Chapter 19 The Kinetic Theory of Gases
Avogadro’s number
NA = 6.02 x 1023 mol-1
Number of moles in a sample:
n = N/NA
Universal Gas Constant
R = 8.31 J/mol . K
Boltzmann constant
k = R/NA = 1.38 x 10−23 J/K
Ideal Gases
At low gas densities, all gases can be treated as
ideal gases. Ideal gases obey the relation:
pV = nRT
IF n is constant, then
(ideal gas law)
p i Vi p f Vf
=
= nR
Ti
Tf
p: absolute (not gauge) pressure.
V: volume of the gas
n: number of moles of gas present.
T: the temperature in Kelvin.
R: gas constant ( same for all gases) R = 8.31 J/mol . K
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Ideal Gases
pV = nRT
Three variables in the ideal gas law (4 if you count n -but let n be constant for now).
Consider special cases
Pressure:
Isobaric -- constant pressure
Volume:
Isochoric (or isovolumic) -- constant volume
Temperature:
Isothermal -- constant temperature
Ideal Gases
Three variables in the ideal gas
law (with n being constant).
Consider special cases
Pressure
pV = nRT
i
f
Pressure:
Isobaric -- constant pressure
Volume
W = ∫ p dV = p ∫ dV = p (Vf − Vi ) = p∆V
Vf
Vf
Vi
Vi
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Ideal Gases
Three variables in the ideal gas
law (with n being constant).
Consider special cases
Pressure
pV = nRT
f
i
Volume:
Isochoric -- constant volume
Vf
Vi
Vi
Vi
Volume
W = ∫ p dV = ∫ p dV = 0 since the integral limts are equal
Ideal Gases
Three variables in the ideal gas
law (with n being constant).
Consider special cases
Pressure
pV = nRT
i
f
Temperature:
Isothermal -- constant temperature
Gas expands from Vi to Vf,
Vf
Vf
Vi
Vi
W = ∫ pdV = ∫
= nRT ∫
Vf
Vi
p = nRT/V
Volume
nRT
dV
V
dV
V
= nRT[ ln V]VVif = nRT ln f
V
Vi
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Summary of work
Work done at constant pressure
p is constant, W = p (Vf – Vi) = p ∆V
Work done at constant volume
dV = 0, so W = 0
Work done by ideal gas at constant temperature
Vf
Vf
Vi
Vi
W = ∫ pdV = ∫
Vf dV
V
nRT
=nRT[ln V]VVif = nRT ln f
dV =nRT ∫
Vi V
V
Vi
Molar specific heat of an ideal gas
Molar specific heat:
Q = c n ( Tf – Ti )
The specific heat c is a value that depends
on the ability of a substance to absorb
energy. As such, c depends on both the
type of material and whether the process
is a constant volume process or a constant
pressure process.
1) Constant-volume process
2) Constant-pressure process
3) Arbitrary process
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Molar specific heat of an ideal gas
Molar specific heat at constant volume: CV
Q = n CV ∆T(constant V process) since V = constant, so
W = 0, so ∆Eint = Q – W = n CV ∆T or Eint = n CV T
For ideal gas, the change in
internal energy depends only
on the change in gas
temperature.
Molar specific heat at constant pressure: Cp
Q = n Cp∆T
(constant Pressure process)
∆Eint = Q – W,
since p = constant, W = p∆V = nR∆T
∆Eint = n CV ∆T = n Cp∆T – nR∆T = n(Cp – R)∆T
Therefore
CV = Cp – R
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Degrees of Freedom
Particles can absorb energy depending on their structure.
A degree of freedom is the way the particle can move.
All particles can move in x-, y-, and z-directions.
3 degrees of freedom
diatomic molecules (N2, O2, etc.) have two rotatational
energies.
2 more degrees of freedom
polyatomic molecules (CH4, SiO2 etc.) have three rotational
energies.
3 more degrees of freedom
Degrees of Freedom
Particles can absorb energy depending on their structure.
A degree of freedom is the way the particle can move.
Molecule Translational Rotational
monotomic
3
0
diatomic
3
2
polyatomic
3
3
Total
3
5
6
CV
3/2
5/2
6/2 = 3
Each degree of freedom contributes 1/2 to the CV.
3
2
monotomic ∆E int = nR∆T
5
2
diatomic ∆E int = nR∆T
polyatomic ∆E int = 3nR∆T
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Degrees of Freedom
Particles can absorb energy depending on their structure.
A degree of freedom is the way the particle can move.
3
2
monotomic ∆E int = nR∆T
5
2
diatomic ∆E int = nR∆T
polyatomic ∆E int = 3nR∆T
Remember Quantum physics corrects
this for low temperatures!
Adiabatic expansion of an ideal gas.
For an adiabatic process, Q = 0.
pVγ = a constant
γ = Cp/CV treat γ as a constant that
depends on the type of the gas molecules.
Free expansions
Q = W = 0,
So ∆Eint = 0,
Therefore Ti = Tf
or PiVi = PfVf
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Isobaric --- constant pressure process
Q = n Cp∆T,
W = p ∆V
Isothermal --- constant temperature process
∆Eint = 0,
Q = W = nRT ln(Vf/Vi)
Isochoric process --- constant volume process
Q = ∆Eint = n CV ∆T,
W=0
Adiabatic expansion of an ideal gas.
Q = 0, pVγ = a constant
Free expansion
∆Eint = Q − W
for all processes.
Q = W = 0 => ∆Eint = 0, => Ti = Tf => piVi = pfVf
Daily Quiz, November 22, 2004
The figure here shows five paths
traversed by a gas on a p-V diagram.
Which path has the greatest change in
internal energy.
1, 2, 3, 4, 5,
or 0) all equal
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Daily Quiz, November 22, 2004
∆Eint = Q – W = n CV ∆T
Path 5 has the greatest ∆T
The figure here shows five paths
traversed by a gas on a p-V diagram.
Which path has the greatest change in
internal energy.
1, 2, 3, 4, 5,
or 0) all equal
Tomorrow, the Microscopic World
What causes pressure?
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