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Physical Principles in Biology
Biology 3550
Fall 2016
Lecture 25
Thermodynamics II: Reversible Processes and Entropy
Friday, 28 October
c
David
P. Goldenberg
University of Utah
[email protected]
Important Thermodynamic Concepts and Quantities
Introduced so Far
Energy: ability to do work, with units of joules or calories.
E , internal energy. Directly related to temperature and motion of
molecules.
∆E , change in internal energy. ∆E = Efinal − Estart
w , work done by the surroundings on the system during a process.
q, heat flowing into the system from the surroundings during a process.
Adiabatic: without heat flow in or out of the system, q = 0.
Isothermal: Without temperature change, ∆T = 0 and ∆E = 0.
The first law of thermodynamics: ∆E = q + w
Clicker Question #1
Which of the following are true?
1
∆E < 0
4
q<0
7
w <0
2
∆E = 0
5
q=0
8
w =0
3
∆E > 0
6
q>0
9
w >0
Perfect
insulation
Up to 3 answers accepted.
Correct answers are worth
1
2
point. Wrong answers cost
Total guaranteed to be ≥ 0.
1
2
point.
Clicker Question #2
Weight
Which of the following are true?
Weight
1
∆E < 0
4
q<0
7
w <0
2
∆E = 0
5
q=0
8
w =0
3
∆E > 0
6
q>0
9
w >0
Up to 3 answers accepted.
Correct answers are worth
1
2
point. Wrong answers cost
Total guaranteed to be ≥ 0.
1
2
point.
Clicker Question #3
Weight
Which of the following are true?
Weight
Thermal Reservoir
(Constant T)
1
∆E < 0
4
q<0
7
w <0
2
∆E = 0
5
q=0
8
w =0
3
∆E > 0
6
q>0
9
w >0
Thermal Reservoir
(Constant T)
Up to 3 answers accepted.
Correct answers are worth
1
2
point. Wrong answers cost
Total guaranteed to be ≥ 0.
1
2
point.
State Functions and Path-dependent Functions
State functions of a system depend only on the current state of the
system and do not depend on history.
Examples:
• Temperature, T
• Pressure, P
• Volume, V
• Internal energy, E
For any change in a system, the change in a state function depends
only on the beginning and ending states.
∆T = Tfinal − Tstart
∆P = Pfinal − Pstart
∆V = Vfinal − Vstart
∆E = Efinal − Estart
Work, w , and heat, q, are not state functions.
The Maximum-work Path for Gas Expansion
Weight
Weight
Weight
Thermal Reservoir
Thermal Reservoir
Thermal Reservoir
Piston is allowed to move upward in infinitesimally small steps.
Temperature is never allowed to drop.
Pressure drops as gas expands, so less work is done per step.
If larger steps are ever taken:
• The temperature drops.
• The pressure drops more than it would in an infinitesimal step.
• Less work is produced.
The Minimum-work Path for Gas Compression
Weight
Weight
Weight
Thermal Reservoir
Thermal Reservoir
Thermal Reservoir
Piston is pushed down in infinitesimally small steps.
Energy is transferred from piston to gas molecules.
Temperature is never allowed to increase.
Excess energy flows to the reservoir as heat.
P increases as gas is compressed, so more work is required per step.
If larger steps are ever taken, more work is required.
A Reversible Cycle of Compression and Expansion
Weight
Weight
Weight
Thermal Reservoir
Thermal Reservoir
Thermal Reservoir
Steps in both directions are infinitesimal.
Compression and expansion are exactly the reverse of one another.
wcomp = −wexp = −qcomp = qexp
For the complete cycle:
∆E = 0
w =0
q=0
Either compression or expansion can be reversed at any point by an
infinitesimal force in the opposite direction.
Another Kind of State Function
w and q are not state functions.
BUT, we can define the value of w (or q) for a specific process linking
two states to be a change in a state function.
We define the work for the reversible (infinitely slow) conversion of one
state to the another, wrev , to be the change in state function ∆F .
• ∆F is called the change in “free energy.” (Helmholtz free energy)
• ∆F is the maximum amount of work that can be obtained from the
change in state.
• ∆F is “free energy” in the sense that it is energy that is available to do
work.
• “Free energy” is really the most expensive kind of energy.
It’s the kind that we pay for to do work.
The heat for a reversible process, qrev is also a state function that we
will return to shortly.
Calculating the Work for the
Reversible Isothermal Expansion of a Gas
Weight
Weight
Weight
Thermal Reservoir
Thermal Reservoir
Thermal Reservoir
The work is calculated by integrating force with respect to distance.
Z x2
fdx
wrev =
x1
f is force, and x represents the position of the piston.
The force can be expressed in terms of pressure, P and the area that
the piston presents to the gas, A:
f = −P/A
Negative sign indicates that the pressure force is outwards.
Calculating the Work for the
Reversible Isothermal Expansion of a Gas
Weight
Weight
Weight
Thermal Reservoir
Thermal Reservoir
Thermal Reservoir
From the previous slide:
Z x2
Z x2
wrev =
fdx = −
P · Adx
x1
x1
For each small increment in x, there is a corresponding increment in
volume, dV .
dV = dxA
dx = dV /A
Calculating the Work for the
Reversible Isothermal Expansion of a Gas
Weight
Weight
Weight
Thermal Reservoir
Thermal Reservoir
Thermal Reservoir
Substituting dV /A for dx in the work integral:
Z x2
Z V2
PdV
P · Adx = −
wrev = −
x1
V1
(Recall that the product PV has the units of energy or work!)
From the ideal gas law, P = nRT /V . Substituting:
Z V2
Z V2
nRT
dV
wrev = −
dV = −nRT
V
V1
V1 V
Calculating the Work for the
Reversible Isothermal Expansion of a Gas
Weight
Weight
Weight
Thermal Reservoir
Thermal Reservoir
From the previous slide:
Z V2
dV
wrev = −nRT
V1 V
From calculus:
V2
V2
wrev = −nRT ln(V ) = −nRT ln
V1
V1
wrev depends on n, T , V1 and V2 .
wrev represents a change in a state function.
Thermal Reservoir
Calculating the Work for the
Reversible Isothermal Expansion of a Gas
Weight
Weight
Weight
Thermal Reservoir
Thermal Reservoir
Thermal Reservoir
Since the number of moles remains constant, the change in volume
can also be expressed in terms of a change in concentration:
C1 V1 = C2 V2
V2 /V1 = C1 /C2
Substituting into the result from the previous slide:
C1
wrev = −nRT ln
C2
qrev for Isothermal Expansion
For an isothermal process:
∆E = q + w = 0
q = −w
For reversible isothermal expansion:
C2
qrev = −wrev = −nRT ln
C1
Sign check:
For expansion, C2 < C1 and qrev > 0, and the flow is into the system, as
we expect.
For any two states, there is, in principle, a reversible (infinitely slow)
process separating them, and we can define qrev and wrev .
Reconsider the adiabatic expansion without work
q=0
Insulation prevents heat flow
w =0
No work is done
∆E = q + w = 0
Perfect
insulation
Consistent with no temperature
change.
We know that something has been lost, because restoring the original
condition would require work.
We suspect that this has to do with the loss of order, or increase in
entropy.
But, what is entropy? How do we give it a number?
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.
What does this have to do with randomness?
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
∆S =
= nR ln
= nR ln
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 log W
k = Boltzmann’s constant, with
correct units for entropy (J/K).
S = k ln Ω
What is W (or Ω)?
Ludwig Boltzmann, 1844–1906
Tombstone in Vienna, Austria
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.