Lecture 6: Intro to Entropy
• Reading: Zumdahl 10.1, 10.2, 10.3
• Outline:
– Why enthalpy isn’t enough.
– Statistical interpretation of entropy
– Boltzmann’s Formula
– Heat engine leads to Entropy as a state function
• Problems:Z10.1-4, Z10.7, Z10.16-17, Z10.19-22,
Z10.25-26.
1
Enthalpy and Spontaneous Rxns
• Early on in the development of thermodynamics, it was
believed that if a reaction was exothermic, it was
spontaneous.
• But there are plenty of reactions that go that do not give off
heat.
•
Consider the following reaction:
H2O(s)
H2O(l)
ΔH°rxn = +6.02 kJ
•
Endothermic…..yet spontaneous!
Above 0C, but not spontaneous below 0C. Why?
•Consider the hot packs and cold packs again as well.
2
Enthalpy and Spontaneous Reactions:
Joule’s original experiment
• Consider the following Experiment: Mixing of a
gas inside a bulb adiabatically (q = 0).
•
•
•
•
q = 0, The bulbs are insulated
w = 0, External Pressure is zero.
ΔΤ=0, Experimental Result.
ΔE = 0, and ΔH = 0
….but it still happens, the gas
expands (spontaneously) to fill the
two bulbs. Why?
3
Entropy and Statistical Probability
• The connection between weight (Ω) and
entropy (S) is given by Boltzmann’s Formula:
S = k B ⋅ ln Ω
Boltzmann’s constant:
kB = R
NA
= 1.38 ⋅10−23 J
K
• For a given state (where E is fixed) Ω is the number
of ways of arranging the molecules (or microstates).
•The difference in entropy between two states then is:
⎛ Ω2 ⎞
ΔS = k B ln Ω 2 − k B ln Ω1 = k B ln ⎜
⎟
⎝ Ω1 ⎠
4
Molecules Move by Chance Between Bulbs
•
•
•
•
•
•
•
•
•
•
What is the chance that a molecule will be on the left: ½
What is the chance that two molecules will be on the left: ¼
N
And N molecules on the left:
⎛1⎞
For Large N, very small chance: ⎜ 2 ⎟
⎝ ⎠
The molecules move in a statistically independent way.
How does the entropy change for the gas if we double the volume?
How many ways to put a molecule in either side vs 1? Twice as many.
Ω is the number of ways of arranging a molecule in one bulb, then 2Ω is
the number of ways of arranging it in both bulbs.
All particles are statistically independent so get a product of
arrangements.
Apply Boltzmann’s definition of entropy:
⎛ Ω2 ⎞
⎛ N ⎞
⎛ 2Ω ⎞
ΔS = k B ln ⎜
⎟ = k B ln ⎜
⎟ ln 2 = nR ln 2
⎟ = ( kB N A ) ⎜
⎝ Ω ⎠
⎝ Ω1 ⎠
⎝ NA ⎠
N
5
Expansion by any amount:
For each independent particle
Original Weight = Ω
Final Weight = 2Ω
In the previous example the statistical weight was directly
proportional to volume. Now generalizing to any two
different volumes:
⎛ Ωf
ΔS = k B ln ⎜
⎜Ω
⎝ I
N
⎞
⎛ Ωf
⎟⎟ = nR ln ⎜⎜
⎠
⎝ ΩI
⎞
⎛ Vf
⎟⎟ = nR ln ⎜⎜
⎠
⎝ VI
⎞
⎟⎟
⎠
6
The absolute entropy for a gas in a box
•
•
•
Avagadro’s principle: 1 mole occupies 22.4 liters at STP.
From this we will compute the absolute entropy for a gas in a box.
Imagine the the box of volume V (22.4L) is divided up (conceptually anyway)
into a number of sub-boxes each of which is capable of holding one molecule at a
time. Assume a sub-box is about 0.05 nanometer in length (which is the Bohr
radius of an atom). This give us a number of sites (sub-boxes) to put molecules
in, one at a time.
V
22.4 ⋅10−3 m3
Λ=
Vsub
=
( 5 ⋅10−11 m )
3
= 1.8 ⋅1029
Now count the way to put Avagadro’s number of molecules in there:
⎛ Λ ⎞
Ω=⎜
⎟
N
⎝ A⎠
•
NA
This is the number of sites each molecule can occupy, and
they are all statistically independent. So the total Weight is
the product of individual weights.
Compute the entropy:
⎛ Λ ⎞
⎛ 1.8 ⋅ 1029 ⎞
S = k B ln Ω = k B N A ln ⎜
= 12 R = 105 J
⎟ = R ln ⎜
23 ⎟
K
⎝ 6.02 ⋅ 10 ⎠
⎝ NA ⎠
•
This is close to the experimental values for simple gasses (like He or H), and
7
shows that the entropy is proportional to the log of the Volume.
Example: Crystal of CO
• Each arrangement of CO
is possible.
One can flip
• Consider the depiction of
crystalline CO. There are
two possible arrangements
for each CO molecule.
• For a mole of CO:
Ω = 2NA
• Same result as for the
Joule Experiment where
the volume doubled.
S = k B ln Ω = k B ln ( 2 N A ) = k B N A ln ( 2 ) = R ln ( 2 ) = 0.7 ⋅ R = 5.68 J
K
What is Entropy Good For?
• All other thing being equal, the most probable way to
distribute the energy is the most likely way the system will
arrange itself.
• Entropy is a measure of what is probable.
– The larger the entropy, the more probable the event.
– An increase in entropy is a more probably event.
– An increase in volume results in an increase in entropy
• For a chemical reaction, if entropy increases it ought be
more probably (i.e. spontaneous). (See problems 1-4)
– In general then a reaction that creates a gas is going to
increase its entropy, and that can make a reaction happen.
– Temperature can help if the entropy (change) is positive
– If can’t get a gas, going to a liquid from solid is next best.
– In a reaction A2+B2 goes to 2AB, what is entropy change?
9
Heat Engines (and Entropy)
• We will study the gas in a box, and determine what the
entropy change is from a thermodynamic point of view,
and will find that we arrive at the same answer as
Boltzmann, and that entropy is a state function and a
measurable quantity as well.
• We have examined isochoric and isobaric changes in a gas
in a box. Both of these are done by raising or lowering the
temperature (which can be done slowly).
• Now we consider isothermal processes, many of which
cannot be done slowly. We need isothermal, reversible
expansion/contraction process because this tells us what
entropy is from thermodynamics.
10
Example: Isothermal Expansion
• Consider a mass connected to a ideal gas
contained in a “piston”. Piston is
submerged in a constant T bath such that
ΔT = 0 during expansion.
11
Isothermal Expansion: Start
Δh
ΔV = AΔh
PV = PV
1 1 = PV
2 2 = nRT
Δh
• Initially,
V = V1
P = P1
Finally V = V2
P = P2
• Pressure of gas is equal to that created by mass:
P1 = force/area = M1g/A
where A = piston area
g = gravitational acceleration (9.8 m/s2)
12
Isothermal Expansion: One Step
(Not Reversible; it is Irreversible)
• One-Step Expansion. We change the weight to
M2=M1/4, then
Pext = P2=(M1/4)g/A = P1/4
Internal pressure is larger than the external pressure
so the gas will expand. The work done is defined
by the external pressure.
• The mass will be lifted until the internal pressure
equals the external pressure. In this case
V2 = 4V1
• Because
nRT = PV = PV
1 1 = PV
2 2
• q=-w = PextΔV = P1/4 (4V1 - V1) = 3/4 P1V1
13
PV Diagram: One Step vs Reversible Work
at Constant Temperature
1
• Expansion (green line) raises mass M 2 = M 1 in one step.
4
• Compression (red line) lowers mass M 1 in a single step
and returns the system to state 1.
w = 3PV
1 1 = 3nRT
3
3
w = − PV
=
−
nRT
1 1
4
4
14
One Step Compression
State of the system after the 1 Step Expansion:
Vacuum
1M
4 1
Vinitial = 4V1
Pinitial = 14 P1
3M
4 1
What do we have to do to get the system back to its original
state before 1-step expansion? [Pick up Weight]
System did work: Raised M/4; to restore it we have to raise
3M/4 the same distance. We must do more work to
restore the system than we got from it in the first place.
15
Reversible Isothermal Expansion:
(Infinite number of steps; can go back and forth)
• Add up areas (integrate) from V1 to V2
⎛ V2 ⎞
nRT
V2
= − ∫ PdV = − ∫
dV = −nRT(ln(V )) |V1 = −nRT ln⎜ ⎟
⎝ V1 ⎠
V1
V1 V
V2
w total
V2
qrev = − wrev
⎛ V2 ⎞
= nRT ln ⎜ ⎟
⎝ V1 ⎠
16
A Thermodynamic Engine: Defining Entropy
Imagine doing all steps reversibly, then compression is simpler to
understand. The system variables (P,T) are always only
incrementally different from the external values.
Use two different heat reservoirs at two different temperatures.
Let’s consider the four-step cycle
illustrated:
1
4
2
3
–
–
–
–
1:
2:
3:
4:
Isothermal expansion
Isochoric cooling
Isothermal compression
Isochoric heating
V2
V1
Volume
17
First Step of Four Cycle
• Step 1: Isothermal Expansion
at T = Thigh from V1 to V2
1
4
2
3
• Do expansion reversibly. Then:
V2
V1
Volume
• Now ΔT = 0; therefore, ΔE = 0
and q = -w
⎛ V2 ⎞
q1 = − w1 = nRThigh ln⎜ ⎟
⎝ V1 ⎠
18
Step 2
1
4
2
3
V2
V1
Volume
• Step 2: Isochoric Cooling to T = Tlow.
• Now ΔV = 0; therefore, w = 0
• q2 = qV=ΔE = nCvΔT = nCv(Tlow-Thigh)
19
Step 3
• Step 3: Isothermal Compression
at T = Tlow from V2 to V1.
1
4
2
• Now ΔT = 0; therefore, ΔE = 0
and q = -w
3
V2
V1
Volume
• Do compression reversibly, then
⎛ V1 ⎞
q3 = − w3 = nRTlow ln⎜ ⎟
⎝ V2 ⎠
20
Step 4
1
4
2
3
V2
V1
Volume
• Step 4: Isochoric Heating to T = Thigh.
• Now ΔV = 0; therefore, w = 0
• q4 =qV= ΔE = nCvΔT = nCv(Thigh-Tlow) = -q2
21
Add together: All 4 steps of the engine
ΔETotal = ΔH Total = 0
(State Function)
1
qtotal = q1 + q2 + q3 + q4
4
2
3
V2
V1
Volume
qtotal
qtotal
qtotal
qtotal = q1 + q3
⎛ V2 ⎞
⎛ V1 ⎞
= nRThigh ln ⎜ ⎟ + nRTlow ln ⎜ ⎟
⎝ V1 ⎠
⎝ V2 ⎠
⎛ V2 ⎞
⎛ V2 ⎞
= nRThigh ln ⎜ ⎟ − nRTlow ln ⎜ ⎟
⎝ V1 ⎠
⎝ V1 ⎠
⎛ V2 ⎞
= nRΔT ln ⎜ ⎟ = − wtotal ≠ 0
⎝ V1 ⎠
22
Joule’s Experiment (Z10.26)
• One mole of and ideal gas at 1 liter, and 5 Atm. expands
isothermally into an evaluated bulb to reach a total volume of
2 liters. Calculate q, w and qrev for this change of state.
• How do we approach this? Write down first law(s) and the
formula for work.
ΔE = nCV ΔT = q + w but ΔT = 0
w = − Pext ΔV
and Pext = 0
• Now that q=w=0, how do we determine the reversible heat?
It is still isothermal, and we must find a reversible path.
ΔE = nCV ΔT = 0 = qrev + wrev
⇒ qrev = − wrev
⎛ V2 ⎞
wrev = − nRT ln ⎜ ⎟ = − ( PV
1 1 ) ln 2 = −5 ⋅ 101 ⋅ ln 2 = −500 ⋅ 0.7 J = −350 J
⎝ V1 ⎠
q
qrev = +350 J
ΔS = rev = nR ln 2 = 8.3 ⋅ 0.7 = 5 J
K
T
23