H.2 Euler-Maclaur in Sum Formula
4.2 Implications of the Third Law
The third law has certain implications regarding heat capacities and other properties of materials asT →0. From Eq. (3.47) with dV=0, we obtainCV dT=TdSwhereCV is the heat capacity at constant volume. The change in entropy at constant volume from one temperature to another is given by
CV/TdT. Thus S(T1,V)= T1
0
CV(T,V)
T dT. (4.2)
In order for this integral to converge, it is necessary forCV to depend onT in such a way thatCV → 0 asT → 0. Recall thatCV was taken to be a constant for an ideal gas; clearly such an ideal gas becomes impossible asT → 0. For insulating solids, one finds both theoretically and experimentally thatCV ∝T3asT →0. For metals, nearly free electrons contribute to the heat capacity andCV ∝TasT →0. Similar considerations apply to the heat capacity at constant pressure. From Eq. (3.62) with dp=0, we obtainCpdT=TdS, whereCpis the heat capacity at constant pressure. Thus
S(T1,p)= T1 0
Cp(T,p)
T dT (4.3)
and it is necessary2forCp→0 asT →0.
An interesting experimental verification of the third law has been discussed by Fermi [1, p. 146]. At temperatures below T0=292 K, gray tin (α, diamond cubic) is the stable form and above this temperature, white tin (β, tetragonal) is stable. These are allotropic forms of pure tin. It turns out, however, that white tin can exist (in internal equilibrium) below 292 K, even though it is unstable with respect to transformation to gray tin. It is also
1This is the changeGof the Gibbs free energy of the reaction, whose definition and properties we explore later.
2In order for the integrals in Eqs. (4.2) and (4.3) to converge atT=0, it will suffice forCV orCpto go to zero very weakly asT→0, for instance∝Twhere >0.
S
Gray
White
292 K
0 K T
FIGURE 4–1 EntropiesSof gray and white tin as a function of absolute temperatureT. BelowT0=292 K, gray tin is stable and above this temperature white tin is stable. The full curves denote stable phases and the dashed curves denote unstable phases. White tin can be supercooled belowT0so its heat capacity can be measured and its entropy can be calculated. The jump in entropy atT0between gray tin and white tin is due to the latent heat of transformation.
possible to measure the heat capacities of both forms of tin down to very low tempera- tures. One can therefore evaluate the entropy of white tin at 292 K in two different ways, the first by integrating its heat capacity from absolute zero and the second by integrating the heat capacity of gray tin from absolute zero and then adding the entropy associated with transformation to white tin at 292 K. SeeFigure 4–1for a graphic illustration. Thus (with subscriptsgandwfor gray and white), we have
Sw(292K)= 292K 0
Cw(T)
T dT=12.30 cal/mol K, (4.4)
and
Sg(292K)= 292K 0
Cg(T)
T dT =10.53 cal/mol K. (4.5)
The heat of transformation from gray to white tin isHg→w=535 cal/mol so the entropy of transformation is Sg→w=Hg→w/T0=535/292=1.83 cal/mol K. Adding this to the result of Eq. (4.5) gives 12.36 cal/mol K, in reasonable agreement with Eq. (4.4).
The third law can also shed light on the behavior of the coefficient of thermal expan- sion,α, and the compressibility,κT, asT →0. SinceS→0 asT →0 independent ofVor p, one has
∂S
∂V
T=0=0;
∂S
∂p
T=0=0. (4.6)
Through a Maxwell relation (see Eq. (5.90)), it can be shown that ∂S
∂p
T
= − ∂V
∂T
p
= −Vα, (4.7)
where α is the coefficient of isobaric thermal expansion. Indeed, it has been verified experimentally thatα→0 asT →0. By means of another Maxwell relation (see Eq. (5.86))
52 THERMAL PHYSICS
∂S
∂V
T = ∂p
∂T
V =α/κT, (4.8)
whereκTis the coefficient of isothermal compressibility. Thus,κTmust either remain non- zero asT →0 or go to zero more slowly thanα.
See Lupis [5, pp. 21-23] for further discussion of experimental verification of the third law as well as a discussion of some of its other consequences, particularly consequences concerning chemical reactions. See Fermi [1, p. 150] for an excellent discussion of the entropy of mercury vapor.
5
Open Systems
Until now we have dealt with chemically closed thermodynamic systems in which there is no exchange of chemical components with the environment. Such chemically closed systems can receive heatQfrom the environment and do workWon their environment.
Their change in internal energy is given byU =Q−W, which for infinitesimal changes is dU = δQ−δW. For reversible changes in a simple isotropic system, the (quasistatic) work isδW = pdV, wherepis the pressure andV is the volume. The heat received in a reversible change isδQ=TdS, whereT is the absolute temperature andSis the entropy.
If the mole numbers of each chemical component are constant (no chemical reactions), the combined first and second laws (see Chapter 3) lead to
dU=TdS−pdV. (5.1)
Open systems can exchange chemical components with their environment. Conse- quently the number of moles of each chemical component,Ni, fori = 1, 2,. . .,κ, are variables. This requires several modifications. The first law must be amended to read
U=Q−W+Ech, (5.2)
whereEchis the energy (sometimes called chemical heat) that is added to the system when chemical components are exchanged with its environment. Moreover,Unow becomes a function ofS,Tand all of theNi, so additional terms are needed in Eq. (5.1). This also sets the stage for changes ofNidue to chemical reactions within the system, which can even occur for a chemically closed system for whichEch =0. We shall first treat an open system having a single component and then go on to treat multicomponent systems.