Physical Principles in Biology
Biology 3550
Fall 2016
Lecture 26
Thermodynamics III:
The Statistical Definition of Entropy and the Second Law
Monday, 31 October
c
David
P. Goldenberg
University of Utah
[email protected]
The “Classical” Definition of Entropy, S
For two states for which temperature remains constant during the
reversible process of converting one to the other:
qrev
T
For two states separated by a reversible process for which the
temperature does not stay constant:
∆S =
Z T2
∆S =
T1
qrev
dT
T
Units for entropy: energy/temperature, J/K.
Entropy Change for Isothermal Expansion of a Gas
From before, qrev :
qrev = −wrev
V2
= nRT ln
V1
C1
= nRT ln
C2
Entropy:
qrev
V2
C1
= nR ln
= nR ln
∆S =
T
V1
C2
∆S is positive if volume increases (or concentration decreases).
∆S does not depend on temperature (in this case).
The Statistical Definition of Entropy
S = k ln Ω
For a given state, Ω is the
number of equally probable ways
to arrange the components
making up that state.
The different arrangements are
called microstates.
There is no proof! We believe it
because it works.
Ludwig Boltzmann, 1844–1906
Tombstone in Vienna, Austria
Calculating Ω For Gas Molecules in a Container
The state of the system at any instant is defined by the positions and
velocities of each of the molecules.
For a gas, there are a vast number of different microstates with
different positions and velocities for each molecule.
But, if the temperature isn’t going to change, we can focus just on the
positions.
Calculating Ω For Gas Molecules in a Container
Assume that we have a cubic volume and divide this volume up into M
little cubes, with M much larger than the number of molecules, N.
It is very unlikely that any of the little cubes will have more than one
molecule in it, and most will be empty.
Calculating Ω For Gas Molecules in a Container
N is the number of molecules, and M is the number of little cubes that
can contain a molecule. M N.
How many ways can we place N molecules in M positions?
There are M ways to place the first molecule.
There are M − 1 ways to place the second molecule. But, M − 1 ≈ M
If the molecules were labeled, the number of ways to place all N is:
M · M · M · · · M · M · M = MN
But, all of the molecules are identical, so we have to divide this number
by the number of different orders in which the molecules can be
chosen (the permutations), N!
Ω=
MN
N!
Calculating the Entropy of Gas Molecules in a Container
From Boltzmann:
N
S = k ln Ω = k ln
M
N!
!
What if we divided the volume up into bigger or smaller cubes?
Would the entropy change?
That doesn’t sound right!
But, it’s OK if we limit ourselves to changes in entropy.
Calculating ∆S for a Change in Gas Volume
Define starting and ending volumes in terms of the number of little
cubes they contain, M1 and M2 .
V1 = M1 × Vc
V2 = M2 × Vc
where Vc is the volume of a little cube.
Change in entropy for changing the volume:
∆S = S2 − S1 = k ln Ω2 − k ln Ω1
!
!
M2N
M1N
= k ln
− k ln
N!
N!
Quick Review of Some Rules for Logarithms
ln(a · b) = ln(a) + ln(b)
1
ln
= − ln(a)
a
ln
a
b
= ln(a) − ln(b)
ln(ab ) = b ln(a)
Calculating ∆S for a Change in Gas Volume
Change in entropy for changing the volume:
∆S = S2 − S1 = k ln Ω2 − k ln Ω1
!
!
M2N
M1N
= k ln
− k ln
=k
N!
N!
N
M2 N!
∆S = k ln
·
N! M N
1
M2
= Nk ln
M1
!
M2
= k ln
M1
N
ln
M2N
N!
!
− ln
M1N
N!
!!
Calculating ∆S for a Change in Gas Volume
From the previous slide:
M2
∆S = Nk ln
M1
M2 and M1 (the number of little cubes in the two volumes) are related
to the the total volumes according to:
V1 = M1 × Vc
V2 = M2 × Vc
V2
M2 × Vc
M2
=
=
V1
M1 × Vc
M1
V2
∆S = Nk ln
V1
Calculating ∆S for a Change in Gas Volume
From the previous slide:
V2
∆S = Nk ln
V1
N is the number of molecules. If n is the number of moles, N = nNα
V2
∆S = nNα k ln
V1
k is the gas constant, R, divided by Nα . So, kNα = R, and:
V2
∆S = nR ln
V1
Exactly the same as the result from the classical definition!
Clicker Question #1
Calculate the entropy change for reducing the volume of 0.1 mol of a gas
from 1 L to 0.1 L.
1
-8.3 J/K
2
-1.9 J/K
3
1.9 J/K
4
8.3 J/K
V2
∆S = nR ln
V1
R = 8.314 J/K
What About the Size of the Little Cubes?
Suppose that we divide each of the original cubes into 8 smaller cubes.
If M1 and M2 are the numbers of the original cubes making up V1 and
V2 , with the smaller cubes, we have:
V1 = 8 × M1 × Vc,2
V2 = 8 × M2 × Vc,2
where Vc,2 is the volume of the new, smaller cubes.
What About the Size of the Little Cubes?
Change in entropy for changing the volume:
∆S = S2 − S1 = k ln Ω2 − k ln Ω1
!
!
N
N
(8 × M1 )
(8 × M2 )
− k ln
= k ln
N!
N!
(8 × M2 )N
N!
∆S = k ln
·
N!
(8 × M1 )N
M2
= Nk ln
M1
The size factor (8) cancels out!
!
M2
= k ln
M1
N
Another Example: A Bimolecular Reaction
−−−
−*
A+B)
−− C
Assume a volume made up of M little cubes:
• The number of ways to place a molecule of A and a molecule of B in the
volume:
ΩA,B = M 2
• The number of ways to place a molecule of C in the volume:
ΩC = M
The entropy change
∆S = k ln(ΩC ) − k ln(ΩA,B ) = k ln(M) − k ln(M 2 )
M
= k ln
= −k ln(M)
M2
The entropy decreases!
A Bimolecular Reaction
From previous slide:
∆S = −k ln(M)
In this case, the number of little cubes doesn’t cancel out, and the size
of the cubes matters.
A plausible estimate:
• Make the cubes about the right size to hold one or two small molecules,
3
Vc = (2 nm) = 8 nm
3
• Specify the reaction volume, V in liters.
−3
3
10 m
VL
M=
×
×
3
L
8 nm
9
10 nm
m
!3
≈ V (L) × 1023
A Bimolecular Reaction
From previous slide:
∆S = −k ln(M) = −k ln(V (L) × 1023 )
For one mole each of A and B converted to C:
∆S = −Nα k ln(V (L) × 1023 )
= −R ln(V (L) × 1023 )
For V = 1 L:
∆S = −R ln(1023 ) = −8.31 J/K × 53 = −440 J/K
Clicker Question #2
−−
−*
If the reaction volume is doubled for the reaction A + B −
)
−− C, How will the
entropy change be affected?
1
∆S will be unaffected.
2
∆S will become more negative.
3
∆S will become less negative.
∆S = −k ln(M) = −k ln(V (L) × 1023 )
A Bimolecular Reaction
−−−
−*
A+B)
−− C
∆S = −440 J/K
Recall the classical definition of ∆S for an isothermal process:
∆S =
qrev
T
qrev = heat absorbed by system in reversible process.
(maximum/minimum work)
If the process is isothermal (constant T ) ∆E = 0 and qrev = −wrev .
qrev = T ∆S = −T × 440 J/K
wrev = T × 440 J/K
For T = 298 K, wrev = 130 kJ. What does this represent?
A Bimolecular Reaction
−−
−*
A+B−
)
−− C
wrev = 130 kJ
The minimum work required to reduce the entropy of one mole of A
and one mole of B to form single molecules of C, in a volume of 1 L.
Some caveats:
• This is a highly simplified treatment.
• Choice of Vc is not very well justified.
• Ignores rotational freedom of molecules.
• Ignores internal motions of molecules.
• This is a controversial subject, but this estimate is within the bounds of
the controversy!
The Second Law of Thermodynamics
For a spontaneous process, the total entropy of the universe
(the system and its surroundings) will increase.
∆Suniv = ∆Ssys + ∆Ssurr > 0
What does spontaneous mean?
Process occurs without an input of work to the system: w ≤ 0.
If w < 0, the process produces work.
∆Ssys is the quantity we have been working with so far.
What is ∆Ssurr ?
q
∆Ssurr = −
T
q is the heat absorbed by the system during the specific process being
considered. −q is the heat released to the surroundings.
The Second Law of Thermodynamics
For a spontaneous process:
∆Suniv = ∆Ssys + ∆Ssurr > 0
qrev
q
−
>0
T
T
Ssurr is not a state function!
It depends on the path of the change in the system.
Flow of heat to surroundings represents an increase in entropy of the
surroundings.
Heat flows to surroundings only if the system is warmer than the
surroundings.
Entropy of the system can decrease in a spontaneous process, if there
is a flow of heat to the surroundings.