1. The Kinetic Theory of Gases
Chapter 19
The Kinetic Theory of Gases

2. 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

3. 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)

4. What is one mole? 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

5. Avogadro's Number Formula - number of moles
n = N /NA
n = number of moles
N = number of molecules
NA = Avogadro number
M = Molar mass of a substance
Msample = mass of a sample
n = Msample /M

6. 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

7. Ideal Gas Law

8. Isothermal process p =n R T/V = constant/V isotherm
Isothermal expansion
(Reverse is isothermal
Compression)
Quasi-static equilibrium
(p,V,T are well defined)
p =n R T/V
= constant/V

9. Checkpoint 1

10. Work done at constant temperature
W = n R T Ln(Vf/Vi)

11. Work done at constant pressure
isobaric process
W = p (Vf-Vi)

12. Work done at constant volume
isochoric process
W = 0

13. Root Mean Square (RMS) speed vrms
For 4 atoms having speeds v1, v2, v3 and v4
Vrms is a kind of average speed

14. 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
but pV/n = RT
Equation in the textbook
Vrms =  (3RT)/M

15. At room temperature (300K)
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
Molar Mass
(g/mol)
Vrns
(m/s)
Hydrogen 2
1920
Nitrogen 28
517
Oxygen 32
483

16. 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

17. Continue
At a given temperature, all ideal gas molecules – no matter what their masses – have the same average translational kinetic energy.

18. 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.

19. 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.

20. Change in internal energy
Eint = 3/2 n R T, DEint = 3/2 n R DT

21. The Molar Specific Heat of an Ideal Gas
Heat Q2
Heat Q1
Eventhough Ti and Tf is the same for both processes, but Q1 and Q2 are Different because heat depends on the path!

22. 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 T and W=0. Then
Eint = (3/2)n R T = Q = n cv T
cv = 3R/2
Q = n cV T
where cv is molar specific heat at constant volume

23. Eint = (3/2)n R T = Q - PDV = Q - n R DT
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 T and W=PDV. Then
Eint = (3/2)n R T = Q - PDV = Q - n R DT
Q = (3/2nR+nR) DT = 5/2 n R DT
cp = 5/2 R
Q = n cp T
where cp is molar specific heat at constant pressure

24. 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)

25. Checkpoint 3

26. The Molar Specific Heat of an Ideal Gas
monatomic
diatomic
polyatomic

27. (translational + rotational)
Internal energy of monatomic, diatomic, and polyatomic gases (theoretical values)
Degrees of freedom
(translational + rotational) Cv
Eint=n CVT
Cp=Cv+R
Monatomic gas
(3/2) R
= 12.5
(3/2) nRT
5/2 R 3
Diatomic gas
(5/2) R
= 20.8
(5/2) nRT
7/2 R 5
Polyatomic gas
(6/2) R
= 24.9
(6/2) nRT
8/2 R 6
Eint=n CV T

28. Cv of common gases in joules/mole/deg.C (at 15 C and 1 atm.)
Symbol
     Cv
   g 
g     
(experiment)
(theory)
Helium He
12.5
1.666
Argon Ar
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

29. 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

30. Adiabatic Process
P, V and T are related to the initial and final states with the following relations:
PiVi= PfVf
TiVi-1 = TfVf-1
Also T/( -1) V =constant then
piTi(-1)/ = pfTf(-1)/

31. 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)

33. END