Π.1
The gas equations
A.
Introduction:
Matter can exist in three states - gas, liquid and solid.
Solids are orderly arrays of molecules, atoms or ions having definite geometry
depending on the arrangements of these particles.
Liquids do not have definite shapes but will attain the shape of the container into
which they are placed but they have fixed volume.
Gases do not have definite shapes or volumes.
B.
Gas Laws:
The earliest empirical generalization was a result of the work of Robert Boyle (1662).
Boyle found that when a fixed amount of gas at a constant temperature was
subjected to different pressures, the volume of gas varied inversely as pressure.
V ∝
1
k
, V = T ........(1)
P
P
where T and m (mass of the gas) are constant.
Rearranging equation (1), we have
PV = k T ........(2)
This is the mathematical statement of Boyle's law.
Equation (2) is the equation of hyperbola.
C.
Π.2
Charles' law:
Boyle's law gives the change in volume of a given mass of gas for a change in
pressure at constant temperature. This constant temperature process is known as an
isothermal process.
On the other hand, the constant pressure process is called an isobaric process.
The first experimental study of the isobaric variation of volume with temperature was
made by J. Charles (1787).
Charles' law shows that
V = k p T .......(3)
where m and P are constants.
The volume of a gas (V) is directly proportional to the absolute temperature, at
constant pressure (P) and mass (m).
Graphically,
If the volume of a gas is maintained constant and its temperature is raised, the
pressure will increase according to the equation below:
P = k v T .......(4)
where m and V are constants.
Π.3
Consider the following changes:
Suppose a given mass of a gas has a volume V1, and a pressure P1. if the pressure is
changed to P2 while temperature remains constant (T1). The new volume will be , say,
V'. By Boyle's law:
P1V1 = P2V'
V' =
P1V1
..........(*)
P2
If the pressure is now kept constant at P2, and the temperature is changed from T1 to
T2. The volume will be changed to V2. The by Charles's law:
V ' V2
=
T1 T2
..........(**)
Put equation (*) into (**)
P1V1 P2V2
=
T1
T2
or
PV
= constant .......(5)
T
In 1811, the Italian scientist Amedeo Avogadro suggested an important hypothesis
which later became known as Avogadro's theory:
Equal volumes of all gases at the same temperature and pressure contain equal
number of molecules.
According to the theory, equation (5) can be written as the ideal gas law:
PV = nRT .......(6)
where
n is the number of moles of the gas,
R is the gas constant (0.082 atmdm3K-1mol-1 or 8.31 JK-1mol-1)
For those gases that obey equations (1) to (6) are called ideal or perfect gases.
Classwork :
At stp, a certain mass of gas has a volume of 1.00dm3. At 30 atm pressure,
the volume is 31.2cm3. At 60 atm, the volume is 14.9cm3. Does the gas show
ideal behaviour? Explain your answer.
In practice, real gases obey the equations very closely at certain pressure and
temperature.
The following diagram (on p.4) shows you the deviation of real gases from
the ideal gas behaviour.
Π.4
For a real gas to be regarded as ideal gas, we should have the following assumptions:
(Assumptions of Kinetic Theory of gas)
1.
2.
3.
4.
All gases consist of molecules which are in continuous, random motion.
They travel in straight lines between collisions,
In a given volume, the number of molecules is very large and the volume
occupied by the molecules themselves is negligible compared to the total
volume of the container,
The molecules are small and have no interaction between them,
The impacts of the molecules with each other and with the walls of the
container are perfectly elastic. An elastic collision is one in which the only
energy change is in the kinetic energy of the colliding body upon collision
and energy must be conserved on collision.
Therefore, for any real gases, the ideal gas equation will become the real gas equation
- called van der waals' equation.
Classwork:
1.
Try to think that if the gases compressed (high pressure) what kind of
force which has not been taken into account in Kinetic Theory
becomes important?
2.
How does the volume of the gas molecules come to affect the ideal gas
law when the gas is at high pressure?
Form question 1, we have to modify the pressure (P).
From question 2, we have to modify the volume (V).
After the two modifications, ideal gas law becomes van der waals' equation (out of
syllabus).
(P +
a
V
2
)(V - b) = nRT
Π.5
D.
Determining the relative molecular masses of gases or volatile liquids:
We can measure the relative molecular masses by making use of the ideal gas law.
PV = nRT .......(6)
but
n=
m
Mr
where Mr is the relative molecular mass of the gas,
m is the actual mass of the gas.
Equation (6) becomes,
m
PV =
RT ......(7)
Mr
or
Mr =
mRT
.......(8)
PV
The following set-up is used to determine Mr.
Example:
Calculate the relative molecular mass of gas X according to the following
data.
mass of liquid in hypodermic syringe (m) = 0.16 g
volume of gas X vaporized in the graduated syringe = 46 cm3
temperature of gas = 100 oC
pressure of X = 1 atm (101325 Nm-2)
gas constant = 8.31 JK-1mol-1
Π.6
E.
Avogadro's theory and relative molecular masses:
By Avogadro's theory (equal volume of gas containing equal number of molecules)
mass of x dm3 of gas X
mass of x dm3 of H2
=
mass of n molecules of gas X
mass of n molecules of H2
=
mass of 1 molecules of gas X
mass of 1 molecules of H2
=
mass of 1 molecules of gas X
2 x mass of 1 atom of H
Since relative atomic mass of H = 1
mass of x dm3 of gas X
mass of x dm3 of H2
=
1
relative molecular mass of gas X
2
Usually,
mass of x dm3 of gas X
mass of x dm3 of H2
is called the vapour density of the gas.
Therefore,
2 x (vapour density of a gas) = relative molecular mass of the gas
Classwork :
1.
The density of a certain gas is 0.72g/dm3 at stp. What is the molecular
mass of the gas, if 100cm3 of hydrogen at stp weigh 0.009g?
2.
1g of a gas occupies 420cm3 at stp. What is its molecular mass? (molar
volume of any gas at stp = 22.4dm3)
Π.7
F.
Determination of molecular mass by direct weighing:
From Ideal gas equation,
PV = nRT
=
By
ρ=
m
RT
M
m
V
M =
ρRT
P
The following diagram shows the set-up.
Classwork:
R = 8.31JK-1mol-1, Gas molar volume = 22.4dm3 at stp
1.
2.
3.
4.
Calculate the molar mass of a gas which has a density of 2.615gdm-3 at
298K and 101kNm-2.
At 273K and 1.01x105Nm-2, 6.319g of a gas occupy 2.00dm3.
Calculate the molar mass of the gas.
An ideal gas occupies 225cm3 at 280K and 4.80x105Nm-2. What
amount of gas (in moles) is present?
Calculate the volume occupied by 1.15mole of an ideal gas at
1.01x105Nm-2 and 20oC.
G.
Π.8
Dalton's law of partial pressure
Besides the above equations, an ideal gas also obey Dalton's law of partial pressure
which stated that the total pressure of a mixture of gases is equal to the sum of the
pressures that each gas would exert if it occupied the space alone.
PT = P a + Pb + Pc + ...
where Pa, Pb, Pc, ... are the pressure of the corresponding gases in the mixture (partial
pressure of the gas).
Consider a mixture of gases, a, b, c, ..., occupying a volume V, and contains na, nb,
nc, ... moles of each gas. Therefore,
total no. of moles of gases nt =
Multiplying both sides by RT/V,
nt
RT
=
V
Therefore, total pressure Pt equal
Pt = Pa + Pb + Pc + ... (proved)
On the other hand, partial pressure of individual gas can be calculated separately.
Pa = na
RT
RT
and Pt = nt
V
V
Therefore,
Pa
=
Pt
Rearrangement gives:
Pa =
=
where χa is called the mole fraction of gas a. (χa =
Classwork:
na
)
nt
1.
5 moles of carbon dioxide at 105Nm-2 are mixed with 15 moles of
hydrogen at the same pressure. If the pressure of the mixture is
1.00x105Nm-2, what is the partial pressure of carbon dioxide?
2.
A mixture of gases at a pressure 1.01x105Nm-2 has the volume
composition of 30% CO, 50% O2, 20% CO2.
a.
What is the volume ratio of the three gases? Hence, deduce the
mole ratio of the gases.
b.
What is the partial pressure of each gas?
3.
4.
Π.9
Into a 10.0dm3 vessel are introduced 4.0dm3 of methane at a pressure
of 2.02x105Nm-2, 12.5dm3 of ethane at a pressure of 3.50x105Nm-2 and
15.0dm3 of propane at a pressure of 1.01x105Nm-2. What is the
pressure of the resulting mixture?
A mixture of 20% NH3, 55% H2 and 25% N2, by volume has a pressure
of 9.80x104Nm-2.
a.
What is the partial pressure of each gas?
b.
What changes will take place in the partial pressures of
hydrogen and nitrogen if the ammonia is removed by the
addition of solid phosphorus(V) oxide?