H.2 Euler-Maclaur in Sum Formula
3.4 Combined First and Second Laws
For a chemically closed system, the first law gives
dU=δQ−δW. (3.44)
For a simple7homogeneous isotropic system for whichUdepends only onSandV, dU=
∂U
∂S
V
dS+ ∂U
∂V
S
dV. (3.45)
For a reversible transformation in this system, for which the only work is the quasistatic work, we have
δQ=TdS; δW=pdV; reversible. (3.46)
Substitution of Eq. (3.46) into Eq. (3.44) gives
dU=TdS−pdV. (3.47)
We can therefore identify the derivatives T=∂U
∂S
V
; −p=∂U
∂V
S
. (3.48)
We emphasize that Eq. (3.47) holds for all infinitesimal changes ofU(S,V)within the field of equilibrium states. Equation (3.46), which is only true for reversible processes, was only used to identify the derivatives in Eq. (3.45). Equations that give explicit forms of the functionsT(S,V)andp(S,V)are known asequations of state. If all8equations of state are known, Eq. (3.47) can be integrated to recover the functionU(S,V), except for an additive constant which has to do with the arbitrary zero of energy. If the second partial derivatives ofUare continuous, as we shall assume to be the case for thermodynamic functions, the order of partial differentiation does not matter and we obtain
∂T
∂V
S
= ∂2U
∂V∂S = ∂2U
∂S∂V = − ∂p
∂S
V
. (3.49)
(∂T/∂V)S= − ∂p/∂S
V is an example of aMaxwell relation. In Chapter 5 we will take up Maxwell relations for systems that depend on several variables.
Since Eq. (3.44) holds even for irreversible transformations and Eq. (3.47) is generally true, we can eliminate dUto obtain
pdV−δW=TdS−δQ. (3.50)
7Note that Eqs. (3.45) and (3.47) hold only for a chemically closed system in which no chemical reactions are occurring. If chemical reactions are allowed,Uwould depend on additional variables (progress variables of the reactions). Equation (3.6) would not hold if these reactions were irreversible. See Eq. (5.128) for further clarification.
8For open systems, one must include the numbers of moles of each chemical component,N1,N2,. . .,Nκas additional variables inU, in which case there are more equations of state (see Chapter 5). In general,Udepends on a complete set of extensive state variables.
42 THERMAL PHYSICS
For reversible transformations, Eq. (3.46) holds and both sides of Eq. (3.50) are zero. But for an irreversible process, Eq. (3.46) no longer applies. Instead, Eq. (3.5) applies and Eq. (3.50) leads to an interesting inequality. We divide Eq. (3.50) byTand rearrange to obtain
pdV−δW T +δQ
T =dS. (3.51)
Then we subtractδQ/Tsfrom both sides of Eq. (3.51) and apply Eq. (3.5) to obtain pdV−δW
T +δQ 1
T − 1 Ts
=dS−δQ
Ts >0, natural, irreversible. (3.52) The first term on the left of Eq. (3.52) is due to the process of irreversible work and the second term on the left is due to the irreversible process of heat conduction between the external source and the system. These terms can be regarded [16, pp. 95-95] as repre- senting entropy production during independent irreversible processes and areseparately positive. A positive value of the first term leads to the inequalityW <pdV, in agreement with Eq. (2.5). If we substituteW = pextdV wherepext is an effective pressure of purely mechanical origin as in Eq. (2.3), this work inequality becomes (p−pext)dV > 0. The second term is the same as in Eq. (3.7), derived for the case in which the system was considered to be a heat source that could do no work.
We can rearrange Eq. (3.47) in the form dS= 1
T dU+p
T dV (3.53)
from which it follows that 1 T =
∂S
∂U
V
; p T =
∂S
∂V
U
. (3.54)
Equations that give 1/T andp/T as functions ofU andV are also equations of state. If we know these functions, Eq. (3.53) can be integrated to recoverS(U,V). We also have the Maxwell relation(∂(1/T)/∂V)U = ∂(p/T)/∂U
V.
Since the entropy is postulated to be a monotonically increasing function of the internal energy, the internal energy is also a monotonically increasing function of the entropy. The inverse transformation between S(U,V)and U(S,V) is therefore unique, and either of these functional forms can be chosen to give a complete representation of the thermodynamic system.9 One speaks of the entropy representation S(U,V) or the energy representation U(S,V). Either of these equations can be regarded as afun- damental equation of the systemand either contains complete information about the system.
9For more complicated systems, bothSandUdepend on an additional set of extensive variables, but these behave just likeV.
Example Problem 3.4. For a hypothetical thermodynamic system,T = (4/A)(U/V)3/4and p=3U/V, whereAis a constant. Find the fundamental equation in the entropy representation.
Solution 3.4. We readily calculate 1/T = (A/4)(V/U)3/4 and p/T = (3A/4)(U/V)1/4 so Eq. (3.53) takes the form
dS=(A/4)(V/U)3/4dU+(3A/4)(U/V)1/4dV, (3.55) which integrates to giveS=AU1/4V3/4+S0, whereS0is a constant.
Example Problem 3.5. This problem concerns one mole of an ideal monatomic gas that obeys the equation pV = RT, where p is the pressure, V is the volume, T is absolute temperature, andRis the universal gas constant. The gas has a heat capacity (per mole) at constant volume ofCV =(3/2)R. In its initial state, it is in equilibrium at temperatureT1and volumeV1in the left chamber of a box, as shown inFigure 3–2. The right chamber of the box, which has volumeV2−V1, is initially evacuated. The two chambers are surrounded by exterior walls that are rigid and impenetrable. The chambers are separatedinitiallyby an interior wall that is rigid, impenetrable, and insulating. Under various conditions detailed below, the gas is allowed to expand and finally comes to equilibrium in the total volumeV2.
Apply the first and second laws of thermodynamics, the definition ofCV, the ideal gas equation of state, and integration to answer the following questions.
(a) Suppose, by whatever means, that the gas expands into the total volumeV2and comes to equilibrium at temperatureT2. What is the change,S, in entropy of the gas from its initial to its final state?
(b) The entire system is maintained at constant temperatureT1by contact with a heat reser- voir. The gas is allowed to expand by means of an external agent that moves the internal wall separating the chambers very slowly (such that the gas is practically in equilibrium at each stage of the process) until the gas occupies the entire volumeV2. What is the change,U, in its internal energy? How much external work,W, does the system do on the external agent that moves the wall? How much heat,Q, is added to the system during this process? CompareQto the relevantSand deduce whether this process is reversible or irreversible.
V1 V2−V1
Ideal gas Vacuum
FIGURE 3–2 A monatomic ideal gas at temperatureT1initially occupies the left chamber of the box. The right chamber of the box, which has volumeV2−V1, is evacuated. The interior wall that separates the gas from the evacuated chamber is rigid, impenetrable and insulating, but can be moved or ruptured.
44 THERMAL PHYSICS
(c) The entire system is insulated and the wall separating the chambers is suddenly ruptured, allowing the gas to fill the entire volumeV2. How much external work,W, does the system do? What is the final temperature of the gas? CompareQto the relevantSand deduce whether this process is reversible or irreversible.
(d) The entire system is insulated. The gas is allowed to expand by means of an external agent which moves the internal wall separating the chambers very slowly (such that the gas is practically in equilibrium at each stage of the process) until the gas occupies the entire volume V2. What is the final temperature, T2, of the gas?
Compare Q to the relevant S and deduce whether this process is reversible or irreversible.
Solution 3.5.
(a) SinceSis a state function,Sdepends only on the initial and final states of the system, irrespective of how the system gets from the initial state to the final state. We substitute the ideal gas law and the equation dU=CVdTinto Eq. (3.53) to obtain
dS=CV
dT T +RdV
V , (3.56)
which integrates to give
S=CVln(T2/T1)+Rln(V2/V1), one mole of ideal gas. (3.57) (b) Udepends only onTfor an ideal gas, soU=0. Thus from the first law,W=Q. Since the
work is quasistatic,W=
pdVwhere the integral is to be carried out along an isothermal pathT = T1. Therefore we can usep = RT1/V and take the constantsRT1outside the integral to obtain
Q=W=RT1
V2 V1
dV
V =RT1ln(V2/V1). (3.58) SinceT2=T1for this process, part (a) givesS=Rln(V2/V1)so
S=Q/T1 (3.59)
and the process is reversible (as expected for quasistatic work). Note that the entropy increases for this reversible process. In this case, entropy increase does not automatically imply irreversibilitybecause the system is not isolated. Similarly, for reversible adiabatic contraction, bothQandSare negative, and the entropy of the system decreases. This does not violate Eq. (3.1)because the system is not isolated.
(c) W = 0 because the outer wall is rigid and there is no way to do mechanical work on the environment of the system. SinceQ=0, we conclude from the first law thatU=0.
SinceUdepends only onT, we haveT2=T1. (During the process itself, which we shall see is irreversible,T is at best inhomogeneous and probably undefined.) The change in entropy, from part (a), is againS=Rln(V2/V1) >0. Therefore, sinceδQ=0 at every stage of the process,
S>
δQ
T =0, (3.60)
so the process is irreversible as expected.
(d) Q = 0 because the system is insulated. The work is quasistatic soδW = pdV, and since δQ=0 at each stage of the process, the first law gives dU+pdV =0. Since dU =CVdT, this becomesCVdT+RTdV/V =0. Division byT(which isnotconstant in this process) yieldsCVdT/T+RdV/V=0 which integrates to give
CVln(T2/T1)+Rln(V2/V1)=0. (3.61) Thus,S=0 and
S=
δQ T =0,
so the process is reversible and isentropic, as expected for this quasistatic process with adiabatic walls. By means of Eq. (2.27), the final temperature can be written more succinctly asT2= T1(V1/V2)2/3, so the temperature drops, as expected, for this reversible adiabatic expansion.
3.4.1 Latent Heat
When a substance melts or evaporates, heat must be supplied to partially or totally break atomic bonds and rearrange structure, and hence to change the phase to a state of higher disorder, which we shall see later is a state of higher entropy. Melting and vaporization pro- cesses are generally carried out at constant pressure, for example, atmospheric pressure.
The heat needed to change the phase reversibly at constant pressure and temperature is known aslatent heat. Heat must be supplied when a solid melts to become a liquid (heat of melting); the same amount is given up when a liquid freezes to become a solid (latent heat of fusion). When a liquid becomes a gas, it is necessary to supply heat (heat of vaporization); when a gas condenses to become a liquid, the same amount of heat is given up (latent heat of condensation). These are generally reported as positive quantities, usually per mole or per unit mass.
Consider, for example, the melting of ice, which takes place at atmospheric pressure at a temperature of 0◦C = 273.15 K. As we supply heat to cold ice, it is warmed from below its melting point to 273.15 K where melting occurs and water begins to form. As heat continues to be supplied, the ice-water mixture remains at 273.15 K until all of the ice melts. This requires 80 calories of heat per gram of ice, the latent heat of fusion. Further heating causes the temperature of the water to rise.
Processes such as this, which take place at constant pressure, may be analyzed conve- niently in terms of the enthalpy,H=U+pVpreviously introduced in connection with the first law (see Section 2.5). We saw that dH=dU+pdV+Vdpwhich in view of Eq. (3.47) becomes
dH =TdS+Vdp. (3.62)
But at constant pressure we have
dH=CpdT, (3.63)
46 THERMAL PHYSICS
whereCpis the heat capacity at constant pressure. Equation (3.63) applies in the absence of phase change, say forTI ≤ T < TMand also forTM < T ≤ TW, whereTIis the initial temperature of the ice,TMis the melting point andTWis the final temperature of the water.
AtT =TM,Hincreases by the amountHM, the latent heat of fusion. The total change in His therefore
H= TM
TI
Cp(ice)dT+HM+ TW
TM
Cp(water)dT. (3.64)
H as a function ofT is shown inFigure 3–3a. Formally, the effective heat capacity at the melting point can be represented as a delta function (the formal derivative of a step function) as shown in Example Problem 2.4.
By combining Eq. (3.62) with Eq. (3.63) at constant pressure, we obtain dS=Cp
T dT, (3.65)
which can be integrated to find the entropy change that occurs prior to melting and subsequent to melting. During the melting itself, we integrate10Eq. (3.62) at constantpto obtainSM=HM/TM, which is called the entropy of fusion. Therefore, the total change of entropy is given by
S= TM TI
Cp(ice)
T dT+HM
TM + TW TM
Cp(water)
T dT. (3.66)
Sas a function ofTis shown inFigure 3–3b.
If the range of temperature is not large,Cp(ice)andCp(water)can be considered to be practically independent ofT, so we have the simplifications
H≈Cp(ice)(TM−TI)+HM+Cp(water)(TW−TM) (3.67)
265 270 275 280 285 290 295 250
500 750 1000 1250 1500 1750
ΔHM
(a)
265 270 275 280 285 290 295 1
2 3 4 5 6 7
ΔSM
(b)
FIGURE 3–3 (a) Enthalpy changeHin cal/mol and (b) entropy changeSin cal/(mol K) as a function of temperature Tin K for melting of ice. The curvature of the logarithms inSis not apparent on this scale. The jumps are related byHM=TMSM. (a) EnthalpyHversusTand (b) EntropySversusT.
10We assume that the whole process is done slowly and carefully so that it is reversible.
and
S≈Cp(ice)lnTM
TI +HM
TM +Cp(water)lnTW
TM
. (3.68)
To get an idea of the magnitudes involved, we approximateCp(ice)≈Cp(water)≈1 cal/g K, takeTI= −10◦C andTW=20◦C. Then for every mole of H2O (18 g/mol) we have
H=(189+1440+351)cal/mol=1980 cal/mol (3.69) and
S=(0.67+5.27+1.27)cal/K mol=7.21 cal/K mol. (3.70) For many monatomic substances,SM=HM/TM ∼R≈2 cal/K mol, an empirical rule known as Richard’s rule. For ice, the entropy of fusion is much larger (5.27 cal/K mol) because of the complexity of the H2O molecule. For vaporization, a similar empirical rule known asTrouton’s ruleleads to the estimateSV = HV/TV ∼ 10.5R≈ 21 cal/K mol, as compared to 26 cal/K mol for water. The fact that the entropy of vaporization is larger than the entropy of fusion is because essentially all atomic bonds must be broken for evaporation and because of the large volume change from liquid to gas.