Chapter 19
The Kinetic Theory of
Gases
Chapter-19 The Kinetic Theory
of Gases
Topics to be covered:
Avogadro number
Ideal gas law
Internal energy of an ideal gas
Distribution of speeds among the atoms in
a gas
Specific heat under constant volume
Specific heat under constant pressure.
Adiabatic expansion of an ideal gas
Kinetic theory of gases
It relates
the macroscopic property of
gases
(pressure - temperature volume - internal energy)
to
the microscopic property - the
motion of atoms or molecules
(speed)
http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm
What is one mole?
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Avogadro’s Number NA = 6.02 X 1023
One mole of any element contains Avogadro’s number of atoms
of that element.
One mole of iron contains 6.02 X 1023 iron atoms.
One mole of water contains 6.02 X 1023 water molecules.
From experiments:
12 g of carbon contains 6.02 X 1023 carbon atoms.
Thus, 1 mole carbon = 12 g of carbon.
4 g of helium contains 6.02 X 1023 helium atoms.
Thus, 1 mole helium = 4 g of helium.
Avogadro’s Number NA = 6.02 X 1023 per mole = 6.02 X 1023
mol-1
Avogadro's Number
Formula - number of moles
n = N /NA
n = Msample /M
n = number of moles
N = number of molecules
NA = Avogadro number
M = Molar mass of a substance
Msample = mass of a sample
Ideal Gas Law
At low enough densities, all gases tend to obey the
ideal gas law.
 Ideal gas law
p V=n R T
where R= 8.31 J/mol.K (ideal gas constant), and T
temperature in Kelvin!!!
 p V= n R T = N k T; N is the number of molecules
and K is Boltzman constant k = R/NA
Ideal Gas Law
Isothermal process
isotherm
Isothermal expansion
(Reverse is isothermal
Compression)
Quasi-static equilibrium
(p,V,T are well defined)
p =n R T/V
= constant/V
Checkpoint 1
Work done at constant
temperature
The work W done by the ideal gas is given by the equation
W 
Vf
 pdV .
From the ideal gas law we have that
Vi
V
V
f
f
nRT
nRT
dV
p
W  
dV  nRT 
 nRT  lnV
V
V
V
Vi
Vi
W = n R T Ln(Vf/Vi)
V
Vf
i
;
Work done at constant
pressure

isobaric process
Consider process i  f . During this process
the pressure is kept constant at p and the volume
changes from Vi to V f .
Vf
The work W done by the gas is W 
 pdV  p  dV  p V
Vi
W = p (Vf-Vi)
Vf
Vi
f
 Vi .
Work done at constant
volume

isochoric process
Consider process i  f .
During this process the volume of the ideal gas is kept constant.
Thus the work W done by the gas is W   pdV  0.
W=0
Root Mean Square (RMS)
speed vrms

For 4 atoms having speeds v1, v2, v3 and v4
v rms
v v 2 v 3 v 4

4
2
1
2
2
2
Vrms is a kind of average speed
http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm
Pressure, Temperature, RMS
Speed
The pressure p of the gas is
related to root-mean -square
speed vrms, volume V and
temperature T of the gas
p=(nM vrms 2)/3V
Equation 19-21 in the textbook
Vrms =  (3RT)/M
but pV/n = RT
Continue…
Vrms =  (3RT)/M
R is the ideal gas constant
T is temperature in Kelvin
M is the molar mass (mass of one mole of the gas)
At room temperature (300K)
Gas
Hydrogen
Nitrogen
Oxygen
Molar Mass
(g/mol)
2
28
32
Vrns
(m/s)
1920
517
483
Translational Kinetic
Energy K
Average translational kinetic energy of one molecule
Kavg=(mv2/2)avg=m(vrms2)/2
Kavg=m(vrms2)/2=(m/2)[3RT/M]
=(3/2)(m/M)RT=(3/2)(R/NA)T
Kavg=(3/2)(R/NA)T=(3/2)kT
Continue
3kT
K

avg
2
At a given temperature, all ideal gas
molecules – no matter what their
masses – have the same average
translational kinetic energy.
Checkpoint 2
A gas mixture consists of molecules of
type 1, 2, and 3, with molecular
masses m1>m2>m3.
Rank the three types according to
average kinetic energy, and rms
speed, greatest first.
The Molar Specific Heat of an Ideal
Gas
Internal energy of an ideal gas Eint
For a monatomic gas (which has individual atoms
rather than molecules), the internal energy Eint is the
sum of the translational kinetic energies of the atoms.
Eint = N Kavg= N (3/2) k T = 3/2 (N k T)
= 3/2 (n R T)
Eint = 3/2 n R T
The internal energy Eint of a confined ideal gas is a function of
the gas temperature only, it does not depend on any other
variable.
Change in internal energy
Eint = 3/2 n R T, DEint = 3/2 n R DT
The Molar Specific Heat of an Ideal
Gas
Heat Q1
Heat Q2
Eventhough Ti and Tf is the same for both processes, but Q1 and
Q2 are Different because heat depends on the path!
Heat gained or lost at
constant volume
For an ideal gas process at constant
volume pi,Ti increases to pf,Tf and heat
absorbed
Q = n cv DT and W=0. Then
DEint = (3/2)n R DT = Q = n cv DT
cv = 3R/2
Q = n cV DT
where cv is molar specific heat at constant volume
Heat gained or lost at constant pressure
For an ideal gas process at constant
pressure Vi,Ti increases to Vf,Tf and heat
absorbed
Q = n cp DT and W=PDV. Then
DEint = (3/2)n R DT = Q  PDV = Q  n R DT
Q = (3/2nR+nR) DT = 5/2 n R DT
Q = n cp DT
cp = 5/2 R

where cp is molar specific heat at constant pressure
The Molar Specific Heats
of a Monatomic Ideal Gas
Cp = CV + R;
specific heat ration = Cp/ CV
For monatomic gas Cp= 5R/2, CV= 3R/2
and = Cp/ CV = 5/3
(specific heat ratio)
Checkpoint 3
The Molar Specific Heat of
an Ideal Gas
monatomic
diatomic
polyatomic
Internal energy of monatomic,
diatomic, and polyatomic gases
(theoretical values)
Monatomic
gas
Diatomic gas
Polyatomic
gas
Degrees of
freedom
Cv
Eint=n CVTCp=Cv+R (translational
+ rotational)
(3/2) R (3/2) nRT
5/2 R
3
= 12.5
(5/2) R
= 20.8
(5/2) nRT
7/2 R
5
(6/2) R
= 24.9
(6/2) nRT
8/2 R
6
Eint=n CV T
Cv of common gases in joules/mole/deg.C
(at 15 C and 1 atm.)
Gas
Symbol
Cv


(experiment)
(experiment)
(theory)
Helium
He
12.5
1.666
1.666
Argon
Ar
12.5
1.666
1.666
Nitrogen
N2
20.6
1.405
1.407
Oxygen
O2
21.1
1.396
1.397
Carbon Dioxide
CO2
28.2
1.302
1.298
Adiabatic Expansion for an
Ideal Gas
In adiabatic processes, no heat transferred to
the system Q=0
Either system is well insulated, or process
occurs so rapidly
In this case
DEint = - W
Adiabatic Process
P, V and T are related to the initial
and final states with the following
relations:


PiVi = PfVf
TiVi
/( -1)

-1
= TfVf
-1
Also T
V =constant then
(-1)/
(-1)/
piTi
= pfTf
Free Expansion of an Ideal Gas
An ideal gas expands in an
adiabatic process such that no
work is done on or by the gas
and no change in the internal
energy of the system i.e. Ti=Tf
Also in this adiabatic process
since
( pV=nRT),
piVi=pfVf ( not


PiVi = PfVf )
END