Download Lecture 9

Thermodynamics
Boltzmann (Gibbs) Distribution
Maxwell-Boltzmann Distribution
Second Law
Entropy
Lana Sheridan
De Anza College
May 8, 2017
Last time
• modeling an ideal gas at the microscopic level
• pressure, temperature, and internal energy from microscopic
model
• equipartition of energy
• rms speed of molecules
• heat capacities for ideal gases
• adiabatic processes
Overview
• the Boltzmann distribution (distribution of energies)
• the Maxwell-Boltzmann distribution (distribution of speeds)
• the Second Law of thermodynamics
• entropy
Adiabatic Process in Ideal Gases
For an adiabatic process (Q = 0):
PV γ = const.
and:
TV γ−1 = const.
Adiabatic Process in Ideal Gases
For an adiabatic process (Q = 0):
PV γ = const.
and:
TV γ−1 = const.
(Given the first one is true, the second follows immediately from
the ideal gas equation, P = nRT
V .)
Example
Based on problem 28, Chapter 21.
How much work is required to compress 5.00 mol of air at 20.0◦ C
and 1.00 atm to one-tenth of the original volume in an adiabatic
process? Assume air behaves as an ideal diatomic-type gas.
1
Serway & Jewett, page 647.
Example
Based on problem 28, Chapter 21.
How much work is required to compress 5.00 mol of air at 20.0◦ C
and 1.00 atm to one-tenth of the original volume in an adiabatic
process? Assume air behaves as an ideal diatomic-type gas.
Another way:
One way:
Ti Viγ−1 = Tf Vfγ−1
PV γ = Pi Viγ
Z
and
W = − P dV
1
Serway & Jewett, page 647.
and
0
7 = nCV ∆T
W = ∆Eint − Q
Example
Based on problem 28, Chapter 21.
How much work is required to compress 5.00 mol of air at 20.0◦ C
and 1.00 atm to one-tenth of the original volume in an adiabatic
process? Assume air behaves as an ideal diatomic-type gas.
Another way:
One way:
Ti Viγ−1 = Tf Vfγ−1
PV γ = Pi Viγ
and
Z
and
0
7 = nCV ∆T
W = ∆Eint − Q
W = − P dV
W = 46.0 kJ
1
Serway & Jewett, page 647.
Weather and Adiabatic Process in a Gas
On the eastern side of the Rocky Mountains there is a
phenomenon called chinooks.
These eastward moving wind patterns cause distinctive cloud
patterns (chinook arches) and sudden increases in temperature.
Weather and Adiabatic Process in a Gas
As the air rises from the ocean it expands in the lower pressure at
altitude and cools. The water vapor condenses out of the air and
falls as precipitation.
As the air passes over the mountain it absorbs the latent heat from
the water condensation, then it stops cooling. As it descends, it is
compressed (nearly) adiabatically as the ambient pressure
increases. The air temperature rises!
Temperature and the Distribution of Particles’
Energies
In a gas at temperature T , we know the average translational KE
of the molecules.
However, not all of the molecules have the same energy, that’s just
the average.
How is the total energy of the gas distributed amongst the
molecules?
Temperature and the Distribution of Particles’
Energies
Ludwig Boltzmann first found the distribution of the number of
particles at a given energy given a thermodynamic system at a
fixed temperature.
Assuming that energy takes continuous values we can say that the
number of molecules per unit volume with energies in the range E
to E + dE is:
Z E +dE
N[E ,E +dE] =
nV (E ) dE
E
Where
nV (E ) = n0 e −E /kB T
and n0 is a constant setting the scale: when E = 0, nV (E ) = n0 .
The Boltzmann Distribution
This particular frequency distribution:
nV (E ) ∝ e −E /kB T
is called the Boltzmann distribution or sometimes the Gibbs
distribution (after Josiah Willard Gibbs, who studied the behavior
of this distribution in-depth).
This distribution is even easier to understand for discrete energy
levels.
The Boltzmann Distribution
This particular frequency distribution:
nV (E ) ∝ e −E /kB T
is called the Boltzmann distribution or sometimes the Gibbs
distribution (after Josiah Willard Gibbs, who studied the behavior
of this distribution in-depth).
This distribution is even easier to understand for discrete energy
levels.
The probability for a given particle to be found in a state with
energy Ei drawn from a sample at temperature T :
p(Ei ) =
1 −Ei /kB T
e
Z
where Z is simply a normalization constant to allow the total
probability to be 1. (The partition function.)
The Boltzmann Distribution
p(Ei ) =
1 −Ei /kB T
e
Z
If we know the energies of two states E1 and E2 , E2 > E1 , we can
find the ratio of the number of particles in each:
nV (E2 )
= e −(E2 −E1 )/kB T
nV (E1 )
States with lower energies have more particles occupying them.
The Boltzmann Distribution
Lower temperature
1
Higher temperature
Figure from the website of Dr. Joseph N. Grima, University of Malta.
Aside: Lasers
Lasers emit coherent light. One photon interacts with an atom and
causes another to be emitted in the same state.
This starts a cascade.
Inside a laser cavity there are atoms that are in a very strange
state: a higher energy level is more populated than a lower one.
This is called a “population inversion”.
Aside: Lasers
This is necessary for the photon cascade. Since:
nV (E2 )
= e −(E2 −E1 )/kB T , E2 > E1
nV (E1 )
we can associate a “negative temperature”, T , to these two energy
states in the atoms.
Maxwell-Boltzmann speed distribution
The Boltzmann distribution for energy can be leveraged to find a
distribution of the speeds of the molecules.
This is the Maxwell-Boltzmann speed distribution.
The number of molecules with speeds between v and v + dv is
Z v +dv
Z v +dv
Nv dv =
v
4πN
v
m0
2πkB T
3/2
v 2 e −m0 v
2 /2k T
B
dv
Maxwell-Boltzmann speed distribution
The number of molecules with speeds between v and v + dv is
3/2
Z v +dv
Z v +dv
2
21.5 Distribution of Molecular Speeds m0
641
Nv dv =
4πN
v 2 e −m0 v /2kB T dv
2πkB T
v
v
of speeds of N gas
(21.41) N
v
stant, and T is the
mann factor e 2E/k BT
The number of molecules
having speeds ranging from v
to v ! dv equals the area of
the tan rectangle, Nv dv.
vmp
v avg
vrms
hat lower than the
distribution curve
Nv
(21.42)
dv
v
Maxwell-Boltzmann speed distribution
The energy of a molecule can be written:
E = Ktrans + + U
where
2
p
• translational kinetic energy, Ktrans = 2m
0
• includes any rotational or oscillational energy
• U is potential energy (if relevant) that depends on the
location of the molecule
Since we only want to know about the distribution of speeds, we
will need to get rid of any dependence on and U.
Aside: Reminder about probability distributions
Suppose I have a probability distribution over two variables, x and
y:
p(x, y )
If the two variables are independently distributed then:
p(x, y ) = p(x)p(y )
Aside: Reminder about probability distributions
Suppose I have a probability distribution over two variables, x and
y:
p(x, y )
If the two variables are independently distributed then:
p(x, y ) = p(x)p(y )
We can eliminate the dependence on x by just summing over x:
X
X
X
1
>
p(x, y ) =
p(x)p(y ) = p(y )
p(x)
= p(y )
x
x
x
This is how we eliminate rotational and vibrational motion.
Maxwell-Boltzmann speed distribution
Put this expression in the Boltzmann distribution:
p(r, p, ) d3 r d3 p d = Ae −E /kB T d3 r d3 p d
Eliminate dependence on position, rotation, and oscillation:
Z Z
p(r, p, ) d3 r d3 p d
p
Z
Z
−p2 /2m0 kB T
0
−/kB T
00
−U/kB T 3
=Ce
dp C
e
d
C
e
d r
r
Maxwell-Boltzmann speed distribution
Put this expression in the Boltzmann distribution:
p(r, p, ) d3 r d3 p d = Ae −E /kB T d3 r d3 p d
Eliminate dependence on position, rotation, and oscillation:
Z Z
p(r, p, ) d3 r d3 p d
p
=Ce
−p2 /2m0 kB T
Z
:1
:1 Z
00
0
−/k
T
−U/k
T
3
B
B
dp C e
d r
d C e
r
Maxwell-Boltzmann speed distribution
Put this expression in the Boltzmann distribution:
p(r, p, ) d3 r d3 p d = Ae −E /kB T d3 r d3 p d
Eliminate dependence on position, rotation, and oscillation:
Z Z
p(r, p, ) d3 r d3 p d
p
=Ce
−p2 /2m0 kB T
Z
:1
:1 Z
00
0
−/k
T
−U/k
T
3
B
B
d C e
d r
dp C e
2 /2m k T
0 B
p(p) d3 p = Ce −p
d3 p
r
Maxwell-Boltzmann speed distribution
Put this expression in the Boltzmann distribution:
p(r, p, ) d3 r d3 p d = Ae −E /kB T d3 r d3 p d
Eliminate dependence on position, rotation, and oscillation:
Z Z
p(r, p, ) d3 r d3 p d
p
=Ce
−p2 /2m0 kB T
Z
:1
:1 Z
00
0
−/k
T
−U/k
T
3
B
B
d C e
d r
dp C e
2 /2m k T
0 B
p(p) d3 p = Ce −p
r
d3 p
replace momentum with velocity components:
2
2
2
p(v) d3 v = Ce −m0 (vx +vy +vz )/2kB T dvx dvy dvz
Maxwell-Boltzmann speed distribution
We can find C .
The total probability must equal one.
ZZZ
2
2
2
Ce −m0 (vx +vy +vz )/2kB T dvx dvy dvz = 1
Using the identity:
Z∞
2
e −x dx =
√
π
−∞
the three integrals can be evaluated separately:
r
Z∞
2πkB T
−m0 vx2 /2kB T
e
dvx =
m0
−∞
There are three integrals, so
C=
m0
2πkB T
3/2
Maxwell-Boltzmann speed distribution
Now our distribution is:
3/2
2
2
2
m0
3
p(v) d v =
e −m0 (vx +vy +vz )/2kB T dvx dvy dvz
2πkB T
Lastly, we want an expression for how many molecules have speeds
between v and v + dv.
This means we need to get rid of the direction dependence –
transform to spherical coordinates.
dvx dvy dvz = v 2 sin θ dv dθ dφ
Maxwell-Boltzmann speed distribution
Z Z
p(v) d3 v
φ θ
=
m0
2πkB T
= 4π
3/2
m0
2πkB T
2 −m0 v 2 /2kB T
v e
Zπ
Z 2π
dφ
dv
0
3/2
v 2 e −m0 v
2 /2k T
B
sin θdθ
0
dv
This is the probability density for 1 molecule. For N molecules:
Nv dv = 4πN
m0
2πkB T
3/2
v 2 e −m0 v
2 /2k
BT
dv
both on mass and on temperature. At a given temperature, the fraction of molecules with speeds exceeding a fixed value increases as the mass decreases. Hence,
The Shape of the Maxwell-Boltzmann distribution
The total area under either curve is
equal to N, the total number of
molecules. In this case, N " 105.
Nv [molecules/(m/s)]
200
Note that vrms # vavg # vmp.
T " 300 K
160
vmp v
120
avg
vrms
80
T " 900 K
40
0
0
200
400
600
800
1000 1 200 1 400 1600
v (m/s)
3
From
previous
For the derivation
of thislecture:
expression, see an advanced textbook on thermodynamics.
r
r
3kB T
kB T
vrms =
= 1.73
m0
m0
Fig
but
mo
The Shape of the Maxwell-Boltzmann distribution
Average speed, can find by integrating over all speeds, then
dividing by the number of particles.
vavg
Z
1 ∞
v Nv dv
=
N 0
3/2
Z∞ 2
m0
4π
=
v 3 e −m0 v /2kB T dv
2πkB T
0
3/2 Z ∞
2
m0
= 4π
v 3 e −m0 v /2kB T dv
2πkB T
0
3/2 m0
1
m0 −2
= 4π
2πkB T
2 2kB T
vavg
r r
r
8 kB T
kB T
=
= 1.60
π
m0
m0
The Shape of the Maxwell-Boltzmann distribution
To find the most probable speed (peak of the distribution), can
find the value of v for which the derivative of the particle number
distribution is zero.
The Shape of the Maxwell-Boltzmann distribution
To find the most probable speed (peak of the distribution), can
find the value of v for which the derivative of the particle number
distribution is zero.
Set
dNv
dv
=0
Then we find:
r
vmp =
r
2kB T
kB T
= 1.41
m0
m0
The Shape of the Maxwell-Boltzmann distribution
To find the most probable speed (peak of the distribution), can
find the value of v for which the derivative of the particle number
distribution is zero.
Set
dNv
dv
=0
Then we find:
r
vmp =
r
2kB T
kB T
= 1.41
m0
m0
So,
vrms > vavg > vmp
641
The Shape of the Maxwell-Boltzmann distribution
21.5 Distribution of Molecular Speeds
of speeds of N gas
(21.41) N
v
stant, and T is the
mann factor e 2E/k BT
The number of molecules
having speeds ranging from v
to v ! dv equals the area of
the tan rectangle, Nv dv.
vmp
v avg
vrms
hat lower than the
distribution curve
Nv
(21.42)
dv
v
> vavg
vmp distriFigurevrms
21.10
The>speed
0
bution of gas molecules at some
(21.43)
Graph from Serwaytemperature.
& Jewett, page
641.
The
function Nv
Speed Distribution and Evaporation
We can understand evaporation as a change of some of our system
from the liquid to the gaseous state at the surface of the liquid.
Even well below the boiling point there are some molecules with
very high translational KE.
These molecules move fast enough to overcome the strength of the
liquid bonds.
Slower moving molecules are left behind, so the remaining liquid is
cooler.
The Second Law of Thermodynamics
We will state this law in several different ways. First an intuitive
statement:
2nd Law
Unless work is done on a system, heat in the system will flow from
a hotter body in the system to a cooler one.
This is obvious from experience, but it’s not obvious why this
should happen.
It also indicates there is are processes in the physical world that
seem not to happen in the same way if time is reversed.
The Second Law of Thermodynamics and
Reversibility
Scientists and engineers studying and designing steam engines
wanted to make them as efficient as possible.
They noticed there were always losses.
There seemed to be more to it. Energy seems to always spread
out. Heat goes from hotter to colder objects. Energy is lost as
heating in friction.
These things do not happen in reverse.
Isolated, Closed, and Open Systems
Isolated system
does not exchange energy (work, heat, or radiation) or matter with
its environment.
Closed system
does not exchange matter with its environment, but may exchange
energy.
Open system
can exchange energy and matter with its environment.
Reversible and Irreversible Processes
Reversible process
a process that takes a system from an initial state i to a final state
f through a series of equilibrium states, such that we can take the
same system back again from f to i along the same path in a PV
diagram.
Irreversible process
any process that is not reversible.
In real life, all processes are irreversible, but some are close to
being reversible. We use reversible processes as an idealization.
20-3 CHANGE IN ENTROPY
Irreversible Process Example
not obey a conservation law.
always remains constant. For
m always increases. Because of
called “the arrow of time.” For
orn kernel with the forward
he backward direction of time
to the exploded popcorn rerd process would result in an
hange in entropy of a system:
nergy the system gains or loses
atoms or molecules that make
proach in the next section and
by looking again at a process
free expansion of an ideal gas.
m state i, confined by a closed
ed container. If we open the
r, eventually reaching the final
n irreversible process; all the
f of the container.
ows the pressure and volume
System
537
Stopcock closed
Vacuum
Insulation
(a) Initial state i
Irreversible
process
Stopcock open
(b) Final state f
S is
and
Irreversible Process Example
i
Pressure
n of
ure,
ve a
the
volnd a
This process has well-defined initial and final equilibrium states,
but during the expansion of the gas is not in equilibrium.
f
opy
way
Volume
the
Fig. 20-2 A p-V diagram showing the
It cannot be plotted on a PV diagram. Also, no work is done on
ects the gas
initial
i and the final state f of the free
in thisstate
process.
h on
Reversible Counterpart
Allow gas to expand very slowly through equilibrium states at20.5
constant temperature.
The hand
reduces its
downward force,
allowing the
piston to move
up slowly. The
energy reservoir
keeps the gas at
temperature Ti .
The gas is
initially at
temperature Ti .
Energy reservoir at Ti
a
The First Law
Th
in
te
Ti
co
by
m
wi
ab
Energy reservoir at Ti
b
c
Figure 20.7 Gas in a cylinder. (a) The gas is in contact with an energy reservoir. The walls of the
base in contact with the reservoir is conducting. (b) The gas expands slowly to a larger volume. (c) T
Reversible Counterpart
606
Chapter 20
We can plot this isothermal expansion:
The First Law of Thermo
Isothermal Ex
P
Isotherm
Pi
i
Suppose an ideal
This process is d
hyperbola (see A
stant indicates th
Let’s calculate
The work done on
process is quasi-st
PV = constant
The curve is a
hyperbola.
f
Pf
Vi
Vf
V
Figure 20.9 The PV diagram
Because
Negative work is done
the gas, expansion
heat is transferred
in,
and theT is con
for anon
isothermal
of an
n and R:
ideal
from an initial
stateconstant.
to a
internal energy and
thegas
temperature
remain
final state.
Comparing the Processes
In both of these processes the gas expands into a region it was not
in previously.
The energy of the system spreads out.
This corresponds to a change of state, but it is not captured by the
internal energy of the gas system, which does not change in either
process.
Comparing the Processes
In both of these processes the gas expands into a region it was not
in previously.
The energy of the system spreads out.
This corresponds to a change of state, but it is not captured by the
internal energy of the gas system, which does not change in either
process.
Something does change in these processes and we call it entropy.
State Variables
State variables of a thermodynamics system are variables that are
determined if the system is in thermodynamic equilibrium and you
know the system’s state.
Examples: pressure, volume, internal energy, temperature. Also,
entropy.
State Variables
State variables of a thermodynamics system are variables that are
determined if the system is in thermodynamic equilibrium and you
know the system’s state.
Examples: pressure, volume, internal energy, temperature. Also,
entropy.
Each variable on it’s own is not enough to determine the state of
the system. (Many systems in different states might have the same
volume.)
Once the current state is known, we do know all of these variables’
values.
How the system arrived at its current state, does not affect these
values. Heat and work are not state variables.
Entropy
Sadi Carnot discovered that the most efficient possible engine must
be reversible (more on this to come).
Rudolph Clausius interpreted this as being due to the behavior of a
new quantity (entropy).
The change in entropy moving between two states i and f is:
Zf
∆S =
i
dQr
T
where dQr is an infinitesimal heat transfer when the system follows
a reversible path.
(T can be a function of Q!)
Entropy
When a reversible path is followed:
Zf
∆Sr =
i
dQ
T
Can we find the entropy change for an irreversible process?
Entropy
When a reversible path is followed:
Zf
∆Sr =
i
dQ
T
Can we find the entropy change for an irreversible process?
Yes! Since entropy is a state variable, we can consider the entropy
change in any reversible process with the same initial and
final states.
Then:
Zf
∆Sirr =
i
dQr
T
The entropy change in that process will give us the entropy
difference between those two states, regardless of the process.
Summary
• Boltzmann distribution (energies)
• Maxwell-Boltzmann distribution (speeds)
• the second law
• entropy
Test Monday, May 15.
Homework Serway & Jewett:
• new: Ch 21, onward from page 644. Probs: 33, 37, 41, 42,
43, 52, 57, 73
• new: Ch 22, onward from page 679. OQs: 9, 11; CQs: 9;
Probs: 39, 43, 45, 47, 49, 53