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CHEMISTRY FORM 5-PHYSICAL CHEMISTRY – Gases

CHEMISTRY FORM 5-PHYSICAL CHEMISTRY - Gases

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CHEMISTRY FORM 5-PHYSICAL CHEMISTRY – 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 ( D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image001.gif ).

 

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 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image002.gif Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image003.gif
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
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif = Means square speed of the molecules
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif = Is given by
D:\..\..\thlb\cr\tz\__i__images__i__\P36.jpg

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 .
D:\..\..\thlb\cr\tz\__i__images__i__\2227.PNG

i.e The ratio of volume of temperature is constant.
Thus,
D:\..\..\thlb\cr\tz\__i__images__i__\dg10.PNG
From kinetic theory of gases
PV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif

For any gas to obey charle’s law
D:\..\..\thlb\cr\tz\__i__images__i__\33310.PNG

V = k D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image011.gif
From kinetic theory of gases
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif = T

V = KT
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image012.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image013.gif
D:\..\..\thlb\cr\tz\__i__images__i__\P114.jpg

 

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.
D:\..\..\thlb\cr\tz\__i__images__i__\44410.PNG
From kinetic of gases
PV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif

From gas to obey Boyle’s law
D:\..\..\thlb\cr\tz\__i__images__i__\5554.PNG

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
D:\..\..\thlb\cr\tz\__i__images__i__\P210.jpg

For two gases which are G1 and G2
For Gas1
D:\..\..\thlb\cr\tz\__i__images__i__\P37.jpg

Comparison
For the ratio of their rates
D:\..\..\thlb\cr\tz\__i__images__i__\p47.jpg
In terms of their density
If the volume is constant
D:\..\..\thlb\cr\tz\__i__images__i__\P63.jpg
In terms of their masses

4.  DEDUCTION OF GRAHAM’S LAW

From Graham’s law
D:\..\..\thlb\cr\tz\__i__images__i__\P54.jpg
From kinetic theory of gases
Pv = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif

3PV = Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image011.gif
But Nm = m
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image038.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image039.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image040.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image038.gif

For any gases to obey graham’s law

3P = k
constant = 1

3P = 1
Now,
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image041.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image042.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif =1/ D:\..\..\thlb\cr\tz\__i__images__i__\chem24.JPG

But,

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif = r2

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image044.gif 2 =  D:\..\..\thlb\cr\tz\__i__images__i__\chem9.JPG

D:\..\..\thlb\cr\tz\__i__images__i__\P54.jpg
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 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif
Since the volume is common.
For gas A
PAV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif NAmA D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image041.gif A
Since the volume is common.
For gas B
PBV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image048.gif NBmB D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image041.gif 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
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif is the sum of D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image011.gif of A and B
PtV= D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif Nt m D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif
Nt = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image049.gif
D:\..\..\thlb\cr\tz\__i__images__i__\chem10.JPG

D:\..\..\thlb\cr\tz\__i__images__i__\chem11.JPG

But Nt = NA + NB
D:\..\..\thlb\cr\tz\__i__images__i__\DG12.PNG
Since K. E is conserved ( It is the same before and after )
D:\..\..\thlb\cr\tz\__i__images__i__\G35.PNG = MA D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image056.gif = MB D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image057.gif
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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image015.gif T

ii)  Boyle’s law
V D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image015.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image058.gif

Combining

V D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image059.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image017.gif

V = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image060.gif

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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image061.gif
ii)  R = 8.314 -1 mol -1
Some questions  need  just general gas equation.
Recall
D:\..\..\thlb\cr\tz\__i__images__i__\P73.jpg
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
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image066.gif = 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   D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image066.gif   and this bring about the deviation of real gases from ideal behaviour. This deviation is graphically represent when the graph of  D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image068.gif  v/s  Pressure is plotted.
The graph of       D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image070.gif       v/s   P (Amagat curve)

D:\..\..\thlb\cr\tz\__i__images__i__\DG5.PNG




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 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image074.gif
PV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image075.gif RT × D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image076.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image077.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image078.gif
Mr  =  D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image079.gif

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
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif = 0.97 atm
Mr = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image078.gif
Mr = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image081.gif
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 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image082.gif

PV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image083.gif RT

Mr D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image084.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image085.gif

MrP = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image085.gif

But, D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image086.gifD:\..\..\thlb\cr\tz\__i__images__i__\chem25.JPG

D:\..\..\thlb\cr\tz\__i__images__i__\p83.jpg

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 = ?
D:\..\..\thlb\cr\tz\__i__images__i__\chem27.JPG  =  D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image090.gif

T  =  D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image091.gif

= 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif Mr = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image078.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif PV

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image093.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image094.gif

V = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image094.gif

V = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image095.gif

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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 105 NM-2
T2 = 250
P2 = ?
V2 = 1.88
So.

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image096.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image097.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image098.gif   =  D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image099.gif

Note:
P2 = 1.746 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif 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
D:\..\..\thlb\cr\tz\__i__images__i__\p92.jpg

From Graham’s law

D:\..\..\thlb\cr\tz\__i__images__i__\P102.jpg

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

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image117.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image118.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image119.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image029.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image120.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image118.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image121.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image122.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image123.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image118.gif

D:\..\..\thlb\cr\tz\__i__images__i__\P116.jpg
D:\..\..\thlb\cr\tz\__i__images__i__\P124.jpg

t0   = 623 sec        MH = 2

tH   = 130 sec
D:\..\..\thlb\cr\tz\__i__images__i__\P133.jpg

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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image143.gif NH4

D:\..\..\thlb\cr\tz\__i__images__i__\chem19.JPG

D:\..\..\thlb\cr\tz\__i__images__i__\P143.jpg

D:\..\..\thlb\cr\tz\__i__images__i__\7772.PNG

21407 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image174.gif 429.4X + 2.147X2 = X2
2.147 X2 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image175.gif
Solving quadratically
The solid deposit 59.4 cm away from N D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image176.gif

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)

D:\..\..\thlb\cr\tz\__i__images__i__\dg111.PNG

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

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image194.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image195.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image196.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image197.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image198.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image199.gif

D:\..\..\thlb\cr\tz\__i__images__i__\dg6.PNG

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image202.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image204.gif

 

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image205.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image204.gif

28 = 9.75 + 4.62n
0.165 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image206.gif

28 = 9.75 + 4.62n
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image207.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image208.gif

n = 3.9 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image116.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image209.gif ( D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image210.gif )
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
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image211.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image212.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image213.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image214.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image215.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image216.gif

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 %
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image218.gif

In decimals
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image219.gif
Ie D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image220.gif

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 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image221.gif
= D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image222.gif

= 0.25

No of moles of CO2 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image223.gif

= 0.1

Total number of moles = 0.1 + 0.25

= 0.35

% composition of m moles = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image224.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image225.gif

= 0.174 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 100

= 71.4%

% composition of CO2 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image226.gif 100
= 0.285 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image227.gif
Volume I is common rate of mixture = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image229.gif tare of Br2

but

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image230.gif = rate for mixture

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image231.gif = rate Br2

Mr Br2 = 80 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image232.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image230.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image233.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image231.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image234.gif
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image231.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 2 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image235.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image234.gif
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image237.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image234.gif

D:\..\..\thlb\cr\tz\__i__images__i__\DG7.PNG
4 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image240.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image241.gif

M = 40
Mr of CO = 28
Mr of CO2 = 44
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO Mr D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO2 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image242.gif Mr CO2 = 40
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image243.gif CO D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image122.gif 28 + D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image244.gif = 40
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image242.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO = 1
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO2 = 1 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image169.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 28 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image242.gif 1 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image246.gif
28 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image247.gif CO = 40
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image248.gif = 40 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image169.gif 44
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image249.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image250.gif
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image066.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image251.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image169.gif 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

D:\..\..\thlb\cr\tz\__i__images__i__\dg8.PNG
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image254.gif

But n = 1 for 1mole

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image256.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image257.gif
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image258.gif     This is correction in pressure.

From ideal gas equation

PV = nRT      (n = 1)

PV = RT
D:\..\..\thlb\cr\tz\__i__images__i__\DG9.PNG
For one mole of the gas
Where “a” is a van der Waal constant
Note
In the van der Waal equation D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image261.gif 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

D:\..\..\thlb\cr\tz\__i__images__i__\S_8.PNG

PV D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image266.gif
PV = RT D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image267.gif
The equation normally work under two assumption these is
If pressure is small then the volume is maximum
The terms Pb and D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image268.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image271.gif
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




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 ( D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image001.gif ).

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 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image002.gif Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image003.gif
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
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif = Means square speed of the molecules
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif = Is given by
D:\..\..\thlb\cr\tz\__i__images__i__\P36.jpg

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 .
D:\..\..\thlb\cr\tz\__i__images__i__\2227.PNG

i.e The ratio of volume of temperature is constant.
Thus,
D:\..\..\thlb\cr\tz\__i__images__i__\dg10.PNG
From kinetic theory of gases
PV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif

For any gas to obey charle’s law
D:\..\..\thlb\cr\tz\__i__images__i__\33310.PNG

V = k D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image011.gif
From kinetic theory of gases
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif = T

V = KT
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image012.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image013.gif
D:\..\..\thlb\cr\tz\__i__images__i__\P114.jpg

 

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.
D:\..\..\thlb\cr\tz\__i__images__i__\44410.PNG
From kinetic of gases
PV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif

From gas to obey Boyle’s law
D:\..\..\thlb\cr\tz\__i__images__i__\5554.PNG

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
D:\..\..\thlb\cr\tz\__i__images__i__\P210.jpg

For two gases which are G1 and G2
For Gas1
D:\..\..\thlb\cr\tz\__i__images__i__\P37.jpg

Comparison
For the ratio of their rates
D:\..\..\thlb\cr\tz\__i__images__i__\p47.jpg
In terms of their density
If the volume is constant
D:\..\..\thlb\cr\tz\__i__images__i__\P63.jpg
In terms of their masses

4.  DEDUCTION OF GRAHAM’S LAW

From Graham’s law
D:\..\..\thlb\cr\tz\__i__images__i__\P54.jpg
From kinetic theory of gases
Pv = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif

3PV = Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image011.gif
But Nm = m
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image038.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image039.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image040.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image038.gif

For any gases to obey graham’s law

3P = k
constant = 1

3P = 1
Now,
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image041.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image042.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif =1/ D:\..\..\thlb\cr\tz\__i__images__i__\chem24.JPG

But,

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif = r2

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image044.gif 2 =  D:\..\..\thlb\cr\tz\__i__images__i__\chem9.JPG

D:\..\..\thlb\cr\tz\__i__images__i__\P54.jpg
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 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif Nm D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif
Since the volume is common.
For gas A
PAV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif NAmA D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image041.gif A
Since the volume is common.
For gas B
PBV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image048.gif NBmB D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image041.gif 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
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif is the sum of D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image011.gif of A and B
PtV= D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image009.gif Nt m D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image004.gif
Nt = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image049.gif
D:\..\..\thlb\cr\tz\__i__images__i__\chem10.JPG

D:\..\..\thlb\cr\tz\__i__images__i__\chem11.JPG

But Nt = NA + NB
D:\..\..\thlb\cr\tz\__i__images__i__\DG12.PNG
Since K. E is conserved ( It is the same before and after )
D:\..\..\thlb\cr\tz\__i__images__i__\G35.PNG = MA D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image056.gif = MB D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image057.gif
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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image015.gif T

ii)  Boyle’s law
V D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image015.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image058.gif

Combining

V D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image059.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image017.gif

V = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image060.gif

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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image061.gif
ii)  R = 8.314 -1 mol -1
Some questions  need  just general gas equation.
Recall
D:\..\..\thlb\cr\tz\__i__images__i__\P73.jpg
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
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image066.gif = 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   D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image066.gif   and this bring about the deviation of real gases from ideal behaviour. This deviation is graphically represent when the graph of  D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image068.gif  v/s  Pressure is plotted.
The graph of       D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image070.gif       v/s   P (Amagat curve)

D:\..\..\thlb\cr\tz\__i__images__i__\DG5.PNG

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 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image074.gif
PV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image075.gif RT × D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image076.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image077.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image078.gif
Mr  =  D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image079.gif

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
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif = 0.97 atm
Mr = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image078.gif
Mr = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image081.gif
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 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image082.gif

PV = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image083.gif RT

Mr D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image084.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image085.gif

MrP = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image085.gif

But, D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image086.gifD:\..\..\thlb\cr\tz\__i__images__i__\chem25.JPG

D:\..\..\thlb\cr\tz\__i__images__i__\p83.jpg

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 = ?
D:\..\..\thlb\cr\tz\__i__images__i__\chem27.JPG  =  D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image090.gif

T  =  D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image091.gif

= 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif Mr = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image078.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif PV

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image093.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image094.gif

V = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image094.gif

V = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image095.gif

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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 105 NM-2
T2 = 250
P2 = ?
V2 = 1.88
So.

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image096.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image097.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image098.gif   =  D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image099.gif

Note:
P2 = 1.746 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif 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
D:\..\..\thlb\cr\tz\__i__images__i__\p92.jpg

From Graham’s law

D:\..\..\thlb\cr\tz\__i__images__i__\P102.jpg

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

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image117.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image118.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image119.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image029.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image120.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image118.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image121.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image122.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image123.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image118.gif

D:\..\..\thlb\cr\tz\__i__images__i__\P116.jpg
D:\..\..\thlb\cr\tz\__i__images__i__\P124.jpg

t0   = 623 sec        MH = 2

tH   = 130 sec
D:\..\..\thlb\cr\tz\__i__images__i__\P133.jpg

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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image143.gif NH4

D:\..\..\thlb\cr\tz\__i__images__i__\chem19.JPG

D:\..\..\thlb\cr\tz\__i__images__i__\P143.jpg

D:\..\..\thlb\cr\tz\__i__images__i__\7772.PNG

21407 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image174.gif 429.4X + 2.147X2 = X2
2.147 X2 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image175.gif
Solving quadratically
The solid deposit 59.4 cm away from N D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image176.gif

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)

D:\..\..\thlb\cr\tz\__i__images__i__\dg111.PNG

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

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image194.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image195.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image196.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image197.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image198.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image199.gif

D:\..\..\thlb\cr\tz\__i__images__i__\dg6.PNG

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image202.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image204.gif

 

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image205.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image204.gif

28 = 9.75 + 4.62n
0.165 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image206.gif

28 = 9.75 + 4.62n
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image207.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image208.gif

n = 3.9 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image116.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image209.gif ( D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image210.gif )
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
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image211.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image212.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image213.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image214.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image215.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image216.gif

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 %
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image218.gif

In decimals
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image219.gif
Ie D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image220.gif

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 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image221.gif
= D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image222.gif

= 0.25

No of moles of CO2 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image223.gif

= 0.1

Total number of moles = 0.1 + 0.25

= 0.35

% composition of m moles = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image224.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image225.gif

= 0.174 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 100

= 71.4%

% composition of CO2 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image226.gif 100
= 0.285 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image227.gif
Volume I is common rate of mixture = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image229.gif tare of Br2

but

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image230.gif = rate for mixture

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image231.gif = rate Br2

Mr Br2 = 80 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image232.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image230.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image233.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image231.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image234.gif
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image231.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 2 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image235.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image234.gif
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image237.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image234.gif

D:\..\..\thlb\cr\tz\__i__images__i__\DG7.PNG
4 = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image240.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image241.gif

M = 40
Mr of CO = 28
Mr of CO2 = 44
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO Mr D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO2 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image242.gif Mr CO2 = 40
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image243.gif CO D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image122.gif 28 + D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image244.gif = 40
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image242.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO = 1
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO2 = 1 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image169.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image092.gif 28 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image242.gif 1 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image246.gif
28 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif CO D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image247.gif CO = 40
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image248.gif = 40 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image169.gif 44
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image249.gif = D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image250.gif
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image080.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image066.gif D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image251.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image169.gif 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

D:\..\..\thlb\cr\tz\__i__images__i__\dg8.PNG
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image254.gif

But n = 1 for 1mole

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image256.gif

D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image257.gif
D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image258.gif     This is correction in pressure.

From ideal gas equation

PV = nRT      (n = 1)

PV = RT
D:\..\..\thlb\cr\tz\__i__images__i__\DG9.PNG
For one mole of the gas
Where “a” is a van der Waal constant
Note
In the van der Waal equation D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image261.gif 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

D:\..\..\thlb\cr\tz\__i__images__i__\S_8.PNG

PV D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image266.gif
PV = RT D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image267.gif
The equation normally work under two assumption these is
If pressure is small then the volume is maximum
The terms Pb and D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image268.gif 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 D:\..\..\thlb\cr\tz\PHYSICAL CHEMISTRY - Gases_files\image271.gif
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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GENERAL CHEMISTRY



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