statistics Connecting and entropy
Thermal expansion and contraction
Ron Reifenberger
Birck Nanotechnology Center
Purdue University
April 4, 2012
2012
S = kB
ln (w)
Lecture 11
1
Review: A reversible process does NOT produce entropy
Processes that are idealized as reversible include:
Frictionless movement
Restrained compression or expansion
Energy transfer as heat due to infinitesimal
temperature gradients
Electric current flowing through a zero resistance
Restrained chemical reaction
Mixing of two samples of the same substance in the
same state
Processes that are irreversible include:
Movement with friction
Unrestrained expansion
Energy transfer as heat due to large temperature
gradients
Electric current flowing through a non zero resistance
Spontaneous chemical reaction
Mixing of matter of different composition or state
2
There is a strong connection between the
2nd Law of Thermodynamics, Probability
and Statistics
You can devise countless simple experiments to
illustrate that movement toward more disorder is a
law that nature follows.
Highly ordered systems become improbable as the
number of objects in a system increases. Essentially,
Essentially
every observed configuration of a large number of
objects is highly disordered.
This leads to the conclusion that nature has a
strong tendency to move toward maximum
multiplicity (and maximum entropy).
3
A simple example of states and multiplicities
(pulling 4 marbles out of a hat)
Large number
of marbles
Equal number of
red & green
Multiplicity or
number of
microstates
microstates (w)
Macro
States
4R
1G,3R
2G,2R
3G,1R
4G
,
,
,
,
,
,
,
,
,
,
1
4
6
4
1
4
16
The multiplicity of possible states
when
when pulling N marbles from a jar
Total
Microstates
Microstates
Multiplicity w
N=1
N=2
N=3
N=4
N=5
N=6
1
1
1
1
1
1
4
5
6
2
3
6
1
3
10
15
2
4
1
20
4
10
1
15
1
5
8
16
32
1
6
1
64
etc.
a macrostate
Multiplicity of the
5
Central Configuration
1
6
15
15
20
15
6
1
These numbers mean:
1 way to get (6R, 0G);
0G);
6 ways to get (5R, 1G);
15 ways to get (4R, 2G);
20 ways to get (3R,3G);
15 ways to get (2R, 4G);
6 ways to get (1R, 5G);
1 way to get (0R, 6G).
In general:
wn1 ,n2
N!
N!
6!
6 5 4 3 2 1
for example, w4,2
15
n1 !n2 ! 4!2! 4 3 2 1 2 1
n1 !n2 !
wi
15
probability pi
0.234
wi 64
6
Which
Which state has greatest multiplicity?
has
Which state has highest entropy?
Which state has lowest entropy?
S = kB
ln (w)
e.g., the entropy of the (4R, 2G) state would be
the
2G)
S=1.38x10-23 J/K x ln (15)
= 3.7x10-23 J/K
This simple thought experiment can be mapped into
many other situations
7
Consider 10 Gas Atoms in a Container
left half
right half
3
7
2
5
order
increases
4
1
10
6
8
9
probability
decreases
Probability that any one particular atom is in left side of box =
Probability that any two particular atoms are in left side of box = x
5
Probability that any five particular atoms are in left side of box =( ) = 1/32
10
Probability that ALL atoms are in left side of box = ()
= 1/1024
8
Working it out Multiplicities or Probabilities?
Sinit = kB
ln (w
Sfin = kB
init)
Probinit =Pinit = winit / Wtot
ln (w
fin)
Probfin = Pfin = wfin / Wtot
S= Sfin Sinit
ln (w )- k ln (w )
= k ln (P ) + k ln (Wtot) [k ln (P
[k
= k ln (P ) k ln (P )
= kB
fin
B
fin
B
fin
B
init
B
B
B
init)
+ kB
ln (Wtot)]
init
9
left half
Initial:
right half
3
7
2
5
6
4
9
1
left half
Final:
2
9
5
1
10
8
right half
36
7
4
10
8
S= S10
left side
S5
left side =
kB
ln (1/1024) - k ln (1/32)
B
= kB [ -6.93 -3.46 (-3.46)]
= kB
10
Is this transition spontaneously possible for N gas atoms?
Initial
N/2
N/2
V1
N
Final
V2
1
V2 V1
2
11
Calculate the probability that N molecules
appear on one side
The probability that a gas of N molecules spontaneously
compress from original volume V1 to final volume V2 = V1 is
1
2
..
......N
Pfin = (V2/V1) x (V2/V1) . . . . x (V2/V1)
= (V2/V1)
(V
N
Pinit (for V2=V1)=1;
ln (1)=0
So.S= kB ln (Pfin)
ln (Pfin) = N ln (V2/V1) = nNA ln (V2/V1); n=# moles
so
so S = kBnNA
Reif, pg. 126
ln (V2/V1) = nR ln (V2/V1)
12
For instance, suppose we have three moles of an
ideal gas. To spontaneously find all atoms in left
To
half of a container, we have
V2/V1 =
ln (Pfin)
and
= nNA ln () = 3 x 6.02 x1023 x (-.693)
= -1.25 x 1024
then S = kB
ln (Pfin)
-17.28 J/K
(same answer as in last lecture!)
Since the entropy change is negative, the
final entropy is LESS than initial entropy, so
the process will not occur spontaneously
because entropy must always increase.
13
Why Free Energy and Enthalpy?
So far, we have discussed systems with known or zero energy
exchange between boundaries processes like gas expansions
and the conversion of heat to work in heat engines. Under
these circumstances, entropy is a good thermodynamic
predictor of what processes are possible.
For
For chemical processes inside test tubes inserted in a
laboratory heat bath, the work or heat flow is not a
controlled quantity, but rather the temperature and pressure.
This slight change in conditions requires the definition of new
thermodynamic functions like free energy and enthalpy rather
than entropy. Systems held at constant temperature do not
tend toward their state of maximum entropy. Instead, they
Instead they
tend toward states of minimum free energy. This is why
chemists and biologists tend to focus on free energy rather
than entropy.
14
The 2nd Law has both practical and
philosophical implications
Practical: Energy locked into nonrenewable energy sources is constantly
gy
being used up.
Philosophical: Nature has a built in hierarchy
of more useful and less useful forms of
energy. As useful energy is converted into
less useful forms of energy, we no longer can
use it to do work.
15
The 2nd Law and Work
Assuming
Assuming you have an isolated system:
If you take a system from a low entropy state to a
higher entropy state, the system will produce a
the
certain amount of work - Wout.
If you then can somehow take the system from the
higher entropy state back to the low entropy state,
you will need to supply a certain amount of work Win
The 2nd Law guarantees that Win > Wout
This ends our discussion of Entropy!
16
Thermal Expansion
L+L
L
T
T+T
L
L
= T
is a number the coefficient of
linear
linear expansion
17
How much longer will a 1 cm rod
of aluminum become if the
o
temperature increases by 1 C?
For aluminum,
2.4 x 10-5 K-1
L = T L
= 2.4 x 10-5 K-1 x 1 K x 0.01 m
= 240 nm
18
Bimetallic Strip
Unpinned end
Pinned end
T-T
colder
T
1
brass
steel
T+T
2
1 > 2
hotter
19
Thermal Expansion of a
Plate with a Hole
Area of hole = Ao
T
T+T
A
2T
Ao
Expansion
Expansion occurs in every dimension, and every
and
dimension is increased in the same proportion
20
Volume expansion for
isotropic solid
L1
L3
V = L1L2L3
L2
dV
dT
1
V
dL3
dL2
dL1
+ L1 L3
+ L2 L3
= L1 L2
dT
dT
dT
dV
dT
1
=
L3
dL3
+
dT
1
L2
dL2
dT
+
1
L1
dL1
dT
= 3 =
21
Example
V=V T
= 10 gal x (10 x
o
10-4/ C)
o
x (-25 C)
o
T=0 C
= - 0.25 gal
G
A
S
o
T=25 C
To
South
Park
G
A
S
22
Water is anomalous!
23