to be either NO or NO2 occupy the volume of
1.07 dm3 at 250c and pressure of 737 mmHg. Explain what oxide is it and why? R = 0.0821 atm
mol -1 k -1 L.
Solution
Mass = 1. 27g
Volume = 1.07 dm3
Temperature = 737 mmHg
R = 0.0821
Unit conversion
1 dm3 = 1.0 L
T = 273 + 25 = 298k
1 atm = 760 mmHg
= 0.97 atm
Mr =
Mr =
Mr = 29.9g/mol
Theoretical molar mass
NO2 = 14 + 32 = 46
NO = 30

11. ∴ The gas is NO because the molar mass is 30 g/mol.
2. Determination of densities of gaseous materials
From ideal gas equation
PV = nRT
But n =
PV = RT
Mr =
MrP =
But, =
Example 1
At what temperature would the oxygen gas be if the pressure is kept constant at 745 mmHg . If
the density of the gas is 1.00g/dm3. Given that
R is 0.0821 atm mol -1 k-1 L
Solution
From O2
Mr = .32
Pressure = 745
R = 0.0821
Density = 1.00g/dm3
T = ?
=
T =
= 381.9k
... The temperature required = 381.9k

12. Example 2
What is the volume occupied by 13.7g of chlorine gas at 45oc and 745 mmHg
R = 0.0821 atom mol-1 k-1 L
Solution
M = 13.7g
T = 45oc
P = 745mmHg
R = 0.0821
Mr = 71
Conversion of units
T = 45 + 273 = 318K
P = 1 atm = 760mmHg
T = 745
P = 0.98 atm
From
PV Mr = PV
=
V =
V =
V = 5.14 Liters
∴ The volume of chlorine gas is 5.14dm3
EXAMPLES ON GAS LAWS
Example1
A certain mass of an ideal gas has a volume of 3.25dm3 at 25oc and 1.01 105 NM -2 . What
pressure is required to compress it to 1.88dm3 at the some temperature.
Solution
V1 = 3.25dm
T1 = 25OC
P1 = 1.01 105 NM-2

13. T2 = 250
P2 = ?
V2 = 1.88
So.
=
=
Note:
P2 = 1.746 105NM-2
1.01 105NM-2 = 1atm
1.01 105NM-2 = 760 mmHg
1 NM-2 = 1 Pa
Example 2.
It takes 54.4 sec. for 100cm3 of a gas to effuse through an aperture and 36 . 5 sec for 100cm3 of
O2 to effuse through the same aperture
i) What is the molar mass of gas X
ii) Suggest the gas X
Solution
Rate
From Graham’s law
MrX = 70.9

14. MrX≈71
i) The molar mass of gas X is 71g/mol
ii) The gas X is chlorine gas
Example 3
a) Define the following terms
i. Effusion
ii) Critical temperature
b) State Grahm’s law of diffusion
A certain volume of hydrogen takes 2min and 10 sec to diffuse through a porous plug and an
oxide of nitrogen takes 10min 223 sec.
What is:
i) Molar mass of an oxide
ii) Give the following of an oxide
Solution
V be the volume of hydrogen
V be the volume of oxide
=
=
=
t0 = 623 sec MH = 2
tH = 130 sec

15. M0 = 46
i) The moles mass of the oxide is 46g/mol
ii) The formula of the oxide is NO2
iii) Effusion is the escaping of a gas through a porous without molecular collision between
molecules of gas.
iv) Graham’s law of diffusion state that. At the same temperature and pressure the rate of
diffusion of different is inversely proportional to the square roots their Mr mass.
v) Critical temperature is the temperature above which the gas cannot be liquefied without
further cooling.
Example 4
a) State the kinetic theory of gases
b) Write down the equation of the gas which do obey all assumption in (a) above and define the
terms
c) A plug of cotton wool one soaked in conc. Hcl where inserted into opposite ends of a
horizontal glass tube
A disc of solid ammonium chloride formed in the tube plug is the 1m long how for from the
ammonia plug is the deposited
Solution
Reaction
NH3 + Hcl NH4

16. 21407 429.4X + 2.147X2 = X2
2.147 X2
Solving quadratically
The solid deposit 59.4 cm away from N
a) The kinetic theory of gases states that
i) Molecules of a gas are far way such that force of attraction between the individual molecule is
neglible
ii) The collision of gas molecules are perfectly elastic
iii) The pressure of a gas inside the container is due to the collision between the molecules of the
gas and the wall of the container
iv) Gases container a large number of molecules which exist in continuous random motion
v) Kinetic energy of molecules is directly proportional to the absolute temperature
vi) The volume of individual molecules of the gas is negligible compared to the volume of the
container.
b) Equation of the gases that obey kinetic theory of gases is as follows:
PV = nRT
Where by
P = Is the pressure of the gas
V = Is the volume of the gas
n = I s the number of molecules of the gas
R = is the universal constant
T = Is the absolute temperature
Example 5.
A) A certain volume of SO2 diffuses through a porous plug in 10.0 min and the same volume of
second gas takes 15.8 min. Calculate the relative molecules mass of the second gas
B) Nicked from a carbonyl, Ni (CO) n Deduce the value of n from the fact that carbon monoxide
diffuse 2.46 times faster than the carbon compound .
Solution
Let V be the volume of SO2
V be the volume of gas x (gx)

17. Solution
Mr of Ni (CO)n = 59 +(28)n
Mr of CO = 28
Let time takes for carbonyl to diffuse = x
Time taken for carbon monoxide to diffuse = 2.46x
From Graham’s law
=
=

18. =
=
=
28 = 9.75 + 4.62n
0.165 =
28 = 9.75 + 4.62n
=
n = 3.9 4
The volume of n in Nl (CO) is 4
MOLE FRACTION OF A GAS
Mole fraction of a gas is defined as the ratio of number of mass of such gas per total moles of
gases present
The moles fraction is denoted by ( )
Example
If there are two gas ie. Gas A and gas B in the container. Show how can you find the mole
fraction each gas
Solution
Gases present
Gas A and gas B
Let n A be number of moles of gas A
nB be number of moles of gas B
nT = nA + nB
=
= =

19. Note
When mole are expressed in % will sum up to 100%
And when expressed in decimal will not exceed 1.00
From the example above In %
In decimals
Ie
Example
The container was found to have 8g of oxygen and 4.4g of CO2. What is the percentage
composition of O2 and CO2 in the container
Solution
Mass of oxygen = 8g
Mass of carbon dioxide = 4.4g
No of mole of O2 =
=
= 0.25
No of moles of CO2 =
= 0.1
Total number of moles = 0.1 + 0.25
= 0.35
% composition of m moles =
= 0.174 100
= 71.4%
% composition of CO2 = 100
= 0.285 100
= 29%
The percentage of composition of O2 and CO2 in the container are 71.4% and 29%
Critical pressure is the pressure at which the gas starts to liquify

20. Example
A mixture of CO and CO2 diffusion through a porous diagram in on half of the time taken for the
same volume of Bromine vapour. What is the composition by volume of the mixture?
Solution
Let time taken by Br2 be t
Time taken by the mixture CO + CO2 be
Volume I is common rate of mixture = tare of Br2
but
= rate for mixture
= rate Br2
Mr Br2 = 80
=
2 =
=
4 =
M = 40
Mr of CO = 28
Mr of CO2 = 44
CO Mr CO2 Mr CO2 = 40
CO 28 + CO = 40
CO CO = 1

21. CO2 = 1 CO
CO 28 1
28 CO CO = 40
= 40 44
=
CO = 0.25
% Of CO = 25%
% Of CO2 = 75%
VAN DER WAAL EQUATION
The deviation of real gases from ideal behaviour has been stuglied by different scientist such as
Amagat who deter minal different curves in P
Later on van der wall explained the deviation and gave his equation which tries to work / take
particle volume and attraction into account
In his equation
i) He subtracted the particle volume from the volume of the container. If the volume of container
is V and that of particular be presented as “b”
Then the total volume of the system will be
V b
This is called “Volume correction”
ii) He also added change in pressure which is caused by the existence of intermolecular forces of
attraction among the gaseous molecules
This
But n = 1 for 1mole
This is correction in pressure.

22. From ideal gas equation
PV = nRT (n = 1)
PV = RT
For one mole of the gas
Where “a” is a van der Waal constant
Note
In the van der Waal equation is added to the total pressure in order to cancel the effect of
intermolecular forces of attraction
Expansion of the Van der waal equation
From Van der Waal’s equation
PV
PV = RT
The equation normally work under two assumption these is
If pressure is small then the volume is maximum
The terms Pb and can be neglected
PV = RT - a/V
ii) When the pressure is very low that is P is approaching O P O
So V will be infinity V is
We can calculate that the three term for correction can be neglected
PV = RT
This is for one mole
General conclusion
At very low pressure the real gases normally behave as ideal gases