7-1 Chapter 7 - Thermodynamic equilibrium 0) A few words about second derivatives of thermodynamic potentials: Postulate 2 implies that S is maximized at constant U: dS=0, d 2 S < 0. The second inequality is required to make sure that the point where the slop is zero has negative curvature, so it is indeed a maximum. Postulate 2’ similarly implies that U is minimized at constant S: dU=0, d 2 U > 0. The second inequality ensures that U has a positive curvature where its slope vanishes, yielding a minimum. Similarly, d 2 G >0 at constant T and P, d 2 A >0 at constant T, and d 2 H >0 at constant P. These potentials are derived from energy and minimize U tot for the open system with the appropriate intensive variable. For Massieu functions, d 2 f < 0, as they are derived from entropy. There is a bewildering array of second derivatives from all of these potentials. Fortunately, only three of them (not counting derivatives with respect to mole numbers) are independent, including all the possible potentials. So (∂P/∂T) S , (∂μ/∂n) V , (∂G/∂S)| V , (∂U/∂T) P etc. cannot all be independent of one another. The reason is that thermodynamic potentials are state functions. For a potential z=Xdx+Ydy+Σμ i dn i , where X=∂z/∂x| y and Y=∂z/∂y| x , only the following are independent second derivatives: 1) ∂X/∂x = ∂ 2 z/∂x 2 2) ∂Y/∂y = ∂ 2 z/∂y 2 3) ∂X/∂y = ∂ 2 z/∂x∂y ∂Y/∂x = ∂ 2 z/∂y∂x = ∂ 2 z/∂x∂y is not independent of 3. Thus there is no fourth second derivative. In addition, Maxwell’s relations and Jacobians can be used to calculate the derivatives of any other potential, once three have been picked for potential z. This is so because other potentials are related to z by Legendre transforms, and do not contain independent information about stability. They are merely alternative formulations of U tot for the closed system, cleverly written in terms of only variables of the open subsystem of interest. In practice, chemical problems usually involve constant T and P, so the derivatives of the Gibbs free energy are usually used: 1) ∂S/∂T = = ∂ 2 G/∂T 2 : T/n (∂S/∂T) P = dq rev /dT| P = c P (heat capacity at constant P) 2) ∂V/∂P = ∂ 2 G∂P 2 : -1/V (∂V/∂P) T = κ isothermal compressibility 3) ∂V/∂T = ∂ 2 G/∂P∂T: 1/V (∂V/∂T) P = α isobaric thermal expansion coefficient
Transcript
7-1
Chapter 7 - Thermodynamic equilibrium
0) A few words about second derivatives of thermodynamic potentials:
Postulate 2 implies that S is maximized at constant U: dS=0, d2S < 0. The secondinequality is required to make sure that the point where the slop is zero has negativecurvature, so it is indeed a maximum.
Postulate 2’ similarly implies that U is minimized at constant S: dU=0, d2U > 0. Thesecond inequality ensures that U has a positive curvature where its slope vanishes,yielding a minimum.
Similarly, d2G >0 at constant T and P, d2A >0 at constant T, and d2H >0 at constant P.These potentials are derived from energy and minimize Utot for the open system with theappropriate intensive variable. For Massieu functions, d2f < 0, as they are derived fromentropy.
There is a bewildering array of second derivatives from all of these potentials.Fortunately, only three of them (not counting derivatives with respect to mole numbers)are independent, including all the possible potentials. So (∂P/∂T)S, (∂µ/∂n)V, (∂G/∂S)|V,(∂U/∂T)P etc. cannot all be independent of one another.
The reason is that thermodynamic potentials are state functions. For a potentialz=Xdx+Ydy+Σµidni, where X=∂z/∂x|y and Y=∂z/∂y|x, only the following are independentsecond derivatives:
1) ∂X/∂x = ∂2z/∂x2
2) ∂Y/∂y = ∂2z/∂y2
3) ∂X/∂y = ∂2z/∂x∂y
∂Y/∂x = ∂2z/∂y∂x = ∂2z/∂x∂y is not independent of 3. Thus there is no fourth secondderivative. In addition, Maxwell’s relations and Jacobians can be used to calculate thederivatives of any other potential, once three have been picked for potential z. This is sobecause other potentials are related to z by Legendre transforms, and do not containindependent information about stability. They are merely alternative formulations of Utotfor the closed system, cleverly written in terms of only variables of the open subsystem ofinterest.
In practice, chemical problems usually involve constant T and P, so the derivatives of theGibbs free energy are usually used:
The principle: A system is in stable equilibrium if a perturbation induces processes thatrestore equilibrium. (d2S < 0 or d2U > 0)
Example:Let a small amount of heat Q = δU be added to or subtracted from one side of adiathermal rigid wall. The resulting sub-system “2” is now slightly out of equilibrium.T2 = T + δT ≠ T1 on that side. We also know that
dS = dU1
T1+dU2
T2=
1T−
1T + δT
⎛⎝⎜
⎞⎠⎟dU1 =
dU1
T 2 δT
will be the change of entropy as equilibrium is re-established. In the above formula, weused dU1 = -dU2 by postulate 1. The latter equality holds if δT is small enough.
Clearly,
�
dS ≠ 0 for any small energy flow
�
dU1, so the system is out of equilibrium.Equilibrium is restored spontaneously when T2 = T + δT and T1 = T are equalized again.This occurs through heat flow from 2 to 1 through the diathermal membrane. What is thedirection of that heat flow? To satisfy P2 of thermodynamics, dS>0⇒ sign(dU1) = sign(δT ) since T2 > 0.
If δT > 0, then dU1 > 0 and energy flows from 2 to 1 to increase the energy in “1” so T1
increases towards T2. If δT < 0, then dU1 < 0 and energy flows from 1 to 2 to increase theenergy in “2” so T2 increases towards T1.
Therefore, after a perturbation, changes in the system oppose the perturbation.
3) The low T limit
The Nernst Postulate 3 (Planck version) states that limT→0
S = 0 . Microscopically thisimplies only a single microstate is populated, so kBlnW = kBln(1)=0 (a choice of severalmicrostates introduces disorder).
Ideally, this may be true: if two states nominally have the same energy E, some smallcoupling V to the environment will lower the energy one of them, and as T 0, only E1,not E2, is populated.† In practice, the Nernst postulate is not appropriate for systemswhere barriers > kBT: The system may be trapped in a state of higher energy, E2. Eventhough a lower energy state E1 exists, it cannot be reached from E2. Glasses are a goodexample of such a situation: the glass structure is not the energy minimum (the crystal
† In practice, this could be calculated by diagonalizing the matrix H =
E VV E⎛⎝⎜
⎞⎠⎟
. One of
the resulting eigenvalues, E1, will indeed be lower than the other, E2.
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usually is), but dynamics of units within the glass are so slow that the equilibrium cannotbe reached. Thus S > 0 even when T = 0, especially because things don’t move very fastat low temperature.
Often we can get away treating the system as though Nernst’s postulate were trueanyhow. If an experiment is done on a time scale much smaller than the barrier crossingtime, one may take the minimum one happens to be in as ‘the one.” Effectively, theexperiment knows nothing about the lower state on the other side of the barrier.
Because S = S0 +cvTdT
0
T
∫ at constant volume => cv 0 as T 0 (or S is not finite).
Similarly => cp 0 as T 0
If S 0 as T 0, then the S-change of any isothermal process must also go to zero asT 0:
⇒∂S∂P
⎛⎝⎜
⎞⎠⎟ T
→ 0 ⇒∂V∂T
⎛⎝⎜
⎞⎠⎟ P
→ 0 as T → 0 or α → 0 as T → 0 .
Use these consequences of postulate 3 with caution. If you know that your system is notin a minimal energy state as T approaches 0, at least make sure it really can’t get there, sothat for all practical purposes your system is in its minimal energy state.
4) Chemical equilibrium at constant temperature and pressure
Consider a chemical reaction −ν1C1 −ν2C2 + ... ...+ νk−1Ck−1 + νkCk , where we definethe stoichiometric coefficients
�
ν1,ν 2 ...< 0 for reactants and
�
...ν k−1,ν k > 0 for products.This definition allows us to write the reaction as:
ν iCi = 0, {ni (0)}i∑ .
The curly brackets provide the initial condition, e.g. particle number, pressure, orconcentration of all the reactants and products.
Because of mass balance, the changes in ni are not independent:
�
dni = ν i dξ . Here,
�
ξ is a“progress parameter” or “reaction coordinate.” As ξ increases or decreases from 0 (thereaction begins), the concentrations vary in proportion to the stoichiometric coefficients.Integrating
dni = ν i dξ0∫
0∫ ⇒ ni = ni (ξ = 0) + ν i ξ
Thus our initial condition can be written as
�
ξ =0.
At constant (T,P), equilibrium occurs when G is minimized:
Note that the chemical potentials also vary as the reaction progresses, i.e. they depend onξ. For example, the partial pressures Pi of gaseous reactants and products changedepending on ξ as the reaction proceeds, and this changes their chemical potentials, evenfor ideal gases. Differentiating the equation for G, we have in the Gibbs ensemble
�
dG = −SdT +VdP + µi dnii∑ = µi(T,P,ξ)ν i dξ = 0
i∑
at equilibrium and constant T, P. It therefore follows that
=> ∂G∂ξ
⎛⎝⎜
⎞⎠⎟ T ,P
≡ ΔG = µiν i = 0i∑
at equilibrium. This is illustrated in the figure below. Note that our beloved ΔG is notthe free energy of the system. It is the derivative of the free energy, or change in freeenergy has the reaction proceeds by a unit amount . Hence ΔG ~ dG = 0 at equilibrium,while G is minimized and certainly not necessarily 0.
a
ξ
G(ξ)
ξ
ΔG(ξ)0
0 ξmax
(a reactantused up)
dGdξ = 0
> 0< 0dGdξ
dGdξ
Equil.
G(T,P,n(0))i
ni=ni +νiξeq(0)
Figure 7.1: Relationship between free energy G as a function ofreaction coordinate ξ, and the derivative (or change in) free energywith respect to ξ, usually called ΔG(ξ) = ∂G/∂ξ . When G is at aminimum (equilibrium has been reached) its derivative equals zero(ΔG = 0).
For our particle number ni, or our partial pressure Pi, or our concentration ci , we mustsubstitute ci →γ ici under nonideal conditions. But under ideal conditions, we have
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µi = µi(0) + RT lnci , ci =
ni(0) + ν iξV
as proved earlier in chapter 4. If needed, V can of course be calculated from the freeenergy as (∂G/∂P)|T,ni. Inserting the concentration into our definition of ΔG, we get
=> dGdξ
⎛⎝⎜
⎞⎠⎟ T ,P
= µi(0)ν i
i=1
k
∑ + RT ν i lncii=1
k
∑
=> ΔG = ΔG (0) + RT ln{ ciνi
i=1
k
∏ } = ΔG (0) + RT lnQ; Q =...ck−1
νk−1ckνk
c1ν1 c2
ν2 ...
For given T, P, ci, the derivative ΔG gives the change in free energy/mole when we allowthe reaction to proceed by a small amount. The above is known as the mass action law,and describes how the derivative of the free energy changes with concentration (orpressure, etc.). When the derivative reaches 0, equilibrium has been established atconstant T, P.
In the figure, initially, ΔG≠0. At equilibrium, ΔG=0 ⇒
Qeq = Πi=1
kCi,eq
νi = e−ΔG(0 ) (T )
RT ≡ Keq (T )
defines the equilibrium constant. Because ΔG is additive for different reactions, so ln(K)is additive also.
Because G = H – TS =>
�
dGdξ
= dHdξ
−T dSdξ
, or ΔG = ΔH −TΔS . This formula relates the
change of free energy as the reaction progresses to the change of enthalpy and entropy.We can integrate this equation on both sides to obtain:
Gf −Gi = dGi
f
∫ = dξ dGdξ
≡ dξ ΔG(ξ) = dξ (ΔH (ξ) − TΔS(ξ))i
f
∫i
f
∫i
f
∫ .
Integration allows you to get the free energy difference for a finite progress of thereaction. For example, let’s say ΔG = -20 kJ/mole at the initial condition of a reaction. Ifthe reaction is allowed to proceed by some very small amount, say ξ=0.01 moles, then thefree energy G will indeed drop by –0.2 kJ. But this cannot continue: as equilibrium isapproached, ΔG→0, so the free energy will drop less and less until it stops changing. Toget the actual change in free energy Gf −Gi , ΔG needs to be integrated from “i” to “f”.Since the free energy contains the full information about the system, otherthermodynamic quantities can be obtained from it. For example, the entropy can beobtained from
∂G∂T
⎛⎝⎜
⎞⎠⎟ P
= −S⇒ ∂S∂ξ
= ΔS = −∂∂ξ
∂G∂T
⎛⎝⎜
⎞⎠⎟ P
= −∂∂T
∂G∂ξ
⎧⎨⎩
⎫⎬⎭P
=∂ΔG∂T
⎛⎝⎜
⎞⎠⎟ P
.
Similarly, the enthalpy can be obtained from
⇒ ΔH =∂H∂ξ
⎛⎝⎜
⎞⎠⎟ T ,P
= ΔG + TΔS = ΔG − T ∂ΔG∂T
⎛⎝⎜
⎞⎠⎟ P
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Important: note that ΔH and ΔS derived this way are not expressed in terms of theirnatural variables (S and V, or U and V, respectively). Thus their derivatives are notindependent when they are written in this form. You have obtained equations of state,not fundamental relations.
Example reaction The A B cooperative transition. This is a unimolecular reactionwhere the free energy depends on a tunable parameter d. An example would be theunfolding of a protein when the denaturant urea is added to the protein solution. Expandthe dependence of free energy in a Taylor series
ΔG (0) (d) = ΔG (0) (0) + ∂ΔG (0)
∂d⎛⎝⎜
⎞⎠⎟d + 1
2!∂2ΔG (0)
∂d 2⎛⎝⎜
⎞⎠⎟d 2 + ... ≈ ΔG (0) (0) + md ,
where d is a thermodynamic variable (e.g. T, P, or a solute concentration such as [urea]).Because only A or B are present, the mole fractions are χA = nA / (nA + nB ) andχB = nB / (nA + nB ) ; also χA / χB = nA / nB . We can thus write the equilibrium constantas
K(d) = χB
χA
≈ e−ΔG(0 ) (0)+md
RT .
Since χA + χB = 1 , we can rewrite this as
χB =K(d)
1+ K(d)=
e−ΔG(0 ) (0)+md
RT
1+ e−ΔG(0 ) (0)+md
RT
= 1+ eΔG(0 ) (0)+md
RT⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪
−1
.
The figure below shows a plot of this equation, which is a Sigmoid curve. It makes afairly sharp transition with slope –m/4RT at a transition value d0 = ΔG(0)(0)/m. The curveshows us how the concentration of folded vs unfolded protein switches as denaturant isadded.
K = 1⇒ d = −ΔG (0) /m ∂χB
∂d⎛⎝⎜
⎞⎠⎟ d=−ΔG(0 ) /m
=ΔG (0)
m= −
m4RT
a
χB
d
slope ~ -m/(4RT)
midpoint~ΔG(0)/m
Fig 7.2 Titration of reversible reaction A B withparameter d (e.g. temperature, urea concentration, pH, etc.)
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Example: the Van’t Hoff relation for reference enthalpy
Because the temperature dependence of the entropy and enthalpy terms in the free energyis different, we can use the total derivative of the equilibrium constant K to obtaininformation about the enthalpy directly. This is a case where a total derivative, instead ofa partial derivative, can be useful:
lnK = −ΔG (0) (T ,P)
RT⇒
ddTlnK =
ΔG (0)
RT 2 −1RT
∂ΔG (0)
∂T⎛⎝⎜
⎞⎠⎟ P
Note that G, and hence K, are functions of T, P, and ni as independent variables, so in thesecond term above we do not take further derivatives like ∂P/∂T (P is an independentvariable).
concentration. We can substitute this equation relating ΔS(0) and ΔG(0) to remove thederivative in the equation for ln K, obtaining
d lnKdT
=1RT 2 ΔG (0) + TΔS(0){ } = ΔH (0)
RT 2 .
If we know K(T,P), we immediately know ΔH (0) (T ) . Integrating this derivative weobtain
d lnKT1
T2
∫ = ln K2
K1=
ΔH (0) (T )RT 2
T1
T2
∫ dT ;
Changes in K(T) can be obtained from ΔH (0) only, without explicit use of ΔG (0) or ΔS(0) .This is a nice bonus when you are doing calorimetry because enthalpy is much easier tomeasure calorimetrically than free energy or entropy.
Example A B reaction with Δcp = cpB − cpA ≠ 0 .We will assume that pressure is constant (hence cP), but we want to determine thetemperature dependence of the equilibrium constant. Phase transitions from phase A tophase B often involve a change in heat capacity between states. Using the superscript (P)
to indicate a constant reference pressure (usually 1 atm), we can write for the enthalpyand entropy
Note: ΔS(T ) can indeed be derived from ΔG(T ) by partial derivative, and ΔG = 0 solvesthe free energy minimum:
∂ΔG (P )
∂T= ΔS(P ) (T ) − Δcp ln T
T0
= ΔS(T ); equilibrium: ΔG (P ) = 0
Note: as in the case of ΔG, ΔCP is not the heat capacity of the reaction mixture afterreaction minus the heat capacity before reaction, unless the reaction starts with pure Aand goes to pure B. Rather, ΔcP = ∂CP / ∂ξ just like the case of ΔG. Consider it in detailfor the simple A B reaction. Let’s take as reaction coordinate nB, the number ofmoles of B. Thus nA = n – nB. The heat capacity of the reaction mix is cp = cpAnA + cpBnB= cp = cpAn + (cpB − cpA )nB (ideal mixture case). Taking the derivative,∂cp / ∂ξ = ∂cp / ∂nB = cpB − cpA = ΔcP .
Example Interpretation of ΔG,ΔH , and ΔS for system and environment
The Gibbs free energy is minimized for an open system in contact with a T, P reservoir:
dG = 0, d 2G > 0 at equilibrium ⇒ ∂G∂ξ
= ΔG < 0 for a spontaneous direction of progress ξ
For a moment, let us explicitly write “sys” for the system thermodynamic variables:ΔGsys = ΔHsys − TΔSsys
For a quasistatic heat release, ΔHsys = Qsys = −Qres = −TΔSres .
We can combine the two entropies into⇒ ΔGsys = −T (ΔSsys + ΔSres ) = −TΔStot
As we showed earlier in conjunction with the Helmoltz potential (Ch. 4), when thesystem is connected to a bath that keeps intensive parameters Iij fixed (e.g. IUS = T =∂U/∂S)V,n), then the potential whose natural variables are those Iij is minimized when thetotal energy is minimized (equivalently, when the total entropy is maximized):
ΔStot > 0&T > 0⇒ ΔGsys < 0 .
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Note that postulate P2 (the “Second Law”) is not violated if virtually all the available freeenergy is converted to work: if W
< ΔGsys , ΔStot remains slightly positive. ThereforeΔG is the maximum amount of work available form a constant T, P reaction per mole ofreaction. W > ΔGsys would imply ΔStot < 0 , violating postulate 2.
5) Partial molar quantities: derivatives with respect to ni instead of ξ
The thermodynamic potentials, including the Gibbs potential, are rather unusual functionsbecause not only is dG = −SdT +VdP + µdn , but in fact G =U − TS + PV = µn for thecase of a simple system. It is true that dG / dn = ∂µ / ∂n n + µdn / dn = n∂µ / ∂n + µ , butthis is irrelevant from the point of view of thermodynamic partial derivatives. Therelevant partial derivative is ∂G / ∂n = µ(T ,P,n) , and it is also a function of T, P, and nin the Gibbs ensemble.
We saw when deriving the Gibbs free energy by Legendre transform that for a multi-component system G = Σµini . From this follows for each component at constanttemperature and pressure:
dG = µidni i∑ (not dG = µidni + nidµi
i∑ because of the Gibbs-Duhem relation!)
and dGdni
⎛⎝⎜
⎞⎠⎟ T ,P
= µi .
The partial molar derivative of the Gibbs free energy is the chemical potential ofcomponent “i”, and the Gibbs free energy can be obtained simply by adding thesechemical potentials multiplied by the mole numbers.We can similarly define
∂V∂ni
⎛⎝⎜
⎞⎠⎟ T ,P
= ν i; ∂S∂ni
⎛⎝⎜
⎞⎠⎟ T ,P
= si; ∂H∂ni
⎛⎝⎜
⎞⎠⎟ T ,P
= hi = µi + Tsi
These are the partial molar volumes, entropies and enthalpies. It is important to note thatwhile G = Σµini , it is absolutely not true that V = Σvini , etc. Only the Gibbs free energyis the particular Legendre transform of U for which that simple formula holds. Thinkabout it physically: if you mix two liquids, nonideal behavior could actually cause thetotal liquid volume to shrink (strong attraction between the two substances), so the partialmolar volume under that condition for the substance added would be negative! Clearly,you could not just add up those partial volumes and get the total volume. Rather, thepartial molar volume tells you how much, under a given composition, addition of aninfinitesimal amount of substance ni changes the total volume – including a decrease involume. Only for G (or ΔG) do the partial molar quantities (chemical potentials) simplyadd.
We can rewrite partial molar quantities by using Maxwell’s relations. For example,
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−∂S∂ni
⎛⎝⎜
⎞⎠⎟ T ,P,nj≠i
=∂µi∂T
⎛⎝⎜
⎞⎠⎟ ni ,P
or ∂V∂ni
⎛⎝⎜
⎞⎠⎟ T ,P,nj≠i
=∂µi∂P
⎛⎝⎜
⎞⎠⎟ ni ,T
Thus the partial molar quantities can all be derived from the chemical potential withoutexplicit derivatives with respect to the ni. Only the chemical potential itself needs aderivative of G with respect to the ni. The total variation in a variable Y with respect tomole numbers at constant T, P can always be written as
dYT ,P =∂Y∂ni
⎛⎝⎜
⎞⎠⎟dni = yidni
i∑ =
i∑ ∂µi
∂X⎛⎝⎜
⎞⎠⎟ ni
dnii∑ ,
where X and Y are conjugate variables. (In some cases, the conjugate variable such as Pmay be (-) a derivative, so don’t forget the (-) sign in such cases.)
Example Entropy of mixing of two ideal gasesFor an ideal gas,
µi = µi(0) + RT lnPi ⇒ si = −
∂µi∂T
⎛⎝⎜
⎞⎠⎟= −
∂µi(0)
∂T⎛⎝⎜
⎞⎠⎟− R lnPi = si
(0) − R lnP .
For two components, the total change in entropy upon a small change in mole numbers isdS = s1dn1 + s2dn2 = (s1
(0) − R lnP1)dn1 + (s2(0) − R lnP2 )dn2
For an ideal gas, si is independent of ni and we can integrate. Consider a 2-componentideal gas initially in two containers of volumes V1 and V2. Before and after mixing,
Sinit = n1s1(0) + n2s2
(0) − Rn1 lnP1 − Rn2 lnP2
Sfinal = n1s1(0) + n2s2
(0) − Rn1 ln P1V1
V1 +V2
⎛⎝⎜
⎞⎠⎟− Rn2 ln P2
V2V1 +V2
⎛⎝⎜
⎞⎠⎟
The difference between these two entropies is the entropy of mixing:
ΔSmix = −R n1 lnV1
V+ n2 lnV2
V⎛⎝⎜
⎞⎠⎟> 0 or
ΔSmix / n = −R(χ1 ln χ1 + χ2 ln χ2 ) because nin= χi ,
ViV
= χi
This is always greater than zero because the arguments of the logarithm are less than 1, niis positive, and -R is negative. Unmixing is therefore not a spontaneous process.
A bonus question to think about: let us say we discover after we pull out the thinmembrane between our two gases that the two gases were of the same atom, e.g. Ar. Aratoms are identical particles, so mixing the left and right sides would seem to changenothing detectable about the overall system: even if the left ones moved right and vice-versa, you would never be able to tell. Yet the above formula still says ΔSmix > 0.Granted, the two types of particles corresponding to χ1 and χ2 are now the same, butaren’t you allowing them to move around more, thus increasing the entropy? The answeris no.
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Fittingly, this situation was discovered by Gibbs himself, and is known as the Gibbsparadox. A satisfactory resolution, first provided by John von Neumann, requires theproper quantum mechanical treatment of identical particles, either Bosons or Fermions.We’ll deal with this when we do statistical mechanics. Suffice it to say for now that theclassical description with the wall neglects the possibility of quantum mechanicaltunneling and exchange of identical particles. Classical mechanics always (andincorrectly) treats every particle as unique and distinguishable.
