Chapter 19 The Kinetic Theory of Gases. Chapter-19 The Kinetic Theory of Gases Topics to be...

Chapter 19

The Kinetic Theory of Gases

Chapter-19 The Kinetic Theory of Gases

Topics to be covered:Avogadro numberIdeal gas law Internal energy of an ideal gasDistribution of speeds among the

atoms in a gasSpecific heat under constant volumeSpecific heat under constant pressure.Adiabatic expansion of an ideal gas

Kinetic theory of gases

It relatesthe 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? 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 NumberFormula - number of moles

n = N /NAn = number of moles

N = number of molecules

NA = Avogadro number

M = Molar mass of a substance

Msample = mass of a sample

n = Msample /M

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

Isothermal expansion

(Reverse is isothermal

Compression)

isotherm

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 done by the ideal gas is given by the equation

. From the ideal gas law we have that

ln ;

f

i

f f

f

i

i i

V

V

V VV

VV V

W

W pdV

nRT nRT dVp W dV nRT nRT V

V V V

W = n R T Ln(Vf/Vi)

Work done at constant pressure isobaric

process

Consider process . During this process

the pressure is kept constant at and the volume

changes from to .

The work done by the gas is .f f

i i

i f

V V

f i

V V

i f

p

V V

W W pdV p dV p V V

W = p (Vf-Vi)

Work done at constant volume isochoric

processConsider process .

During this process the volume of the ideal gas is kept constant.

Thus the work done by the gas is 0.

i f

W W pdV

W = 0

Root Mean Square (RMS) speed vrms

For 4 atoms having speeds v1, v2, v3

and v4

2 2 2 21 2 3 4

4rms

v v v vv

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)

GasMolar Mass (g/mol)

Vrns

(m/s)

Hydrogen 2 1920

Nitrogen 28 517

Oxygen 32 483

Kavg=(3/2)(R/NA)T=(3/2)kT

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

Continue

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

3avg 2

kTK

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

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.

Internal energy of an ideal gas Eint

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!

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

Heat gained or lost at constant volume

where cv is molar specific heat at constant volume

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 RQ = n cp T

Heat gained or lost at constant pressure

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)

(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

(6/2) R

= 24.9

(6/2) nRT 8/2 R 6

Eint=n CV T

Monatomic gas

Polyatomic gas

Cv Eint=n CVT Cp=Cv+R

Degrees of freedom

(translational +

rotational)

Cv of common gases in joules/mole/deg.C (at 15 C and 1 atm.)

Gas Symbol      Cv   g g    

    (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

DEint = - W

In this case

Adiabatic ProcessP, 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)/

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

)

Free Expansion of an Ideal Gas

END