Chem 156. Lecture 3.
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Notes
Proof of maximum entropy principle was updated in the notes for Lecture 2.
Intro
In this lecture we will see how to find thermodynamic equilibrium for closed system (thus far we only looked at an isolated system and found maximum entropy principle).
We'll see how Gibbs and Helmholtz energies come about based on maximum entropy principle which can be applied for isolated systems.
Also we'll learn which thermodynamic potential is relevant for finding equilibrium depending on types of constraints applied to the system.
Enthalpy
Enthalpy denoted by symbol H is amount of energy (E) stored in the system plus amount of work that needs to be done on the athmosphere (i.e. constant pressure environment) to create space for holding the system (that will be pV):
H = E + pV
Let's take a look at the differential form of the First Law:
dE = δQ - pdV
and let's rearrange it to get δQ
δQ = dE + pdV
Notice, that δQ = dH if dp = 0
dH = dE + pdV + Vdp = dE + pdV (if dp = 0)
The fact that &deltaQ = dH if pressure is constant (as it is for ambient pressure most of the time), justifies the name "enthalpy" - heat content in greek.
Why isn't Q = E + pV? Because that would violate the First Law, you can see it by yourself by differentiating both sides and comparing with the First Law.
Did you notice that we used squiggly δ in front of Q? That was to remind ourselves that in general (when p ≠ const)
Q ≠ E + pV.
Recoup on Entropy
Important observation about Entropy
For any process where system evolves from state 1 to state 2 we can split change of entropy into two components:
ΔS = ΔSexch + ΔSirrev
So that ΔSirrev > 0 for the irreversible process and = 0 for the reversible
Where:
| ΔS = S2 - S1 | total change entropy in the system as it evolves from state 1 to state 2 |
| ΔSexch | change of system entropy due to heat exchange |
| ΔSirrev | change of entropy due to any irreversibilities |
This observation is very useful for demonstration of "minimum free energy" principles applicable to systems subject to specific constraints (those are discussed below).
What is irreversibility? Another name for that is a "dissipative process" - any mechanism that generates new entropy by degrading work into heat or other means (we can even generate new entropy just from heat transfer - if there is a finite temperature gradient). So in the context of statement "presence of irreversibility requires a presence of a dissipative process" increase of entropy in isolated systems (principle of maximum entropy) is not automatic. There has to be a mechanism for the production of new entropy to occur.
Section below provides a proof for the above equation
Proof
Here we will prove that ΔSsys = ΔSexch + ΔSirrev. Maybe it looks simple, but it is not a very trivial observation
Comments to these equations follow. Equations (8) and (9) define Sexch and Sirrev. From the course of derivation it is evident that Sirrev is always non-negative.
| Eq. | comment |
| 1 | Clausius inequality in general form |
| 2 | Entropy is a function of state |
| 3 | Substract (2) from (1). (Why cant we add? Wouldn't we arrive at a conflicting result if we were to add (1) and (2)? The answer is no, but why?) |
| 4 | Combine two intergrals "under the same roof". We can do it if we follow the same integration path in (1) and (2). |
| 5 | this is true because we can follow any cycle in (4) and (4) would still hold. We won't analyze this in every mathematical detail, but (5) is true given that (4) is true for any path. |
| 6 | same as (5) |
| 7 | If we add dSirrev > 0 (but just right amount) we can convert inequality (6) to equality (7) |
| 8 | Definition of system entropy change to to heat exchange with the environment. We've seen that it works for reversible systems (and then the extra term dSirrev we've added to (6) would be zero, because (6) would be an equality. |
| 9 | Combine (7) and (8) |
| 10 | It is obvious that dSirrev has to be non-negative and is strictly positive if the element of path corresponding to dS is irreversible. |
Helmholtz Free Energy
How we get it
Let's take the Universe and a System that is a part of our universe.
Let's also assume that temperature of the Universe and the System (T) is constant and volume of the System (V) is constant
dT = 0 dV = 0
Change of total entropy of the Universe is
dStot = dSsys + dSexterior ≥ 0
If there is any heat exchange between the system's exterior and its interior then
δQsys = -δQexterior
Let's write for the exterior
dSexterior = -δQ/T
So we have
dSsys - δQsys/T ≥ 0 TdSsys - δQsys ≥ 0
Let's use First Law
dQ = dE + pdV
For our conditions (dV = 0) we'll get
dE - TdS ≤ 0
We can integrate above inequality under condition that T = const (same as dT = 0) and get the following quantity:
A = E - TS
The quantity written above was given a name "Helmholtz Free Energy".
What is it good for
Let's see what happens to A when T and V are constant (under same conditions we used to derive these properties).
Let's differentiate expression for A
dA = dE - TdS - SdT
Since T = const dT = 0
dA = dE - TdS (*)
We already saw in the derivation that this expression will have to be ≤ 0 under constraints that we chose. But let's get there in a different way.
First Law
dE = δQ - pdV
plug in dE into equation (*)
dA = δQ - pdV - TdS
V = const, so pdV is gone so
dA = δQ - TdS
Change of system entropy dS consists of two parts: amount changed due to heat exchange dSexch and amount generated by any irreversibilities in the system dSirrev and
dS = dSexch + dSirrev δQ = TdSexch
By combning above three equations we'll get
dA = -TdSirrev
Since any irreversibility will generate new positive entropy
dSirrev > 0
We obtain
dA < 0 for irreversible system dA = 0 for reversible system
Therefore Helmholtz energy reaches minimum in the state of equilibrium, when T and V are constant.
So what is it good for? We can find equilibrium condition by minimizing A if T and V are constant.
Gibbs Free Energy
How we get it
Derivation of Gibbs Energy goes the same way as above except that instead of V = const we shall use p = const. So we have a different set of constraints.
dT = 0 or T = const dp = 0 or p = const
Just like in the derivation above we'll get to the point
TdSsys - δQsys ≥ 0
And using
δQ = dE + pdV δQ = dH if dp = 0 (as it is in our case)
obtain
dE + pdV - TdSsys ≤ 0
under constraints (p = const, T = const) we can integrate and get to the quantity
G = H - TS
this one was called "Gibbs Free Energy".
What is it good for
Just like in the case with Helmholtz Free Energy, and using same method we can show that
if T = const and p = const dG < 0 for irreversible system dG = 0 for reversible system and G = minimum at the state of equilibrium
So Gibbs Energy is useful to determine equilibrium condition when p and T are const. G is most commonly used state function because above mentioned constraints well apply to the environment we live.
Why "Free Energy"
Free energy is amount of energy free to perform useful work in a reversible process.
For example let's take a look at Helmholtz energy, when T=const:
dA = dE - TdS
dA = δQexch - pdV - T(dSexch + dSirrev)
δQexch = TdSexch
so
dA = -pdV - TdSirrev
Now if we have a certain amount of dA to spare, then maximum possible amount of -pdV will be exactly equal dA and that is only possible if dSirrev=0 (remember that irreversibility only increases entropy, never decreases), that is when work is done reversibly.
Therefore dA is an amount of energy free to do expansion (an other) work when T=const.
Read Chapter 4-3 in Eisenberg text (it explains well why dG is an amount of energy free to do non-expansion work at T,p=const).
How to pick right potential
Well, there is nothing wrong with any of them at any time. Those are just functions whose values can be calculated by plugging in the numbers if explicit functional dependence is known.
However when you need to find equilibrium condition, you would have to pick the right potential. Below is a summary.
| condition | parameter useful to find state of equilibrium |
| E,V=const (isolated system) | S→max |
| T,S=const | E (Internal Energy)→min |
| p,S=const | H (Enthalpy)→min |
| T,V=const | A (Helmholtz Energy)→min |
| T,p=const | G (Gibbs Energy)→min |
What if several potentials fit given conditions for example T,V,p=const, then pick a simpler function. A = E - TS, while G = E - TS + pV, so Helmholtz wins.
Example 1. Deformation of liquid drop
We have:
- deformed drop of liquid suspended in air
- T=const
- effects of gravity are neglected
- pressure outside of drop is uniform but not necessarily constant
- liquid is nearly incompressible
- surface tension γ > 0, Esurf = γ s, where E - excess surface energy, γ - surface tension, s - surface area
What potential will help us find thermodynamic equilibrium of this system?
At equilibrium liquid will adopt spherical shape because this shape minimizes the surface area and excess surface energy is proportional to the surface area.
Since liquid is nearly incompressible, volume of the drop does not depend on its shape. We
So T,V=const. Helmholtz energy fits the situation.
Let's look at the situation in a little more detail:
Molecules in the bulk on average experience the same cohesive force from all directions. Hovever on the surface net cohesive force will be directed perpendicular to the any local patch of surface and will point away from air (otherwise the drop would immediately evaporate). If surface area increases, that uncompensated force will increase, thus increasing total internal energy.
What about entropy of molecules near the surface? It can and probably will be different from entropy of bulk water.
So at const T
dA = dE - TdS
Since both entropy and internal energy may depend on the surface area changes in both terms due to changes in the surface area will have contribution to the amount of change in A.
Shape equilibria of incompressible or nearly incompressible objects at T=const is best described by the Helmholtz Free Energy.
Example 2. Bullet in oil canister
In this example we have:
- T=const
- bullet sealed in a rigid canister filled with viscous oil
- at time=0 canister is inverted
- heat exchange with environment is freely allowed
- in this example gravity is there, but its effects on liquid should be neglected
Comment: this system is obviously closed because there is heat exchange, but even if the can were thermally isolated, it would still stay closed - work of gravity will still cross the system boundary.
Which potential should we choose here to find equilibrium?
Again, T and V are constant so we can use Helmholtz energy. But let's take closer at the conditions at hand: notice that nothing changes with oil and bullet internally after everything is at the final resting state, also there is no change in surface energy since surface area of bullet is constant, so internal state of the system (excluding the bullet position) stays the same. Therefore entropy will stay unchanged.
Now we have T,V,S=const. We can see that internal energy is also an option. (see an Excersise below)
Obviously minimizing E is easier then minimizing F=E-TS.
So E is the best answer.
Excersise 1
Prove that if S,V=const then E tends to min
dE = δQ - pdV +(any other work of environment onto the system,
not considered now, but in the example above work of gravity would be
in this place)
δQ is amount of heat added from the environment
dE = TdSexch
Remember that dS = dSexch + dSirrev
The above equation says that change of entropy in the system equals amount of entropy received by heat uptake from the environment plus amount of entropy generated by irreversibilities in the process
dE = TdS - TdSirrev
dE = -TdSirrev ≤ 0
So E goes to minimum at equilibrium
Excersise 2
Explain why Q ≠ E + pV
Let's assume that Q = E + pV.
Its full differential
dQ = dE + pdV + VdP
But First Law "says"
dE = δQ - pdV
Let's combine above two equations
dQ = δQ - pdV + pdV +Vdp
dQ and δQ are the same thing - small addition of heat, symbol δ is used instead of d only to remind ourselves that δQ is not an exact differential (optional reading) (http://en.wikipedia.org/wiki/Exact_differential)
So we end up with
Vdp = 0
Which is obviously wrong in general, meaning that Q ≠ E + pV when p ≠ const.
However when p = const it is true (and then Q = H).
Q is not an exact differential
Exact differentials have a property to be zero at a closed cycle.
We know that integral of δQ over the closed path is not zero and heat entering (or) the system is not a function of system's state.
We also know that by multiplying by 1/T we can convert heat into a function of state called entropy (given that system is reversible).
dS in contrast is an exact differential and its intergal over the closed cycle is always 0.
Multiplying factors that convert inexact differential to exact ones called "integrating factor". So 1/T is an integrating factor for δQ for reversible paths.