Chapter 8 Thermodynamic Relations

Transcription

1 Chapter 8 Thermodynami Relations 8.1 Types of Thermodynami roperties The thermodynami state of a system an be haraterized by its properties that an be lassified as measured, fundamental, or deried properties. We want to deelop relationships to relate the hanges in the fundamental and deried properties in terms of the measured properties that are diretly aessible from laboratory measurements. Some of the measured properties are,, T, omposition, p, and. The small letters are used to denote speifi quantities for example is speifi olume. The fundamental properties are internal energy u and entropy s. These properties arrie from the first and seond law of thermodynamis. The first law states that energy is onsered, and the seond law states that entropy of the unierse always inreases. The deried properties are defined to failitate the energy balane of systems in whih the ombination of internal energy and other properties often ours. In open systems, the mass that rosses the boundary between the surroundings and the system always ontributes to two terms in the energy balane: internal energy and flow work (). For onenient we an define an enthalpy (h) as h = u + (8.1-1a) In terms of the total enthalpy H, we hae H = U + (8.1-1b) We an then make an enthalpy balane for an open system in whih the flow work is inluded in the enthalpy term. Figure shows a raindrop reated from the surrounding super saturated apor in the atmosphere. Not only the energy U of the raindrop is needed but also some additional energy, equal to, is required to push the atmosphere out of the way to make room for the drop. raindrop with olume and internal energy U Figure n energy of U + is required to reate a raindrop. Enthalpy is the total energy we would need, to reate a system out of nothing and put it in an enironment with onstant pressure. Or, if we ould ompletely annihilate a system, H is 8-1

2 the energy we ould reoer: the system s energy plus the work done by the ollapsing atmosphere. Howeer, we usually are not interested in the total energy needed or the total energy that an be reoered from a system. We will be more interested in the work inoled in a system. For isothermal surroundings, the system an extrat heat from the surroundings for free, so the work required to reate the system from nothing is equal to the internal energy minus the heat reeied. nd if we annihilate the system, we generally annot reoer all its energy as work sine we hae to dispose of its entropy by dumping some heat into the surroundings. Therefore it is more onenient to define the Helmholtz free energy,, for an enironment at onstant temperature T = U TS (8.1-) is the energy that must be proided as work if we reate the system out of nothing. The heat extrated from the surroundings is T S = T(S f S i ) = TS f where S f is the system final entropy and S i the system zero initial entropy. If we annihilate a system with initial entropy S i, is the amount of reoered work, sine we hae to dump some heat, equal to TS i, into the enironment to get rid of the system s entropy. Equation (8.1-) inludes all work, een the work done by the system s surroundings. If the system is in an isothermal and isobari enironment, it is more onenient to use the Gibbs free energy G = U TS + (8.1-) Gibbs free energy is the work required to reate a system from nothing in an enironment with onstant and onstant temperature T. We usually are more interested in the hange in states of a system rather than its reation or annihilation. We then want to look at the hanges in and G. The hange in at onstant temperature is gien by = U T S = Q + W T S (8.1-4) In this expression Q is the heat added and W is the work done on the system. If the proess is reersible then Q = T S and the hange in is preisely equal to the work done on the system. If the proess is irreersible then Q < T S and < W, the hange in is less than the work done on the system. For an enironment with onstant and onstant temperature T, the hange in G is gien by For any proess we hae G = U T S + = Q + W T S + (8.1-5) Q T S 0 (equal sign for reersible proesses) (8.1-6) 8-

3 The work term W onsists of the work done by the enironment,, and any other work done on the system. W = + W other (8.1-7) Substituting equations (8.1-6) and (8.1-7) into equation (8.1-5) we obtain G W other at onstant T, (8.1-8) Example Determine the eletrial work required to produe one mole of hydrogen in the eletrolysis of liquid water at 98 o K and 1 atm. The hemial reation is H O(l) H (g) + 0.5O (g) Data (at 98 o K and 1 atm): H = 86 kj for this reation, S HO = 70 J/ o K, S H = 11 J/ o K, and S O = 05 J/ o K. Solution G = H TS t onstant T we hae G = H T S The hange in system entropy is gien by The hange in G is then S = S H + 0.5S O S HO = (05) 70 = 16.5 J/ o K G = 86 kj (98 o K)(16.5 J/ o K) = 7 kj This is the amount of energy in terms of eletrial work required to produe one mole of hydrogen by eletrolysis. If we burn one mole of hydrogen, the amount of heat we would get is 86 kj. If we an ombine one mole of hydrogen and half a mole of oxygen in a fuel ell to produe water we an extrat 7 kj of eletrial work. The differene H G = T S = 49 kj is the waste heat that must be expel by the fuel ell to get rid of the exess entropy that was in the gases. Therefore the maximum effiieny, ε fuel ell, of the fuel ell is ε fuel ell = 7/86 = 0.89 This effiieny is higher than the 40% effiieny of eletrial power plants. 8-

4 Example In a hydrogen fuel ell shown in Figure 8.1-, hydrogen and oxygen gas pass through porous eletrodes and reat to form water. Eletrons are released at the anode (negatie eletrode) and deposited at the athode (positie eletrode). The oerall reation is H (g) + 0.5O (g) H O(l) Calulate the oltage of the ell. Data (at 98 o K and 1 atm): G = 7 kj for this reation. - + H O H O Figure 8.1- hydrogen fuel ell Solution In a hydrogen fuel ell 1, the steps of the hemial reation are H + OH H O + e (at eletrode) 0.5O + H O + e OH (at + eletrode) Two eletrons are pushed through the iruit eah time the full reation ours. The eletrial work produed per eletron is 7 kj/( ) = J = 1. e (Note: 1 e = J) Sine 1 olt is the oltage needed to gie eah eletron 1 e of energy, so the fuel ell has a oltage of Shroeder, D.., n Introdution to Thermal hysis, ddision Wesley Longman,

5 8. Equations of State In the alulations of energy, enthalpy, and entropy of a substane we need an aurate representation of the relationship among pressure, olume, and temperature. Besides the tabular and graphial presentations of the p--t relationship, analytial formulations, alled equation of state, onstitute another way of expressing the p--t relationship. The equations of state are onenient for performing the mathematial operations required to alulate u, h, s, and other thermodynami properties. In hapter we mention the ompressibility fator, the irial, and the Soae-Redlik-Kwong equation of states. The irial equation of state an be deried from the priniple of statial mehanis to relate the p--t behaior of a gas to the fores between moleules. irial equation of state expresses the quantity p as a power series in the inerse of molar olume. Z = p = 1 + B( T ) C( T ) + D( T ) + + (8.-1) In this equation, B, C, and D are alled irial oeffiient and are funtions of temperature. For a trunated irial equation with two terms we hae p = 1 + B( T ) (8.-) In this equation, B(T) an be estimated from the following equations: B(T) = p (B 0 + ωb 1 ) (8.-) 0.4 B 0 = T R 0.17, B 1 = T R In equation (8.-), ω is the itzer aentri fator, whih is a parameter refleting the geometry and polarity of a moleule. The aentri fator for oer 1000 ompounds an be obtained from omp4.exe program written by T.K. Nguyen. This program is aailable in the Distribution Folder for CHE0 ourse. In the limiting ase where there are no interations between the moleules, all the irial oeffiients are equal to zero. Eq. (8.-1) beomes Z = p = 1 (8.-4) Eq. (8.-4) is the ideal gas equation of state. We will use the an de Walls equation of state to illustrate the ealuation of thermodynami properties. Both the an de Walls and the SRK equations of state hae two adjustable onstants but the an de Walls equation is simpler. The an de Walls equation of state is 8-5

6 = b a (8.-5) In this equation, the onstant b aounts for the finite olume oupied by the moleules and a the term aounts for the attratie fores between moleules. Figure 8.-1 Isotherms from the an der Waals equation. The an der Waals parameters a and b an be determined from the ritial properties sine there is an infletion point at the ritial isotherm as shown in Figure t the ritial point we hae T = T = 0 (8.-6) The isotherm passing through the ritial point is gien by = b a The first and seond deriaties of with respet to are gien by T = ( ) b a + = 0 (8.-6a) T = ( ) b 6a 4 = 0 (8.-6b) We an sole the two equations (8.-6a) and (8.-6b) for the two unknowns a and b. Multiplying equation (8.-6a) by and equation (8.-6b) by ( ν b) and add them together we get 8-6

7 4a 6a 4 ( ν b) = 0 (8.-7) 4aν 6aν + 6ab = 0 ν = b (8.-8) Substituting b = ν / into equation (8.-6a) and soling for a gies 9 a = ν R T 8 t the ritial point we hae = b a (8.-9) We an use equation (8.-9) to sole for ν in terms of ritial temperature and ritial 9 pressure. Substituting a = ν R T and b = ν / into equation (8.-9) we obtain 8 = 9ν 8 = 9 8 = 8 Soling for ν in terms of and T we hae Hene ν = a = ν R T = 8 64 ( ) Using R = 8.14 J/(mol o K) = m bar/(mol o K) and for propane, T = 69.9 o K, = 4.46 bar, we hae 7 a = 64 ( ) 7 = 64 ( ) 4.46 = m 6 bar/mol 8-7

8 Example For the an Der Waals isotherm shown in the following figure, show that the saturation pressure an be determined by loating the horizontal, two-phase segment of the isotherm so that two equal areas are enlosed between it and the an de Waals ure. B 1 B Solution G H G H T S T S U U From the grouping {G, T, }, we hae G = G(T, ), therefore G dg = T G dt + T d = SdT + d long an isotherm of the equation of state, dt = 0, therefore G = d t the saturation pressure G = G G L = 0, we hae G = G G L = d + d + B d + d = 0 B Sine area (1) = d d, and area () = B d + d, the saturation B pressure an be determined by loating the horizontal, two-phase segment of the isotherm so that two equal areas are enlosed between it and the an Der Waals ure. 1 Kyle, B.G., Chemial and roess Thermodynamis, rentie Hall,