Chapter 19 The Kinetic Theory of Gases. Chapter-19 The Kinetic Theory of Gases Topics to be...
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Transcript of Chapter 19 The Kinetic Theory of Gases. Chapter-19 The Kinetic Theory of Gases Topics to be...
![Page 1: 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.](https://reader033.fdocuments.net/reader033/viewer/2022052317/56649c745503460f94927296/html5/thumbnails/1.jpg)
Chapter 19
The Kinetic Theory of Gases
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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
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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
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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
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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
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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
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Ideal Gas Law
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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
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Checkpoint 1
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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)
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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)
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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
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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
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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
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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
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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
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Continue
At a given temperature, all ideal gas molecules – no matter what their masses – have the same average translational kinetic energy.
3avg 2
kTK
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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.
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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.
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Change in internal energy
Eint = 3/2 n R T, DEint = 3/2 n R DT
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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!
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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
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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
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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)
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Checkpoint 3
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The Molar Specific Heat of an Ideal Gas
monatomic diatomic
polyatomic
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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)
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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
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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
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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)/
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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
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