CHEMISTRY FORM FIVE TOPIC 3: GASES

 

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

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 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.

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

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

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.
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

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
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
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

∴  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

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
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

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

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

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)

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

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 nA be number of moles of gas A
nB be number of moles of gas B
nT = nA + nB
 = 

 =  =

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

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
 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.

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

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