The kinetic theory of gases- physical chemistry

The kinetic theory of gases, physical chemistry

1. Chapter 19: The Kinetic Theory of Gases
Avogadro’s number
N
A
= 6.02 x 10
23
/mole
Number of molesin a sample
:
n = N/N
A
; N = number of atoms or molecules
Universal Gas Constant
R = 8.31 J/mol
.
K
Boltzmannconstant
k = R/N
A
= 1.38 x 10
−23
J/K

2. Ideal Gases
At lowgas densities, allgases can be treated as ideal
gases. Ideal gases obey the relation:
p: absolute(not gauge
) pressure.
V: volumeof the gas
n: number of molesof gas present.
T: the temperaturein Kelvin
. It *MUST*be in Kelvin!
R: gas constant(same for all gases) R = 8.31 J/mol
.
K
nR
T
Vp
T
Vp
f
f f
i
ii
= =
IFn is constant, thenpV= nRT (ideal gas law)

3. Ideal Gases
pV= nRT
Three variables in the ideal gas law (4 if you count n --but
let n be constant for now).
Pressure:
Volume: Temperature:
Consider special cases
Isobaric--constant pressure
Isochoric (or isovolumic) --constant volume
Isothermal--constant temperature

4. Ideal Gases
pV= nRT
Three variables in the ideal gas
law (with n being constant).
Pressure: Consider special cases
Isobaric --constant pressure
()
Vp V Vp dV p dVp W
i f
V
V
V
V
f
i
f
i
∆= − = = =
∫ ∫
Volume
Pressure
i
fpV= const

5. Volume
Pressure
Ideal Gases
pV= nRT
Three variables in the ideal gas
law (with n being constant).
Volume:Consider special cases
Isochoric --constant volume
∫ ∫
= = =
i
i
f
i
V
V
V
V
0 dVp dVp W
i
f
since the integral limtsare equal

6. Volume
Pressure
Ideal Gases
pV= nRT
Three variables in the ideal gas
law (with n being constant).
Temperature: Consider special cases
Isothermal --constant temperature
i
f V
V
V
V
V
V
V
V
V
V
ln nRT V]ln nRT[
V
dV
nRT
dV
V
nRT
pdV W
f
i
f
i
f
i
f
i
= = =
= =
∫
∫ ∫
Gas expands from V
i
to V
f, p = nRT/V
i
f

7. i
f V
V
V
V
V
V
V
V
V
V
ln nRT ]V[ln nRT
V
dV
nRT dV
V
nRT
pdV W
f
i
f
i
f
i
f
i
= = = = =
∫ ∫ ∫
Work done by ideal gas at constant temperature
Work done at constant volume
dV= 0, so W = 0
Work done at constant pressure
p is constant, W = p (V
f
–V
i) = p ∆V
Summary of Work
∫
=pdV W

8. Molar specific heat of an ideal gas
Molar specific heat:
Q = c n ( T
f
–T
i
)
1) Constant-volume process
2) Constant-pressure process 3) Arbitrary process
The specific heat cis 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.

9. Molar specific heat of an ideal gas
Molar specific heat at constant volume: C
V
Q = n C
V
∆T(constantV process)
since V = constant, W = 0,
thus ∆E
int
= Q –W = n C
V
∆T or E
int
= n C
V
T
For ideal gas, the change in
internal energy depends only
on the change in gas
temperature.

10. Molar specific heat at constant pressure: C
p
Q = n C
p
∆T (constant Pressure process)
∆E
int
= Q –W,
since p = constant, W = p ∆V= nR∆T
∆E
int
= n C
V
∆T = n C
p
∆T–nR∆T= n(C
p
–R)∆T
ThereforeC
V
= C
p
–R
or
C
P
= C
V
–R

11. Degrees of Freedom
Particles can absorb energy depending on their structure.
A degree of freedom is the way the particle can move.
All particlescan move in x-, y-, and z-directions.
diatomic molecules(N
2
, O
2
, etc.) have two rotational
axes.
3 degrees of freedom
2 more degrees of freedom
polyatomic molecules
(CH
4
, H
2
O, etc)
have three rotational
axes.
3 more degrees of freedom

12. Degrees of Freedom
Particles can absorb energy depending on their structure.
A degree of freedom is the way the particle can move.
Each degree of freedom contributes 1/2 to the C
V
.
Molecule Translational Rotational Total
monotomic 3 0 3
diatomic325
polyatomic 3 3 6
T nR
2
3
E
int
∆ = ∆
monatomic
T nR
2
5
E
int
∆ = ∆
diatomic
T nR3 E
int
∆ = ∆
polyatomic
C
V
3/2
5/2
6/2 = 3

13. Degrees of Freedom
Particles can absorb energy depending on their structure.
A degree of freedom is the way the particle can move.
T nR
2
3
E
int
∆ = ∆
monotomic
T nR
2
5
E
int
∆ = ∆
diatomic
T nR3 E
int
∆ = ∆
polyatomic
Remember Quantum physics corrects
this for low temperatures!

14. Adiabatic expansion of an ideal gas .
For an adiabatic process, Q = 0.
pV
γ
= a constant
γ= C
p
/C
V
treat γas a constant that
depends on the type of the gas molecules.
Free expansions
Q = W = 0,
So ∆E
int
= 0,
Therefore T
i
= T
f
or P
iV
i
= P
fV
f

15. Isobaric---constant pressure process
Q = n C
p
∆T, W = p ∆V
Isothermal---constant temperature process
∆E
int
= 0, Q = W = nRTln(V
f/V
i)
Isochoric process---constant volume process
Q = ∆E
int
= n C
V
∆T, W = 0
∆E
int
= Q −W
for all processes.
Adiabatic expansionof an ideal gas
Q = 0, pV
γ
= a constant
Free expansion
Q = W = 0 => ∆E
int
= 0, => T
i
= T
f
=> p
iV
i
= p
fV
f

16. A Quiz
Can one mole of an ideal gas at a
temperature of 300K occupy a fixed
volume of 10
−2
m
3
(ten liters) at one
atmosphere (1.01x10
5
N/m
2
) of pressure?
Remember: R = 8.31 J/mol
.
K
1) yes.
2) No. The gas needs to be heated up to occupy that volume
with that pressure.
3) No. The gas needs to be cooled down to occupy that volume
with that pressure.
4) Depends on the type of ideal gas molecules.

17. A Quiz
1) yes.
2) No. The gas needs to be heated up to occupy that volume
with that pressure.
3) No. The gas needs to be cooled down to occupy that volume
with that pressure.
4) Depends on the type of ideal gas molecules.
pV= nRT
pV= (1.01x10
5
N/m
2
)(10
−
2
m
3
)
= 1010 Nm = 1,010 J
nRT= (1.0mol)(8.31 J/mol
.
K)(300K)
= 2,493 J
T= pV/ nR
= 1010J/(1.0)(8.31) = 122 K
Can one mole of an ideal gas at a temperature of 300K occupy a
fixed volume of 10
−2
m
3
(ten liters) at one atmosphere
(1.01x10
5
N/m
2
) of pressure? Remember: R = 8.31 J/mol
.
K

18. The Microscopic World
What causes pressure?

19. Chapter 19: the Microscopic World
All macroscopic(i.e., human scale)
quantities must ultimately be explained on
the microscopicscale.

20. Pressure
Definition of pressure:
Pressure=
Force
Area
But now what is force?
Relate forceto impulseand
change in momentum.
dtF pd
dt
pd
F
v
v
v
v
= ⇒ ≡ ()()
x x x i f x
mv2 mv mv p p p−= − −= − = ∆

21. L
mv
v/L2
mv2
t
p
2
x
x
x x
= =
∆
∆
Take ∆tto be the time that molecule takes in hitting
the wall, bouncing off and then hitting it again. It
will have traveled 2L in that time with speed v
x
.
Area of wall: L
2
Thus the pressure will be:
()
2
xN
2
2x
2
1x 3
2
2
xN
2
2x
2
1x
2
x
v v v
L
m
L
L/ mv L/ mv L/ mv
L
F
p
+ + + 





=
+ + +
= =
K
K

22. Since there are N molecules in the box, N = nN
A
and N is usually a *very* big number, we can use
the average speed instead of the actual speeds.
() ()
avg
2
x 3
2
Nx
2
2x
2
1x 3
2
2
Nx
2
2x
2
1x
2
x
v
L
m
v v v
L
m
L
L/ mv L/ mv L/ mv
L
F
p






=
+ + + 





=
+ + +
= =
K
K
The volume of the box is L
3
, so
()
avg
2
x
A
v
V
nNm
p=

23. v
rms
(v
2
)
avg
is the average of the squared speed --which
makes the speed the (square) rootof the mean(average)
squaredspeed --i.e., root-mean-squaredspeed, v
rms
.
()
A
rms
2
rms
A
mNn
pV3
v v
V3
nNm
p= ⇒ =
Rearrange
(v
2
)
avg
= v
rms
Since there are 3 dimensions, v
x
2
+ v
y
2
+ v
z
2
= v
2
and
each dimension is the same, v
x
2
= v
y
2
= v
z
2
=> v
2
= 3v
x
2
()
()
avg
2 A
avg
2 A
avg
2
x
A
v
V3
nNm
v
3
1
V
nNm
v
V
nNm
p= 





= =

24. gas temperature
mN
A
is the molar mass M of the gas and using the ideal
gas law pV= nRT:
M
RT3
Mn
nRT3
mNn
pV3
v
A
rms
= = =
Thus, the characteristic speed of the gas molecules is
related to the temperature of the gas!
Gas v
rms
(m/s)
Hydrogen 1920
Helium 1370
N
2
517
escape speed
Earth
1120

25. Maxwell-BoltzmannDistribution
Define a function P(v) whose area (integral) equals 1.
RT2/ Mv 2
2/3
2
ev
RT 2
M
4)v(P
−






π
π=
P(v) is the Maxwell-BoltzmannProbability function

26. Maxwell-BoltzmannDistribution
RT2/ Mv 2
2/3
2
ev
RT 2
M
4)v(P
−






π
π=
Math:
∫
=
space all
avg
dx)x(fx x

27. Maxwell-BoltzmannDistribution
RT2/ Mv 2
2/3
2
ev
RT 2
M
4)v(P
−






π
π=
Thus,
M
RT8
dv)v(Pv v
0
avg
π
= =
∫∞
and
M
RT3
dv)v(Pv ) v( v
0
2 2
rms
2
avg
= = ≡
∫∞

28. Kinetic Energies
Average kinetic energy:
()
()
()
kT T
N
R
2
3
N2
RT3
m/M
RT3
M
RT3
m
mv vm K
2
3
A A
2
1
2
1
2
rms 2
1
avg
2
2
1
avg
=














= = = =
= =
kT vm K
2
3 2
rms 2
1
avg
= =
Kinetic energy only depends on the gas’s temperature!
The “3” comes from the three dimensions: x, y, and z!

29. Brownian Motion
But.... not all the molecules are going at that speed.
Some are going fasterand some sloooowerthan v
rms
.
A gas molecule interacting
with neighboring molecules
is like you trying to get to the
concession stand during a
rock concert!
p.s.: Einstein, 1905

30. Mean Free path
Define the mean-free-path λas the distance between
collisions. Molecular sizes
V/Nd2
1
2
π
=λ