THERMODYNAMIC RELATIONS FOR A SYSTEM IN EQUILIBRIUM

REVIEW PROBLEMS

4.4 THERMODYNAMIC RELATIONS FOR A SYSTEM IN EQUILIBRIUM

manner, then dw P dV dwnon-P-V; Eq. (4.23) becomes dG dwnon-P-Vor G  w_non-P-V wby,non-P-V. Therefore

(4.24) For a reversible change, the equality sign holds and w_by,non-P-V G. In many cases (for example, a battery, a living organism), the P-V expansion work is not useful work, but w_by,non-P-Vis the useful work output. The quantity G equals the maximum possi-ble nonexpansion work output w_by,non-P-V done by a system in a constant-T-and-P process. Hence the term “free energy.” (Of course, for a system with P-V work only, dw_by,non-P-V 0 and dG 0 for a reversible, isothermal, isobaric process.) Examples of nonexpansion work in biological systems are the work of contracting muscles and of transmitting nerve impulses (Sec. 13.15).

Summary

The maximization of S_univ leads to the following equilibrium conditions. When a closed system capable of only P-V work is held at constant T and V, the condition for material equilibrium (meaning phase equilibrium and reaction equilibrium) is that the Helmholtz function A (defined by A⬅ U TS) is minimized. When such a system is held at constant T and P, the material-equilibrium condition is the minimization of the Gibbs function G⬅ H TS.

4.4 THERMODYNAMIC RELATIONS FOR A SYSTEM

Chapter 4 Material Equilibrium

116

The heat capacities C_V and C_P have alternative expressions that are also basic equa-tions. Consider a reversible flow of heat accompanied by a temperature change dT. By definition, C_X dqX/dT, where X is the variable (P or V ) held constant. But dq_rev T dS, and we have C_X T dS/dT, where dS/dT is for constant X. Putting X equal to V and P, we have

(4.31)*

The heat capacities C_Pand C_Vare key properties since they allow us to find the rates of change of U, H, and S with respect to temperature [Eqs. (4.29) to (4.31)].

The relation dU T dS P dV in (4.25) applies to a reversible process in a closed system. Let us consider processes that change the system’s composition. There are two ways the composition can change. First, one can add or remove one or more sub-stances. However, the requirement of a closed system (dU dq dw for an open sys-tem) rules out addition or removal of matter. Second, the composition can change by chemical reactions or by transport of matter from one phase to another in the system.

The usual way of carrying out a chemical reaction is to mix the chemicals and allow them to reach equilibrium. This spontaneous chemical reaction is irreversible, since the system passes through nonequilibrium states. The requirement of reversibility (dq T dS for an irreversible chemical change) rules out a chemical reaction as ordi-narily conducted. Likewise, if we put several phases together and allow them to reach equilibrium, we have an irreversible composition change. For example, if we throw a handful of salt into water, the solution process goes through nonequilibrium states and is irreversible. The equation dU T dS P dV does not apply to such irreversible composition changes in a closed system.

We can, if we like, carry out a composition change reversibly in a closed system.

If we start with a system that is initially in material equilibrium and reversibly vary the temperature or pressure, we generally get a shift in the equilibrium position, and this shift is reversible. For example, if we have an equilibrium mixture of N₂, H₂, and NH₃ (together with a catalyst) and we slowly and reversibly vary T or P, the position of chemical-reaction equilibrium shifts. This composition change is reversible, since the closed system passes through equilibrium states only. For such a reversible composi-tion change, dU T dS P dV does apply.

This section deals only with reversible processes in closed systems. Most com-monly, the system’s composition is fixed, but the equations of this section also apply to processes where the composition of the closed system changes reversibly, with the system passing through equilibrium states only.

The Gibbs Equations

We now derive expressions for dH, dA, and dG that correspond to dU T dS P dV [Eq. (4.25)] for dU. From H⬅ U PV and dU T dS P dV, we have

(4.32) Similarly,

where (4.32) was used.

S dT V dP

dG d1H TS2 dH T dS S dT T dS V dP T dS S dT

S dT P dV

dA d1U TS2 dU T dS S dT T dS P dV T dS S dT dH T dS V dP

1T dS P dV2 P dV V dP

dH d1U PV 2 dU d1PV 2 dU P dV V dP C_V T a0S

0Tb

, CP T a0S 0Tb

closed syst. in equilib.

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Collecting the expressions for dU, dH, dA, and dG, we have

These are the Gibbs equations. The first can be written down from the first law dU dq dw and knowledge of the expressions for dwrevand dq_rev. The other three can be quickly derived from the first by use of the definitions of H, A, and G. Thus they need not be memorized. The expression for dG is used so often, however, that it saves time to memorize it.

The Gibbs equation dU T dS P dV implies that U is being considered a func-tion of the variables S and V. From U U(S, V), we have [Eq. (1.30)]

Since dS and dV are arbitrary and independent of each other, comparison of this equa-tion with dU T dS P dV gives

(4.37) A quick way to get these two equations is to first put dV 0 in dU T dS P dV to

give ( _V _S P.

[Note from the first equation in (4.37) that an increase in internal energy at constant vol-ume will always increase the entropy.] The other three Gibbs equations (4.34) to (4.36)

give in a similar manner ( _P _S _V _T P,

and

(4.38) Our aim is to be able to express any thermodynamic property of an equilibrium system in terms of easily measured quantities. The power of thermodynamics is that it enables properties that are difficult to measure to be expressed in terms of easily mea-sured properties. The easily meamea-sured properties most commonly used for this purpose are [Eqs. (1.43) and (1.44)]

(4.39)*

Since these are state functions, they are functions of T, P, and composition. We are considering mainly constant-composition systems, so we omit the composition depen-dence. Note that a and k can be found from the equation of state V V(T, P) if this is known.

The Euler Reciprocity Relation

To relate a desired property to C_P, a, and k, we use the basic equations (4.25) to (4.31) and mathematical partial-derivative identities. Before proceeding, there is another partial-derivative identity we shall need. If z is a function of x and y, then [Eq. (1.30)]

(4.40) dz a0z

0xb

dx a0z 0yb

dy⬅ M dx N dy C_P1T, P2, a1T, P2 ⬅ 1

V a0V 0Tb

, k1T, P2 ⬅ 1 V a0V

0Pb

a0G 0Tb

S, a0G 0Pb

V a0U

0Sb

T, a0U 0Vb

P dU a0U

0Sb

dS a0U 0Vb

dV dU T dS P dV

dH T dS V dP dA S dT P dV dG S dT V dP

u closed syst., rev. proc., P-V work only

Section 4.4 Thermodynamic Relations for a

System in Equilibrium 117

(4.33)*

(4.34) (4.35) (4.36)*

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118

where we defined the functions M and N as

(4.41) From Eq. (1.36), the order of partial differentiation does not matter:

(4.42) Hence Eqs. (4.40) to (4.42) give

if (4.43)*

Equation (4.43) is the Euler reciprocity relation. Since M appears in front of dx in the expression for dz, M is ( _yand since we want to equate the mixed second partial derivatives, we take ( _x.

The Maxwell Relations

The Gibbs equation (4.33) for dU is

The Euler relation ( _x _ygives

Application of the Euler relation to the other three Gibbs equations gives three more thermodynamic relations. We find (Prob. 4.5)

(4.44)

(4.45) These are the Maxwell relations (after James Clerk Maxwell, one of the greatest of nineteenth-century physicists). The first two Maxwell relations are little used. The last two are extremely valuable, since they relate the isothermal pressure and volume vari-ations of entropy to measurable properties.

The equations in (4.45) are examples of the powerful and remarkable relationships that thermodynamics gives us. Suppose we want to know the effect of an isothermal pressure change on the entropy of a system. We cannot check out an entropy meter from the stockroom to monitor S as P changes. However, the relation ( _T

Pin (4.45) tells us that all we have to do is measure the rate of change of the system’s volume with temperature at constant P, and this simple measurement enables us to calculate the rate of change of the system’s entropy with respect to pressure at constant T.

Dependence of State Functions on T, P, and V

We now find the dependence of U, H, S, and G on the variables of the system. The most common independent variables are T and P. We shall relate the temperature and pres-sure variations of H, S, and G to the directly measurable properties C_P, a, and k. For U, the quantity ( _Toccurs more often than ( _T, so we shall find the tempera-ture and volume variations of U.

a0S 0Vb

a0P 0Tb

, a0S

0Pb

a0V 0Tb

a0T 0Vb

a0P 0Sb

, a0T 0Pb

a0V 0Sb

10T>0V2S 3 01P2>0S4V 10P>0S2V

dU T dS P dV M dx N dy where M ⬅ T, N ⬅ P, x ⬅ S, y ⬅ V dz M dx N dy

a0M 0y b

a0N 0xb

0 0ya0z

0xb 0 0xa0z

0yb M⬅ 10z>0x2y, N ⬅10z>0y2x lev38627_ch04.qxd 2/29/08 3:13 PM Page 118

Volume Dependence of U

We want ( _T, which was discussed at the end of Sec. 2.6. The Gibbs

equa-tion (4.33) gives dU _Tcorresponds to

an isothermal process. For an isothermal process, the equation dU T dS P dV becomes

(4.46) where the T subscripts indicate that the infinitesimal changes dU, dS, and dV are for a constant-T process. Since ( _Tis wanted, we divide (4.46) by dV_T, the infinitesi-mal volume change at constant T, to give

From the definition of a partial derivative, the quantity dU_T/dV_Tis the partial derivative ( _T, and we have

Application of the Euler reciprocity relation (4.43) to the Gibbs equation dA

T V

[Eq. (4.45)]. Therefore

(4.47) where ( _V a/k [Eq. (1.45)] was used. Equation (4.47) is the desired expression for ( _Tin terms of easily measured properties.

Temperature Dependence of U

The basic equation (4.29) is the desired relation: ( _V CV. Temperature Dependence of H

The basic equation (4.30) is the desired relation: ( _P CP. Pressure Dependence of H

We want ( _T. Starting with the Gibbs equation dH T dS V dP [Eq. (4.34)], imposing the condition of constant T, and dividing by dP_T, we get dH_T/dP_T T dS_T/dP_T V or

Application of the Euler reciprocity relation to dG _T

P[Eq. (4.45)]. Therefore

(4.48) Temperature Dependence of S

The basic equation (4.31) for C_Pis the desired relation:

(4.49) a0S

0Tb

C_P T a0H

0Pb

T a0V 0Tb

V TVa V a0H

0Pb

T a0S 0Pb

V a0U

0Vb

T a0P 0Tb

P aT k P a0U

0Vb

T a0S 0Vb

P dU_T

dV_T T dS_T dV_T P dU_T T dST P dVT

Section 4.4 Thermodynamic Relations for a

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120

Pressure Dependence of S

The Euler reciprocity relation applied to the Gibbs equation dG S dT V dP gives (4.50) as already noted in Eq. (4.45).

Temperature and Pressure Dependences of G

In dG _P S. In dG S dT

V dP, we set dT _T V. Thus [Eq. (4.38)]

(4.51)

Summary on Finding T,P,and V Dependences of State Functions

To find ( _T, ( _T, ( _V, or ( _Pof U, H, A, or G, one starts with the Gibbs equation for dU, dH, dA, or dG [Eqs. (4.33) to (4.36)], imposes the condition of con-stant T, V, or P, divides by dP_T, dV_T, dT_V, or dT_P, and, if necessary, uses one of the Maxwell relations (4.45) or the heat-capacity relations (4.31) to eliminate ( _T, ( _T, ( _V, or ( _P. To find ( _Vand ( _P, it is faster to simply write down the C_Vand C_Pequations (4.29) and (4.30).

In deriving thermodynamic identities, it is helpful to remember that the tempera-ture dependences of S [the derivatives ( _Pand ( _V] are related to C_Pand C_V [Eq. (4.31)] and the volume and pressure dependences of S [the derivatives ( _T and ( _T] are given by the Maxwell relations (4.45). Equation (4.45) need not be memorized, since it can quickly be found from the Gibbs equations for dA and dG by using the Euler reciprocity relation.

As a reminder, the equations of this section apply to a closed system of fixed com-position and also to closed systems where the comcom-position changes reversibly.

Magnitudes of T,P,and V Dependences of U,H,S,and G

We have (Um/T )V CV,mand (Hm/T )P CP,m. The heat capacities C_P,mand C_V,m are always positive and usually are not small. Therefore U_mand H_mincrease rapidly with increasing T (see Fig. 5.11). An exception is at very low T, since C_P,mand C_V,mgo to zero as T goes to absolute zero (Secs. 2.11 and 5.7).

Using (4.47) and experimental data, one finds (as discussed later in this section) that ( _T(which is a measure of the strength of intermolecular forces) is zero for ideal gases, is small for real gases at low and moderate pressures, is substantial for gases at high pressures, and is very large for liquids and solids.

Using (4.48) and typical experimental data (Prob. 4.8), one finds that (Hm/P)Tis rather small for solids and liquids. It takes very high pressures to produce substantial changes in the internal energy and enthalpy of a solid or liquid. For ideal gases (Hm/P)T 0 (Prob. 4.21), and for real gases (Hm/P)Tis generally small.

From ( _P CP/T, it follows that the entropy S increases rapidly as T in-creases (see Fig. 5.11).

We have (Sm/P)T aVm. As noted in Sec. 1.7, a is somewhat larger for gases than for condensed phases. Moreover, V_mat usual temperatures and pressures is about 10³times as great for gases as for liquids and solids. Thus, the variation in entropy with pressure is small for liquids and solids but is substantial for gases. Since a is positive for gases, the entropy of a gas decreases rapidly as the pressure increases (and the vol-ume decreases); recall Eq. (3.30) for ideal gases.

For G, we have (Gm/P)T Vm. For solids and liquids, the molar volume is rela-tively small, so G_mfor condensed phases is rather insensitive to moderate changes in

a0G 0Tb

S, a0G 0Pb

V a0S

0Pb

a0V 0Tb

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pressure, a fact we shall use frequently. For gases, V_mis large and G_mincreases rapidly as P increases (due mainly to the decrease in S as P increases).

We also have ( _P S. However, thermodynamics does not define absolute entropies, only entropy differences. The entropy S has an arbitrary additive constant.

Thus ( _P has no physical meaning in thermodynamics, and it is impossible to measure ( _P of a system. However, from ( _P S, we can derive ( _P S. This equation has physical meaning.

In summary: For solids and liquids, temperature changes usually have significant effects on thermodynamic properties, but pressure effects are small unless very large pressure changes are involved. For gases not at high pressure, temperature changes usu-ally have significant effects on thermodynamic properties and pressure changes have significant effects on properties that involve the entropy (for example, S, A, G) but usu-ally have only slight effects on properties not involving S (for example, U, H, C_P).

Joule–Thomson Coefficient

We now express some more thermodynamic properties in terms of easily measured quantities. We begin with the Joule–Thomson coefficient m_JT _H. Equa-tion (2.65) gives m_JT _T/C_P. Substitution of (4.48) for ( _Tgives

(4.52) which relates m_JTto a and C_P.

Heat-Capacity Difference

Equation (2.61) gives C_P CV T P. Substitution of

( _T aT/k P [Eq. (4.47)] gives CP CV P. Use of a ⬅ V¹( _Pgives

(4.53) For a condensed phase (liquid or solid), C_Pis readily measured, but C_Vis hard to mea-sure. Equation (4.53) gives a way to calculate C_Vfrom the measured C_P.

Note the following: (1) As T → 0, CP → CV. (2) The compressibility k can be proved to be always positive (Zemansky and Dittman, sec. 14-9). Hence C_P CV. (3) If a 0, then CP CV. For liquid water at 1 atm, the molar volume reaches a minimum at 3.98°C (Fig. 1.5). Hence ( _P 0 and a 0 for water at this temperature. Thus C_P CVfor water at 1 atm and 3.98°C.

EXAMPLE 4.2

C_P CV

For water at 30°C and 1 atm: a 3.04 10⁴K¹, k 4.52 10⁵atm¹ 4.46 10¹⁰m²/N, C_P,m 75.3 J/(mol K), Vm 18.1 cm³/mol. Find C_V,mof water at 30°C and 1 atm.

Division of (4.53) by the number of moles of water gives C_P,m CV,m TV_ma²/k. We find

(4.54) For liquid water at 1 atm and 30°C, there is little difference between C_P,m and C_V,m. This is due to the rather small a value of 30°C water; a is zero at 4°C and is still small at 30°C.

C_V,m 74.2 J>1mol K2 TV_ma²>k 1.14 J mol¹ K¹

TV_ma²

k 1303 K2 118.1 10⁶ m³ mol¹2 13.04 10⁴ K¹2² 4.46 10¹⁰ m²>N

C_P CV TVa²>k

m_JT 11>CP2 3T 10V>0T2P V4 1V>CP2 1aT 12

Section 4.4 Thermodynamic Relations for a

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Chapter 4 Material Equilibrium

122

Exercise

For water at 95.0°C and 1 atm: a 7.232 10⁴K¹, k 4.81 10⁵bar¹, c_P 4.210 J/(g K), and r 0.96189 g/cm³. Find c_V for water at 95.0°C and 1 atm. [Answer: 3.794 J/(g K).]

The use of (4.53) and experimental C_P,mvalues to find C_V,mfor solids and liquids gives the following results at 25°C and 1 atm:

Substance Cu(s) NaCl(s) I₂(s) C₆H₆(l ) CS₂(l ) CCl₄(l )

C_V,m/[J/(mol K)] 23.8 47.7 48 95 47 91

C_P,m/[J/(mol K)] 24.4 50.5 54 136 76 132

C_P,mand C_V,musually do not differ by much for solids but differ greatly for liquids.

Ideal-Gas ( _T

An ideal gas obeys the equation of state PV nRT, whereas a perfect gas obeys both

PV _T _V nR/V, and Eq. (4.47) gives

( _T nRT/V P P P 0.

(4.55) We have proved that all ideal gases are perfect, so there is no distinction between an ideal gas and a perfect gas. From now on, we shall drop the term “perfect gas.”

( _Tof Solids, Liquids, and Nonideal Gases

The internal pressure ( _Tis, as noted in Sec. 2.6, a measure of intermolecular in-teractions in a substance. The relation ( _T aT/k P [Eq. (4.47)] enables one to find ( _Tfrom experimental data. For solids, the typical values a 10^4.5K¹ and k 10^5.5atm¹(Sec. 1.7) give at 25°C and 1 atm

For liquids, the typical a and k values give at 25°C and 1 atm

The large ( _Tvalues indicate strong intermolecular forces in solids and liquids.

EXAMPLE 4.3

( _Tfor a nonideal gas

Estimate ( _Tfor N₂gas at 25°C and 1 atm using the van der Waals equa-tion and the van der Waals constants of Sec. 8.4.

The van der Waals equation (1.39) is

(4.56) We have ( _T _V P [Eq. (4.47)]. Solving the van der Waals equation for P and taking ( _V, we have

(4.57) a0U

0Vb

Ta0P 0Tb

P nRT

V nb a nRT

V nb an²

V² b an² V² P nRT

V nb an²

V² and a0P 0Tb

V nb 1P an²>V²2 1V nb2 nRT

10U>0V2^T⬇110³ K¹2 1300 K2 110⁴ atm2 ⬇ 3000 atm ⬇ 300 J>cm³ 10U>0V2^T⬇110^4.5K¹2 1300 K2 110^5.5 atm2 1 atm ⬇ 3000 atm ⬇ 300 J>cm³

10U>0V2^T 0 ideal gas

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From Sec. 8.4, a 1.35 10⁶cm⁶atm mol²for N₂. At 25°C and 1 atm, the gas is nearly ideal and V/n can be found from PV nRT with little error. We get V/n 24.5 10³cm³/mol. Thus

(4.58) The smallness of ( _Tindicates the smallness of intermolecular forces in N₂ gas at 25°C and 1 atm.

Exercise

Use the van der Waals equation and data in Sec. 8.4 to estimate ( _Tfor HCl(g) at 25°C and 1 atm. Why is ( _Tlarger for HCl(g) than for N₂(g)?

[Answer: 0.0061 atm 0.62 J/L.]

U_intermolof a liquid can be estimated as U of vaporization.

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