Introduction
• Ideal gas will obey all the gas laws (Boyle’s Law, Pressure Law, Charles’ Law and Ideal gas
equation) exactly. Real gas at high pressure and low temperature will behave differently.
14.1 Ideal Gas Equation
Boyle’s law
For a fixed mass of gas, pressure is inversely proportional to volume if the temperature is
constant. Boyle’s law can be written as:
p a V or pV = constant or p1 V1 = p2 V2
Pressure law
For a fixed mass of gas, pressure is proportional to the absolute temperature if the volume is
constant. Pressure law can be written as:
p a T or
= constant or
Charles’s law
For a fixed mass of gas, volume is proportional to the absolute temperature if the pressure is
constant. Charles’ law can be written as:
V a T or
= constant or
Ideal gas equation
• For a fixed mass of gas, when all three variables present (pressure, volume and temperature),
we need to use ideal gas equation
PV = nRT or
= constant or
where R = molar gas constant = 8.31 J K-1 mol-1
n = number of moles of the gas
•The number of moles of the gas can also be expressed as:
or
where
m = mass of gas
M = molar mass of gas
N = number of gas molecules
NA = Avogadro number (number of molecules in 1 mole of gas) = 6.02 x 1023 mol-1
•At s.t.p. (standard temperature and pressure) i.e at 273 K and 101 kPa, the volume of one mole
of any gas = 22.4 x 10-3m3.
Relationship between Boltzmann constant and molar gas constant
Boltzmann constant k = gas constant for 1 molecule of gas
Molar gas constant R = gas constant for 1 mole of gas or
= gas constant for NA molecules of gas
Example 1
An ideal gas has been placed in a tank at 30°C. The gas pressure is initially 450 kPa. What will I
1 be the pressure if the temperature is 60°C?
Solution 1
Example 1
A faulty barometer has some air at the top above the mercury. When the length of the air column
is 250 mm, the reading of the mercury above the outside level is 740 mm. When the length of the
air column is decreased to 200 mm, by depressing the barometer tube further into the mercury,
the reading of the mercury above the outside level is 735 mm. Calculate the atmospheric
pressure.
Solution
Example 3
What is the pressure of 8.0 mole of a gas at temperature 77°C occupying a volume of 0.20 m3?
[R = 8.31 J mol-1 ]
Solution
Example 3
A vessel with volume 2.0 litre contains an ideal gaq at temperature 300 K and the pressure of the
gas is 110 kPa.
(a) How many moles of gas are there in the vessel?
(b) How many molecules of gas are there in the vessel? Solution
Example 4
Two gases occupy two containers, A and B. The gas in A, of volume 0.12 m3, exerts a pressure
of 450 kPa. The gas in B, of volume 0.16 m3, exerts a pressure of 150 kPa. The two containers
are united by a tube of negligible volume and the gases are allowed to intermingle. What is the
final pressure in the container if the temperature remains constant?
Solution
14.2 kinetic Theory of Gases
Kinetic theory of matter
•all matter is made up of small particles.
•these particles are in a state of continuous motion.
Brownian movement and diffusion provide the most direct evidence for the kinetic theory.
Kinetic theory of gases
•Particles in gases are relatively far apart and their interactions are consequently simplest.
However certain assumptions have to be made to simplify the mathematical theory.
•Assumptions made
•The gas molecules are in continuous, free and random motion.
• The gas molecules collide with one another and the walls of the containing vessel elastically.
•Forces between molecules are negligible except during collision.
•The volume of the gas molecules is negligible compared with the volume of the gas.
•The time of collision between molecules is negligible compared with the time between
collision.
14.3 Pressure of A Gas
> Consider a gas molecule of mass m moving with velocity c. The components of the velocity c
along x, y and z directions are u, v and w respectively.
> Consider the component of the velocity u in the x-direction.
Change of momentum on striking surface X, DP = mu – (-mu)
= 2mu
> Assuming there is no collision between molecules as the molecule travels between two
opposite surfaces, the time taken for the molecule to collide with the surface X again is
t=
> Hence the rate of change of momentum due to one molecule
=
=
> Therefore the force exerted by the molecule on surface X, Fx =
> The force exerted by N molecules on surface X,
> The pressure exerted by N molecules on surface X, p =
(u12 + u22 + u32 + …… + uN2)
=
=
But <u2> =
Therefore p =
Since N is large and the molecules have random m
then <u2> = <v2> = <w2>
and <c2> = <u2> + <v2> + <w2> = 3<u2>
Therefore <u2> =
Hence p =
<c2>
<c2>
p=
nm<c2> n =
= number of molecules per unit volume
mn = mass of molecules per unit volume = r
(u12 + u22 + u32+ …… + uN2)
p=
<c2>
Example 6
Calculate the rms speed of the atoms in a sample of argon which has a density of 1.6 kg m-3
and a pressure of 100 kPa.
Solution
Root mean square (rms) speed and mean speed
• Root mean square seed,
• Mean speed, <c> =
Example 7
The speed of 10 particles are distributed as follows:
Speed / ms-1
1.0
2.0
3.0
4.0
5.0
No. of particles 1
4
2
1
1
Using the data in the table, find:
(a) The root mean square speed.
(b) The mean speed of the particles. Solution
14.4 Molecular Kinetic Energy
> From p =
r<c2> and r =
where
M = mass of 1 mole of gas
Vm = volume of 1 mole of gas
NA = Avogadro number
m = mass of 1 molecule
6.0
1
=
p=
<c2>
From ideal gas equation: p Vm = RT
<c2> = RT
Hence
(
<c2>) = RT
<c2> =
=
where k = Boltzmann constant
Mean translational kinetic energy of a molecule =
<c2> =
Mean molecular kinetic energy a T where T is the thermodynamic temperature of the gas.
At T = 0 K,
<c2> = 0 for an ideal gas.
14.5 Rms (root mean square) speed of molecules
From
<c2> =
Rms speed of gas molecules =
In terms of the molar mass of the gas, the rms speed of the molecules is given
by:
Example 8
In a mixture of two monoatomic gases P and Q in thermal equilibrium, the molecules of Q I have
twice the mass of those of P. The mean translational kinetic energy of the molecules of Q is 6.2 x
10-21 J.
(a) What is the equilibrium temperature? I
(b) What is the mean translational kinetic energy of the molecules of P? I
(c) What is the ratio of rms speed of P to rms speed of Q?
[k=1.38x10-23J K-1]
Solution I
Example 9
Free neutrons in the core of a fission reactor are sometimes referred to as a `neutron gas’. These
free neutrons may be assumed to behave as molecules of an ideal gas at a temperature of 37°C.
(a) For a free neutron of mass 1.67 x 10-27 kg, calculate:
(i) Its mean kinetic energy.
(ii) Its rms speed.
(b) Determine the temperature of helium gas, assumed to be an ideal gas, whose molecules have
the same rms speed as the free neutrons. The mass of He atom is 4 times the mass of neutron.
Solution
Example 10
At what temperature is the rms speed of gaseous hydrogen molecules (molecular weight = 2) 1 1
equals to that of oxygen molecules (molecular weight = 32) at 27°C? 1
Solution
14.6 A degree of Freedom
> The degree of freedom of a gas molecule is the number of independent ways by which energy
is absorbed by the gas molecule.
translational energy
• A gas molecule of mass m has mean translational kinetic energy. The 3 independent
components of its translational motion gives it three degrees of freedom.
<c2> =
<u2> +
<v2> +
<w2>
Rotational energy
•In rotational motion, there are maximum three degrees of freedom.
I <w2> = Ix <wx2> +
• Monoatomic
Iy <wy2> +
Iz <wz2>
Molecules of a monoatomic gas is regarded as a point. Its moment of inertia is very small.
Therefore rotational kinetic energy is negligible.
• Diatomic
Molecules of a diatomic gas is regarded as a ‘dumb-bell’. There are 2 degrees of freedom of
rotational motion.
• Polyatomic
Molecules of polyatomic gas have 3 degrees of freedom of rotational motion.
Vibrational energy
• The energy of vibration = kinetic energy + potential energy
Vibrational motion has 2 degrees of freedom.
Number of degrees of freedom, f and atomicity of gases
Number of degrees Number of degrees
Number of degrees
Atomicity of of freedom at very
of freedom at room
freedom at very high
gases
low temperature,
temperature,
temperature,
about 50 K
about 300 K
about 1000K
Monoatomic
3 translational
3 translational
3 translational
gas
3 translational
3 translational +
2 rotational +
Diatomic gas 3 translational
2 rotational =
2 vibrational =
5 degrees of freedom
7 degrees of freedom
3 translational +
3 translational +
3 rotational +
Polyatomic
3 translational
3 rotational =
(n x 2) vibrational
gas
6 degrees of freedom where n depends on
the type of molecules
14.7 Maxwell’s Law of Equipartition of Energy
> From kinetic theory for gases,
Mean translational kinetic energy of a molecule =
<c2> =
Or
<c2> =
<u2> +
<v2> +
<w2> =
i.e energy associated with 3 degrees of freedom =
Therefore, energy associated with 1 degree of freedom =
> The law of equipartition of energy states that the energy supplied to a system is distributed
equally among all the effective degrees of freedom and the energy associated with each degree of
freedom is
.
> In general, for a gas with f degrees of freedom:
The mean energy per molecule of the gas = f (
)=
The mean energy for one mole of the gas = (NA)(
)=
The mean energy for n mole of the gas = n
14.8 Internal Energy of An Ideal Gas
> For real gas, the internal energy of the gas is the total kinetic energy and potential energy of the
molecules of the gas.
> For an ideal gas, there is no intermolecular forces between the molecules of the gas, the
potential energy of the molecules = 0.
> So, the internal energy of an ideal gas = kinetic energy of the molecules of the gas. > The mean
internal energy of an ideal gas,
U = total kinetic energies of the molecules in the gas
= (No. of molecules) x (No. of degrees of freedom per molecule) x
=Nf
> For 1 mole of gas, U =
> For n moles of gas, U-= n
Example 11
Assuming oxygen behaves as an ideal gas,
calculate:
(a) The root-mean-square speed of its molecules.
(b) The average kinetic energy of a molecule at 273 K.
[1 mole of oxygen has a mass of 32 g]
Solution
Example 12
The total translational kinetic energy for a monoatomic gas at a certain temperature is KT. What
is the total translational kinetic energy for a diatomic and polyatomic gas at the same
temperature? 1
Solution:
14.9 distribution of Molecular Speeds
Maxwellian distribution
Figure above
•The graph is not symmetrical.
•The lowest possible speed is zero and the highest is infinity.
•The speed for which the number of molecules is a maximum is called the most probable speed,
vp,
•The mean speed is
.
• Vrms >
> vp
•The area under the graph = total number of molecules in the sample.
Distribution of molecular speeds at different temperatures
• When the temperature of the gas increases, the speeds of its molecules increases. The speed
distribution graph will thus be shifted towards higher speeds.
• Since the total number of molecules in the gas remains constant, the area under the graph
remains the same.
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