Posts Tagged ‘internal energy’

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In terms of simplicity, purely mechanical systems have an advantage over thermodynamic systems in that stability and instability can be defined solely in terms of potential energy. For example the center of mass of the tower at Pisa, in its present state, must be higher than in some infinitely near positions, so we can conclude that the structure is not in stable equilibrium. This will only be the case if the tower attains the condition of metastability by returning to a vertical position or absolute stability by exceeding the tipping point and falling over.

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Thermodynamic systems lack this simplicity, but in common with purely mechanical systems, thermodynamic equilibria are always metastable or stable, and never unstable. This is equivalent to saying that every spontaneous (observable) process proceeds towards an equilibrium state, never away from it.

If we restrict our attention to a thermodynamic system of unchanging composition and apply the further constraint that the system is closed, i.e. no mass transfer takes place between system and surroundings, the fundamental relation of thermodynamics can be written as an exact differential expression exhibiting two independent variables: dU = TdS–PdV. There are therefore 22–1 possible Legendre transformations, each representing a new function. We thus arrive at a set of four state functions

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where the independent variables, shown in parentheses, are the natural (canonical) variables for that function [S = entropy, V = volume, T = temperature, P = pressure].

If the above pairs of natural variables are held constant, the energy function is predictably a minimum at equilibrium. But what if the energy function and one of its natural variables are instead held constant? Which extremum will the other natural variable reach at equilibrium – minimum or maximum? There are eight permutations, giving a grand total of twelve sets of equilibrium conditions.

Of these twelve, only four have straightforward proofs based on the fundamental criteria [derived in my previous blogpost Reversible and Irreversible Change] that

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The proofs of the other eight involve a lengthier train of logical argument which is now rarely seen either in textbooks or on the internet. To redress this issue, CarnotCycle herewith presents all twelve proofs in their entirety.

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The Twelve Conditions of Equilibrium and Stability

Here are the four straightforward proofs

1. For given U and V that S is a maximum

For a closed system, the First Law may be written

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Consider an adiabatic change: dq = 0 so the above criteria for equilibrium (1) and observable change (2) become dS = 0 and dS > 0 respectively. If the volume is also kept constant during the change, it follows from the First Law that dU = 0. The criteria for these conditions may be expressed as follows

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Since observable change always proceeds towards equilibrium, S will increase towards a maximum at equilibrium when U and V are held constant.

2. For given H and P that S is a maximum

For a closed system, the First Law may be written

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Consider an adiabatic change: dq = 0 so the above criteria for equilibrium (1) and observable change (2) become dS = 0 and dS > 0 respectively. If the pressure P of the system is kept constant during the change, and is made to differ only infinitesimally from that of the surroundings, equation (3) becomes

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By the definition of enthalpy, we find

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In this case therefore, H and P are constant, and the criteria for these conditions may be expressed as follows

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Since observable change always proceeds towards equilibrium, S will increase towards a maximum at equilibrium when H and P are held constant.

3. For given T and V that A is a minimum

For a closed system undergoing a change at constant temperature, we may write the differential expression

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But by the first law

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so that

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Suppose the system is capable only of PV work. If the volume is kept constant, no work can be done and therefore

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Applying criteria (1) and (2) for equilibrium (dq = TdS) and observable change (dq < TdS) respectively, we obtain the criteria in terms of A:

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Since observable change always proceeds towards equilibrium, A will decrease towards a minimum at equilibrium when T and V are held constant.

4. For given T and P that G is a minimum

For a closed system undergoing a change at constant temperature, we may write the differential expression (cf. equation 4 above)

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Suppose the system is capable only of PV work. If the pressure P of the surroundings is kept constant during the change, and is made to differ only infinitesimally from that of the system, then

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Applying criteria (1) and (2) for equilibrium (dq = TdS) and observable change (dq < TdS) respectively, we obtain the criteria in terms of G:

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Since observable change always proceeds towards equilibrium, G will decrease towards a minimum at equilibrium when T and P are held constant.

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A word of introduction to the remaining eight not-so-straightforward proofs.

On page 2 of his masterwork On the Equilibrium of Heterogeneous Substances, under the heading Criteria of Equilibrium and Stability, J. Willard Gibbs constructs the foundations of chemical thermodynamics on the equivalence of two propositions relating to an isolated system – that for given U and V that S is a maximum, and for given S and V that U is a minimum.

Gibbs shows by neat argument that the truth (according to Clausius) of the first proposition necessitates the truth of the second proposition. In parallel fashion, the proof already given for the first proposition (condition 1 above) will be used to prove that for given S and V that U is a minimum. This sequitur argument repeats throughout, with condition 2 used to prove condition 6, condition 3 used to prove condition 7 etc.

The train of logical argument for these remaining proofs is somewhat labyrinthine, but the method is same in each case. It gets easier to follow as you go from one to the next.

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5. For given S and V that U is a minimum

Imagine a closed system in some state of equilibrium for which the values of the entropy, volume and internal energy are S1, V1 and U1. Then in any non-equilibrium state with the same values V1 and U1, the entropy S2 must be less than S1 because observable change always proceeds towards equilibrium, and the condition (1) of observable change at constant U and V requires entropy to increase.

Now imagine a second equilibrium state for which the values of the entropy, volume and internal energy are S2, V1 and U2. Compared with the first equilibrium state the volume is unaltered but S2 < S1. Now U and S always change in the same direction since dU = TdS – PdV implies (∂U/∂S)V = T which is always positive. Hence in the second equilibrium state U2 < U1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the entropy and volume are the same but U2 < U1. Since observable change always proceeds towards equilibrium, U will decrease towards a minimum at equilibrium when S and V are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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6. For given S and P that H is a minimum

Imagine a closed system in some state of equilibrium for which the values of the entropy, pressure and enthalpy are S1, P1 and H1. Then in any non-equilibrium state with the same values P1 and H1, the entropy S2 must be less than S1 because observable change always proceeds towards equilibrium, and the condition (2) of observable change at constant H and P requires entropy to increase.

Now imagine a second equilibrium state for which the values of the entropy, pressure and enthalpy are S2, P1 and H2. Compared with the first equilibrium state the pressure is unaltered but S2 < S1. Now H and S always change in the same direction since dH = TdS + VdP implies (∂H/∂S)P = T which is always positive. Hence in the second equilibrium state H2 < H1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the entropy and pressure are the same but H2 < H1. Since observable change always proceeds towards equilibrium, H will decrease towards a minimum at equilibrium when S and P are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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7. For given A and V that T is a minimum

Imagine a closed system in some state of equilibrium for which the values of the Helmholtz free energy, volume and temperature are A1, V1 and T1. Then in any non-equilibrium state with the same values V1 and T1, the Helmholtz free energy A2 must be greater than A1 because observable change always proceeds towards equilibrium, and the condition (3) of observable change at constant T and V requires the Helmholtz free energy to decrease.

Now imagine a second equilibrium state for which the values of the Helmholtz free energy, volume and temperature are A2, V1 and T2. Compared with the first equilibrium state the volume is unaltered but A2 > A1. Now A and T always change in opposite directions since dA = –SdT – PdV implies (∂A/∂T)V = –S which is always negative. Hence in the second equilibrium state T2 < T1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Helmholtz free energy and volume are the same but T2 < T1. Since observable change always proceeds towards equilibrium, T will decrease towards a minimum at equilibrium when A and V are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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8. For given G and P that T is a minimum

Imagine a closed system in some state of equilibrium for which the values of the Gibbs free energy, pressure and temperature are G1, P1 and T1. Then in any non-equilibrium state with the same values P1 and T1, the Gibbs free energy G2 must be greater than G1 because observable change always proceeds towards equilibrium, and the condition (4) of observable change at constant T and P requires the Gibbs free energy to decrease.

Now imagine a second equilibrium state for which the values of the Gibbs free energy, pressure and temperature are G2, P1 and T2. Compared with the first equilibrium state the pressure is unaltered but G2 > G1. Now G and T always change in opposite directions since dG = VdP – SdT implies (∂G/∂T)P = –S which is always negative. Hence in the second equilibrium state T2 < T1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Gibbs free energy and pressure are the same but T2 < T1. Since observable change always proceeds towards equilibrium, T will decrease towards a minimum at equilibrium when G and P are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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9. For given U and S that V is a minimum

Imagine a closed system in some state of equilibrium for which the values of the internal energy, entropy and volume are U1, S1 and V1. Then in any non-equilibrium state with the same values S1 and V1, the internal energy U2 must be greater than U1 because observable change always proceeds towards equilibrium, and the condition (5) of observable change at constant S and V requires the internal energy to decrease.

Now imagine a second equilibrium state for which the values of the internal energy, entropy and volume are U2, S1 and V2. Compared with the first equilibrium state the entropy is unaltered but U2>U1. Now U and V always change in opposite directions since dU = TdS – PdV implies (∂U/∂V)P = –P which is always negative. Hence in the second equilibrium state V2 < V1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the internal energy and entropy are the same but V2 < V1. Since observable change always proceeds towards equilibrium, V will decrease towards a minimum at equilibrium when U and S are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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10. For given H and S that P is a maximum

Imagine a closed system in some state of equilibrium for which the values of the enthalpy, entropy and pressure are H1, S1 and P1. Then in any non-equilibrium state with the same values S1 and P1, the enthalpy H2 must be greater than H1 because observable change always proceeds towards equilibrium, and the condition (6) of observable change at constant S and P requires enthalpy to decrease.

Now imagine a second equilibrium state for which the values of the enthalpy, entropy and pressure are H2, S1 and P2. Compared with the first equilibrium state the entropy is unaltered but H2 > H1. Now H and P always change in the same direction since dH = TdS + VdP implies (∂H/∂P)S = V which is always positive. Hence in the second equilibrium state P2 > P1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the enthalpy and entropy are the same but P2 > P1. Since observable change always proceeds towards equilibrium, P will increase towards a maximum at equilibrium when H and S are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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11. For given A and T that V is a minimum

Imagine a closed system in some state of equilibrium for which the values of the Helmholtz free energy, temperature and volume are A1, T1 and V1. Then in any non-equilibrium state with the same values T1 and V1, the Helmholtz free energy A2 must be greater than A1 because observable change always proceeds towards equilibrium, and the condition (3) of observable change at constant T and V requires the Helmholtz free energy to decrease.

Now imagine a second equilibrium state for which the values of the Helmholtz free energy, temperature and volume are A2, T1 and V2. Compared with the first equilibrium state the temperature is unaltered but A2 > A1. Now A and V always change in opposite directions since dA = –SdT – PdV implies (∂A/∂V)T = –P which is always negative. Hence in the second equilibrium state V2 < V1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Helmholtz free energy and temperature are the same but V2 < V1. Since observable change always proceeds towards equilibrium, V will decrease towards a minimum at equilibrium when A and T are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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12. For given G and T that P is a maximum

Imagine a closed system in some state of equilibrium for which the values of the Gibbs free energy, temperature and pressure are G1, T1 and P1. Then in any non-equilibrium state with the same values T1 and P1, the Gibbs free energy G2 must be greater than G1 because observable change always proceeds towards equilibrium, and the condition (4) of observable change at constant T and P requires the Gibbs free energy to decrease.

Now imagine a second equilibrium state for which the values of the Gibbs free energy, temperature and pressure are G2, T1 and P2. Compared with the first equilibrium state the temperature is unaltered but G2 > G1. Now G and P always change in the same direction since dG = VdP – SdT implies (∂G/∂P)T = V which is always positive. Hence in the second equilibrium state P2 > P1. Comparing this second equilibrium state with the above-mentioned non-equilibrium state, the Gibbs free energy and temperature are the same but P2 > P1. Since observable change always proceeds towards equilibrium, P will increase towards a maximum at equilibrium when G and T are held constant.

The foregoing arguments can be summarized in tabulated form as follows:

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P Mander April 2015

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Historical background*

It was the American physicist Josiah Willard Gibbs (1839-1903) pictured above who first introduced the thermodynamic potentials ψ, χ, ζ which we today call Helmholtz free energy (A), enthalpy (H) and Gibbs free energy (G).

In his milestone treatise On the Equilibrium of Heterogeneous Substances (1876-1878), Gibbs springs these functions on the reader with no indication of where he got them from. Using an esoteric lexicon of Greek symbols he simply states:

Let
ψ = ε – tη
χ = ε + pv
ζ = ε – tη + pv

As with much of Gibbs’ writings, the clues to his sudden pronouncements need to be sought on other pages or – as in this case – another publication.

In an earlier paper entitled A method of geometrical representation of the thermodynamic properties of substances by means of surfaces, Gibbs shows that the state of a body in terms of its volume, entropy and energy can be represented by a surface:

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Gibbs’ thermodynamic surface of 1873, realized by James Clerk Maxwell in 1874

It can be demonstrated from purely geometrical considerations that the tangent plane at any point on this surface represents the U-related function

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Now this is none other than Gibbs’ zeta (ζ ) function. The question is, did he recognize it for what it was – a Legendre transform? A key feature of On the Equilibrium of Heterogeneous Substances is the business of finding an extremum for a multivariable function subject to various kinds of constraint, and it is known that Gibbs was familiar with Lagrange’s method of multipliers – he mentions the technique by name on page 71, immediately after equation 41. The point here is that the Legendre transformation can be phrased in the same terms – for example, the multiplier expression for finding the stationary value of U when T and P are held constant yields the Legendre transform shown above.

But suggestive though this is, it actually gets us no closer to determining whether or not Gibbs was aware that ψ, χ, ζ  were Legendre transforms. Gibbs gave no indication in his writings either that he knew the transformation trick, or that he had discovered it for himself. We can only estimate likelihoods and have hunches.

*Text revised following input from Bas Mannaerts (see comments below)

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In the CarnotCycle thermodynamics library, the first textbook reference to Legendre transformation is by P.S. Epstein in 1937. Epstein was a Russian mathematical physicist who was recruited by Caltech in 1921. He was a renowned commentator on Gibbs’ work, especially in statistical mechanics.

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Thermodynamics and the Legendre transformation

The fundamental relation of thermodynamics dU = TdS–PdV is an exact differential expression

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where the coefficients Ci are functions of the independent variables Xi. By means of Legendre transformations (ℑ) the above expression generates three new state functions whose natural variables contain one or more Ci in place of the conjugate Xi

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The equation of the tangent plane to the thermodynamic surface generates ℑ3, with ℑ1 and ℑ2 following procedurally from

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How the Legendre transformation works

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defines a new Y-related function Z by transforming

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into

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Proof
dZ = dY – d(C1X1)
dZ = dY – C1dX1 – X1dC1
Substitute dY with the original differential expression
dZ = C1dX1 + C2dX2 – C1dX1 – X1dC1
The C1dX1 terms cancel, leaving
dZ = C2dX2 – X1dC1

The independent (natural) variables are transformed from Y(X1,X2) to Z(X2, C1)
The same procedural principle applies to ℑ2 and ℑ3.

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The Legendre Wheel

Since exact differential expressions in two independent (natural) variables can be written for the internal energy (U), the enthalpy (H), the Gibbs free energy (G) and the Helmholtz free energy (A), each of these state functions can generate the other three via the Legendre transformations ℑ1, ℑ2, ℑ3. This is neatly demonstrated by the Legendre Wheel, which executes the transformation functions

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from any of the four starting points:

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 [click on image to enlarge]

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Legendre transformations and the Gibbs-Helmholtz equations

For an exact differential expression

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the transforming function

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can be written in terms of the natural variables of Y

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This Legendre transformation is the means by which we obtain the Gibbs-Helmholtz equations. Taking Y=G(T,P) as an example, ℑ1 executes the clockwise transformation

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while the transforming function

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reverses the positions of the natural variables and executes the counterclockwise transformation

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Setting Y=G(T,P) generates six Gibbs-Helmholtz equations, in each of which one of the two natural variables is held constant. Since there are four state functions – U, H, G and A – the total number of Gibbs-Helmholtz equations generated by this procedure is twenty-four. To this can be added a parallel set of twenty-four equations where U, H, G and A are replaced by ΔU, ΔH, ΔG and ΔA.

These equations are particularly useful since they relate a state function’s dependence on either of its natural variables to an adjacent state function on the Legendre Wheel.

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Who was Legendre?

Adrien Legendre (1752-1833) was a French mathematician. He wrote a popular and influential geometry textbook Éléments de géométrie (1794) and contributed to the development of calculus and mechanics. The Legendre transformation and Legendre polynomials are named for him.

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P Mander September 2014