This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 3 Generalized Thermodynamic Potentials
A theory is the more impressive the greater the simplicity of its premises, the more diﬀerent kinds of things it relates, and the more extended is its area of applicability. Therefore the deep impression which classical thermodynamics made upon me. It is the only physical theory of universal content concerning which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown (for the special attention of those who are skeptics on principle). – Albert Einstein1 From the postulates in Chapter 2, we have already seen that from U (S , V , N1 , . . ., Nm ) or S (U , V , N1 , . . ., Nm ) we can derive all thermodynamic information about a system. For example, we can ﬁnd mechanical or thermal equations of state. However, we also know that it is sometimes convenient to use other independent variables besides entropy and volume. For example, when we perform an experiment at room temperature open to the atmosphere, we are manipulating temperature and pressure, not entropy and volume. In this case, the more natural independent variables are T and P . Then, the question arises: Is there a function of (T, P, N1 , . . . , Nm ) that contains complete thermodynamic information? In other words, is there some function, say G(T, P, N1 , . . . , Nm ) from which we could derive all the equations of state? It turns out that such functions do exist, and that they are useful for solving practical problems.
1 Albert Einstein: PhilosopherScientist, p.33, ed. by P. A. Schilpp, (Cambridge U.P., London, 1970) 61 62 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS In the ﬁrst section we show how to derive such a function for any complete set of independent variables using Legendre Transforms. In this book we introduce three widely used potentials: the enthalpy H (S, P, N1 , . . . , Nm ), the Helmholtz potential F (T, V, N1 , . . . , Nm ), and the Gibbs free energy G(T, P, N1 , . . . , Nm ). These functions that contain complete thermodynamic information using independent variables besides S, V and N1 , . . . , Nm are called generalized thermodynamic potentials. These quantities are essential for engineering or applied thermodynamics. For example, we already know that an isolated system attains equilibrium when the entropy is maximized. However, a system in contact with a thermal reservoir attains equilibrium when the Helmholtz potential F is minimized. We show below (Examples 3.2.1 and 3.2.2), that the work necessary to compress any gas isothermally is just the change in F (T, V, N ). In a fuel cell, G(T, P, N ) is important. Or, in the open (ﬂowing) systems considered in Chapter 5, we see that another such function (called enthalpy) is also important. In other words, we need not consider the entropy of the reservoir explicitly to ﬁnd equilibrium. We will also see in later chapters, that the enthalpy H (S, P, N ) plays an important role in ﬂowing systems and in applications. In §3.2, we show how subsystems in contact with thermal and/or mechanical reservoirs are handled more naturally using the generalized potentials. In §3.6 we show how to express any derivative in terms of the measurable quantities introduced in §3.5. In §3.3, we see that the generalized thermodynamic potentials have a relationship between derivatives of the independent variables called the Maxwell relations. These relations are essential for the thermodynamic manipulations discussed in the next chapter, and are used in the remainder of the text. 3.1 Legendre Transforms Recall Example 2.6.1, where we found U (T, V, N ) from S (U, V, N ). Although we could ﬁnd P (T, V, N ) from the S expression, it is not possible to ﬁnd it from U (T, V, N ). Hence, we lose information when we go from the fundamental relation to the equation of state. In this section we show why that is, and introduce a method to ﬁnd a function F (T, V, N ) that has all the information that was in our original fundamental relation S (U, V, N ). The Na¨ Approach ıve On ﬁrst reﬂection, we might think that ﬁnding a function with complete thermodynamic information that has T instead of S as an independent variable is straightforward. We would na¨ ıvely take the derivative of U (S ) with respect to S to ﬁnd 3.1. LEGENDRE TRANSFORMS 63 ˜ T (S ), invert, and insert S (T ) into U (S (T )) to ﬁnd U (T ). However, we can show that such a na¨ approach would lead to a loss of information along the way. ıve Consider a fundamental energy relation U1 (S ) shown on the left side of Figure 3.1 that we wish to manipulate in the na¨ manner. We take the derivative to ﬁnd ıve the thermal equation of state: T = T1 (S ) := ∂U1 . ∂S (3.1) − We invert this expression (somehow) to ﬁnd S = T1 1 (T ). If we now plug this (inverted) expression into the original fundamental energy relation, we ﬁnd − ˜ U1 (T ) = U1 (T1 1 (T )), (3.2) which is shown on the right side of the same ﬁgure. U U2 (S ) ˜ ˜ U1 (T ), U2 (T ) U1 (S ) U S T Figure 3.1: Left side: Two diﬀerent fundamental energy relations as functions of entropy, where U2 (S ) = U1 (S + C0 ). Right side: Internal energy as a function of temperature as predicted by the two fundamental energy relations. Since these curves are indistinguishable, there is no way to recreate the original fundamental relations from these equations of state. Now suppose that we had a somewhat diﬀerent fundamental energy relation U2 (S ) := U1 (S + C0 ), also shown on the left side of Figure 3.1, where C0 is not a function of S . Note that at a given value for U , both fundamental relations have the ˜ same slope, and, hence, the same temperature. Let us ﬁnd the expression U2 (T ) and ˜ prove that it equals U1 (T ). The thermal equation of state for the new fundamental energy relation is T= ∂U2 (S ) ∂U1 (S + C0 ) ∂ (S + C0 ) ∂U1 (S + C0 ) = = = T1 (S + C0 ). ∂S ∂S ∂S ∂ (S + C0 ) (3.3) − We invert this expression to ﬁnd S (T ) = T1 1 (T ) − C0 , which we can insert into the fundamental energy relation to ﬁnd ˜ U2 (T ) = U2 (S (T )) = U1 (S (T ) + C0 ) ˜ = U1 (T −1 (T )) = U1 (T ).
1 (3.4) 64 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS ˜ ˜ Note that the resulting functions U1 and U2 are identical, as seen on the right side, although the fundamental relations were diﬀerent. In other words, our method does not distinguish between the two diﬀerent fundamental energy relations U1 and U2 since they both lead to the same function of T . Information is lost in the transformation. That is why we call U (T, V ) an equation of state, and not a fundamental relation. Put another way, note from Example 2.6.1 that there is no way to derive the van der Waals P V T equation of state from (2.34) alone—one needs the fundamental relation (2.21). U (S ) F (T1 ) F (T2 ) F (T3 ) T3 T2 T1 0 S
Figure 3.2: Internal energy U (S ) represented as both a curve and a family of tangential lines. The
straight lines each have slope Ti and intercept F (Ti ). The slope is the ﬁrst derivative of U at some point Si . Knowing F for all T is suﬃcient to reconstruct the curve. Correct Approach Now we consider a Legendre transformation, where we wish to convert a relationship of the kind U = U (S ), to ﬁnd some function F (T ) of the slope T T (S ) := ∂U (S ) , ∂S (3.5) that contains the same information as U (S ). In other words, knowing F (T ) we want to be able to ﬁnd U (S ) uniquely. Instead of representing the curve as a family of paired points U = U (S ), we may also represent the curve as a family of tangent lines, as shown in Figure 3.2. Each curve has a tangent line of slope T at any given point on the curve. Using the equation of a line y = mx + b, we can ﬁnd a relationship between the intercept F and the slope T F (S ) := U (S ) − T (S )S. (3.6) 3.1. LEGENDRE TRANSFORMS 65 From Eqs. (3.5) and (3.6), we can eliminate S to ﬁnd F (T ). To reconstruct U (S ) from F (T ), note that ∂F S (T ) = − , (3.7) ∂T so that we can solve Eqn. (3.6) for U U (T ) = F (T ) − T ∂F . ∂T (3.8) From Eqs. (3.7) and (3.8) we can eliminate T to ﬁnd U (S ). Hence, knowing F (T ) allows us to reconstruct the entire curve. Therefore, we deﬁne the Helmholtz potential F to be the Legendre transform of U (S ) by Eqn. (3.6). Similarly, if we perform the Legendre transform on U (V ), we obtain the enthalpy H as the Legendre transform; we can perform the Legendre transform simultaneously on both arguments of U (S, V ) to obtain the Gibbs free energy G. H (S, P, N1 , . . . , Nm ) := U + P V G(T, P, N1 , . . . , Nm ) := U − T S + P V F (T, V, N1 , . . . , Nm ) := U − T S (3.9) (3.10) (3.11) Note that the sign is diﬀerent for the transformation from V to P , because the slope of U (V ) is −P . For each of the three transforms, we can reconstruct the original fundamental energy relation by Eqs. (3.7) and (3.8). The results of the inversion are summarized in Table 3.1. For a given general potential, only one set of independent variables gives complete thermodynamic information. We call the members of this set the canonical independent variables for that potential. Example 3.1.1 Find the Helmholtz potential as a function of temperature and volume for a purecomponent, simple, ideal gas, which has the fundamental entropy relation UcV 1 S = N s0 + NR log , (3.12) u0 v0 N 1+c where s0 is the speciﬁc entropy at reference point (u0 , v0 , N0 ). Solution: From the deﬁnition of the Helmholtz potential Eqn. (3.10), we
see that we need both U and S as functions of T and V . Hence, we need the thermal equation of state. Since we have a fundamental entropy formulation, 66 From CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS Find canonical To ﬁnd Internal Energy independent variables U (S, V )
∂H ∂P S,N ∂F ∂T V ,N ∂G ∂T P ,N ∂G ∂P T ,N Eliminate P T T, P H (S, P ) V (S, P ) = F (T, V ) S (T, V ) = − G(T, P ) S (T, P ) = − V (T, P ) = U (S, P ) = H (S, P ) − P V (S, P ) U (T, V ) = F (T, V ) + T S (T, V ) U (T, P ) = G(T, P ) + T S (T, P ) −P V (T, P ) Table 3.1: How to recover the fundamental energy relation U (S, V ) from the generalized thermodynamic potentials F, G and H as functions of their canonical independent variables: Given a fundamental relation of the quantity in the ﬁrst column, use equations in the second and third columns to eliminate the variables in the fourth column leaving U = U (S, V ).
we use the alternative deﬁnition of temperature Eqn. (2.31) to ﬁnd the thermal equation of state 1 T := = ∂S ∂U cNR , U V ,N (3.13) which we may use to ﬁnd U = U (T ). When Eqn. (3.13) is inserted into the fundamental entropy relation for an ideal gas, Eqn. (3.12) we obtain an expression for S (T, V, N ) S = N s0 + NR log T T0
c V N v0 , (3.14) where T0 := u0 /cR. We now have the soughtafter expressions for U = U (T, V ) and S = S (T, V ). When we insert the expression we found for U (T ), Eqn. (3.13), and Eqn. (3.14) into the deﬁnition for F , Eqn. (3.10), we ﬁnd F = N f0 + (cNR − Ns0 )(T − T0 ) − NRT log T T0
c V N v0 , (3.15) where f0 := u0 − s0 T0 is the speciﬁc Helmholtz potential at the reference state (T0 , v0 ). Eqn. (3.15) contains all thermodynamic information about a simple ideal gas, as does Eqn. (3.12) 2 3.1. LEGENDRE TRANSFORMS Example 3.1.2 The Helmholtz potential for a system is given as
T 67 F = N f0 − N s0 (T − T0 ) + N −NRT log V v0 N T0 T′ − T T′ C (T ′ )dT ′ (3.16) where s0 , T0 , f0 and v0 are constants with straightforward interpretation, and C (T ) is some known function of temperature. Can you ﬁnd the fundamental energy relation? Solution: We use the prescription in Table 3.1. Using the second row (which follows from the diﬀerential for F ), we ﬁnd S (T, V ) from Eqn. (3.16) S =− ∂F ∂T (3.17)
V ,N T T0 = N s0 + N C (T ′ ) ′ V dT + NR log . T′ v0 N We used the Leibniz rule for diﬀerentiating an integral Eqn. (A.25) to obtain the second line. We ﬁnd an expression for U from the second column of Table 3.1 U = F + TS
T (3.18) C (T ′ )dT ′ .
T0 = N f 0 + N so T 0 + N If we could invert Eqn. (3.17) to ﬁnd T (S, V, N ), we could insert this expression into the upper limit of the integral in Eqn. (3.18) to obtain U (S, V, N ), and we would be ﬁnished. However, practically speaking, the inversion is impossible without specifying C (T ), and in practice is diﬃcult to do for nearly all expressions C (T ). This is as close as we can come to ﬁnding a fundamental energy relation, since Eqs. (3.17) and (3.18) are suﬃcient to determine U (S, V, N ). 2 This example shows that it is more natural to use F when T and V are the independent variables; the single equation (3.16) gives complete thermodynamic information, whereas we require both Eqn. (3.17) and (3.18) to express the fundamental energy relation. Note that we have internal energy as a function of temperature only U = U (T ) in the thermal equation of state, Eqn. (3.18), but independent of volume. We see in Example 3.1.3 below that the mechanical equation of state is simply the ideal gas law. Hence, Eqn. (3.16) is the fundamental Helmholtz relation for a general ideal gas, whose internal energy has arbitrary dependence on temperature (speciﬁed by giving C (T )). 68 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS The Gibbs potential has a special relationship to the chemical potential. This relationship is made clear in the Euler Relation, which we derive here. Recall that the ﬁrst property of Postulate III requires that internal energy be a ﬁrstorder, homogeneous function of its arguments λU (S, V, N1 , N2 , . . . , Nr ) = U (λS, λV, λN1 , λN2 , . . . , λNr ). We take the partial derivative with respect to λ of each side of this equation (holding S, V, N1 , . . . , Nm constant), using the chain rule for partial diﬀerentiation Eqn. (A.8) on the right side U (S, V, N1 , N2 , . . . , Nr ) = ∂ (λS ) ∂U (λS, λV, λN1 , λN2 , . . . , λNr ) + ∂λ ∂ (λS ) ∂ (λV ) ∂U (λS, λV, λN1 , λN2 , . . . , λNr ) + ∂λ ∂ (λV )
r i=1 (3.19) ∂ (λNi ) ∂U (λS, λV, λN1 , λN2 , . . . , λNr ) . ∂λ ∂ (λNi ) The ﬁrst derivative of each term on the right side is straightforward to ﬁnd. The second derivative is the conjugate variable of each term, whose deﬁnitions are given by Eqs. (2.24) through (2.26). Therefore, the above equation becomes the Euler relation
r U = TS − PV + µi Ni ,
i=1 Euler Relation. (3.20) The Euler relation just follows from the fact that the internal energy is extensive in entropy, volume and mole number. Recalling the deﬁnition of the Gibbs free energy Eqn. (3.11), we can write
r G=
i=1 µi Ni . (3.21) Hence, for a pure substance, the speciﬁc Gibbs free energy is identical to the chemical potential. If we take the diﬀerential of each side of the Euler relation (3.20), and subtract from it the diﬀerential for internal energy, Eqn. (2.27), we obtain the wellknown GibbsDuhem relation 3.2. EXTREMA PRINCIPLES FOR THE POTENTIALS 69 r 0 = SdT − V dP + Ni dµi ,
i=1 GibbsDuhem Relation. (3.22) Finally, we note that one may also take the Legendre Transform of the entropy. Such functions are called Massieu functions [13, §54]. Problem 3.3.D considers a transform on N , for example. A systematic way to handle any such transform may be found in the textbook by Tester, Modell and Reid [55, 92, §§54,5]. Example 3.1.3 Prove that the Helmholtz potential given by Eqn. (3.16)
T F = N f0 − N s0 (T − T0 ) + N −NRT log V v0 N . T0 T′ − T T′ C (T ′ )dT ′ leads to the idealgas mechanical equation of state Solution: Using Eqn. (3.52) we can ﬁnd the mechanical equation of state
from the Helmholtz potential P = = which is the ideal gas law. 2 − ∂F ∂V T ,N NRT , V 3.2 Extrema Principles for the Potentials Our postulates in Chapter 2 told us that equilibrium states of an isolated system are determined by the maximization of the entropy within the unconstrained, independent variables. In section §3.1 we saw that the generalized thermodynamic potentials F, H, G as functions of their canonical independent variables is equivalent to either the fundamental energy or fundamental entropy relations. Therefore, it would be natural to wonder if there are extremum principles for the internal energy and the generalized potentials—like entropy maximization for an isolated system. In this section, we ﬁnd these principles, and show how they can be very useful for ﬁnding the equilibrium states of systems that are kept isothermal or isobaric. 70 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS Before considering the potentials, we ﬁrst need to prove that entropy maximization for a closed system ∂S ∂X = 0,
U,V,N ∂2S ∂X 2 <0
U,V,N (3.23) is mathematically equivalent to internal energy minimization at constant entropy ∂U ∂X = 0,
S,V,N ∂2U ∂X 2 > 0,
S,V,N (3.24) where X is the unconstrained variable (e.g., U (1) in §2.7). Graphically, we are saying the following: the postulates state that Xeq is the equilibrium state in Figure 3.3 a), and we are now going to prove that Figure 3.3 b) follows. For example, in §2.7, X ≡ U (1) , and the ﬁnal energy in container 1 is Xeq . We proved that Figure 3.3 a) means equal temperature in the two containers. We now show that this state is also an internal energy minimum at ﬁxed entropy.
ﬁxed U
ﬁxed S S
a) U Xeq X
b) Xeq X Figure 3.3: Entropy as a function of the unconstrained variable X at ﬁxed internal energy, a), and internal energy as a function of unconstrained variable X at ﬁxed entropy. To show that an extremum in entropy at constant internal energy is an extremum in internal energy, we begin with the ﬁrst derivative of internal energy ∂U ∂X ∂ (U, S ) ∂ (X, S ) ∂ (U, S ) ∂ (X, S ) = ∂ (U, X ) ∂ (U, X ) ∂S ∂S =− ∂X U ∂U ∂S = −T , ∂X U =
S X (3.25) 3.2. EXTREMA PRINCIPLES FOR THE POTENTIALS 71 where it is implied that V and N are held constant in all derivatives. Since the ∂S entropy is an extremum at equilibrium and ∂X U = 0, then Eqn. (3.25) shows that internal energy is also an extremum. To show that a maximum in S corresponds to ∂U a minimum in U , it is convenient to introduce temporarily the notation ψ := ∂X S . Then, we can write ∂2U ∂X 2 =
S = = = = ∂ψ ∂X S ∂ (ψ, S ) ∂ (X, S ) ∂ (ψ, S ) ∂ (X, U ) ∂ (X, U ) ∂ (X, S ) ∂S ∂ψ 1 ∂ψ − T T ∂X U ∂X U ∂U X ∂ψ , when S is maximum. ∂X U (3.26) Note that the second Jacobian on the third line is just T . We expand out the determinant of the ﬁrst one to get the fourth line. The last line follows because we ∂S are considering the state where S is a maximum, so ∂X U = 0. To ﬁnish the proof, we insert Eqn. (3.25) into the last line above, and obtain ∂2U ∂X 2 =
S ∂ ∂X ∂U ∂X ∂2S ∂X 2 ∂2S ∂X 2 S U = −T = −T U − , ∂S ∂X U ∂T ∂X U when S is maximum. (3.27) U We used Eqn. (3.25) to obtain the second line, and the fact that entropy is an extremum to obtain the last line. Eqn. (3.27) shows that when S is maximized for a closed system, internal energy is minimized at constant entropy. To characterize the potentials, we think about how to make a process isothermal in practice. We do this by splitting an isolated system into our subsystem of interest and the rest of the isolated system that we call a heat reservoir, which is at temperature Tres . The subsystem of interest and the heat reservoir are not in mechanical contact (i.e., the pressure is not necessarily the same as the pressure in the reservoir, and dVres = dVsys = 0). A heat reservoir is a large subsystem whose temperature cannot be changed appreciably by the interactions with the subsystem 72 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS of interest. For example, a reaction taking place in a single living cell in the human body might be a reasonable approximation of a system (the cell) in contact with a heat reservoir (the body). From now on we call this subsystem of interest simply the system. We show that the equilibrium state of the system is an extremum in the Helmholtz potential of the system only . Since the system is at equilibrium, we know from Eqn. (2.69) that the temperature of the system is Tsys = Tres . From our postulates, we have proven that an equilibrium state is determined by the minimization of U at constant S , Eqn. (3.24). Since internal energy is extensive, we can write for our combined system plus reservoir 0= = = ∂U ∂X , + ∂ U res ∂X S,V,N ∂ U sys ∂X ∂ U sys ∂X S,V,N S,V,N + T res
S,V,N ∂ S res ∂X S,V,N −P ∂ V res ∂X +µ
S,V,N ∂ N res ∂X ,
S,V,N where we have used the diﬀerential for U to obtain the third line. However, we now use the fact that T sys = T res is a constant, and that the mole number and volume of the reservoir and system must also be constants. Since the total entropy must be ﬁxed at equilibrium, dS res = −dS sys , we ﬁnd 0= = = ∂ U sys ∂X
T sys ,V sys ,N sys − T sys ∂ S sys ∂X T sys ,V sys ,N sys ∂ [U sys − T sys S sys ] ∂X ∂ F sys ∂X
T sys ,V sys ,N sys T sys ,V sys ,N sys (3.28) The last line follows from the deﬁnition of F . In a similar way, one can prove that this extremum is a minimum by examining the secondorder derivative of F with respect to X . It is worthwhile to reﬂect on the importance of the last result. It means that the equilibrium state of a system in contact with a temperature reservoir can be found simply by studying the Helmholtz potential of the system—independent of what is happening in the reservoir . Similarly, one can show that a system in contact with a pressure reservoir reaches equilibrium when dHsys = 0, and a system in contact with both a pressure and temperature reservoir reaches equilibrium when dGsys = 0. In Chapter 4 we show 3.2. EXTREMA PRINCIPLES FOR THE POTENTIALS Type of system: isolated system 73 Equilibrium attained when: entropy of system is maximized, or energy of system is minimized system + heat bath Helmholtz potential of system is minimized system + pressure bath Enthalpy of system is minimized system + heat and pressure baths Gibbs potential of system is minimized Table 3.2: When considering a system in mechanical or thermal contact with a larger system which
acts as a bath, one needs consider the appropriate potential for the system only. how stability requires that these functions be minimized in the intensive variables. The results are summarized in Table 3.2. Example 3.2.1 Prove that the amount of work done on a system in contact with a heat bath is equal to the change in the system’s Helmholtz potential, if the process is done reversibly. Solution: We can begin with the diﬀerential for F , which is
dF = −SdT − P dV + µdN. (3.29) However, since the system is isothermal, the ﬁrst term on the right side is zero. The third term is also zero, since the system is closed. Since dWrev = −P dV , ¯ we have proven our result dWrev ¯ = dF, isothermal, closed system. (3.30) Therefore, one needs only ﬁnd the change in free energy of an isothermal system to determine the amount of reversible work necessary. This is especially useful if a thermodynamic chart is available. 2 Example 3.2.2 Using the thermodynamic diagram for ammonia on p.205, ﬁnd the work necessary to compress 3l of ammonia at 4MPa and 200◦ C isothermally to 25MPa. Find the work necessary for a compression performed adiabatically to the same pressure. Solution: From Example 3.2.1 we know that the amount of work necessary
to compress a substance isothermally is the change in the Helmholtz potential Wrev = ∆F, isothermal, closed. (3.31) We see from the thermodynamic diagram, we can obtain the enthalpy, entropy and density for known temperature and pressure. This is suﬃcient information to obtain the Helmholtz potential, using its deﬁnition F := U − T S = H − P V − T S = H − P Mw N − T S. ρ (3.32) 74 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS
We also used the deﬁnition of the enthalpy, and converted from volume V to density ρ. From the diagram, we get the initial conditions H (T = 200◦ C, P = 4M P a) = 915.kJ/kg S (T = 200◦ C, P = 4M P a) = 10.55kJ/kg · K ρ(T = 200◦ C, P = 4M P a) = 19.kg/m3 . Therefore, we obtain the initial Helmholtz potential to be fi = = (4 × 103 kP a)(1kJ/kPa·m3) − (273 + 200K )(10.55kJ/kg·K). 19kg/m3 −4286.kJ/kg. (3.34) (915.kJ/kg) − (3.33) Note that the particular choice chosen for zero enthalpy and zero entropy in creating the diagram makes the Helmholtz potential negative here. The (arbitrary) choice for zeroing internal energy cancels out when taking the diﬀerence. From the ﬁfth postulate, we know that one cannot arbitrarily choose entropy to be zero at some state. Therefore, the values for entropy given in the diagram should all have an additive value for S0 at the reference point. Since we take a diﬀerence here at constant temperature , this constant will vanish, so our answer is OK. At the ﬁnal state, T = 200◦C, P = 25MP a, we likewise ﬁnd H (T = 200◦ C, P = 25MPa) = S (T = 200◦ C, P = 25MPa) = ρ(T = 200◦ C, P = 25MPa) = 430.kJ/kg 8.75kJ/kg· K 220.kg/m3. (3.35) Therefore, the ﬁnal speciﬁc Helmholtz potential is ff = = (25 × 103 kPa)(1kJ/kPa·m3) − (273 + 200K)(8.75kJ/kg· K) 220.kg/m3 −3822.kJ/kg, (3.36) (430.kJ/kg) − Wrev = = = ∆F = ∆f N = ∆f V ρ (−3822. + 4286.kJ/kg)(3l)(19.kg/m3 )(10−3 m3 /l) 26.4kJ (3.37) and the reversible work required is Had we assumed that the gas were ideal, we would have obtained the result WIG = = = −N RT log Vf Vi = Pi Vi log ρf ρi 220.kg/m3 19.kg/m3 (3.38) (4000kP a)(0.003m3)(1kJ/kP a · m3 ) log 29.4kJ. 3.2. EXTREMA PRINCIPLES FOR THE POTENTIALS
Hence, the ideal gas assumption introduces an error of only 11%. To ﬁnd the adiabatic compression, we recall that an adiabatic, reversible compression is also isentropic. Hence, the ﬁnal state must have Pf = 25M P a, and speciﬁc entropy equal to the initial, 10.55kJ/kg · K . From the diagram, we see that the ﬁnal state is Tf = 395K , ρf = 80kg/m3, and hf = 1350kJ/kg . The required work is just the change in internal energy. From the deﬁnition of enthalpy, we can ﬁnd ui ui = = = hi − Pi vi = hi − Pi ρi 4000kP a (915.kJ/kg ) − 19.kg/m3 704.5kJ/kg. 75 (3.39) Similarly, we ﬁnd the ﬁnal speciﬁc internal energy to be uf = 1038kJ/kg . Hence, the required work for the adiabatic compression is Wadiab = = = ∆uN = ∆uρi Vi (1037.5 − 704.5kJ/kg )(19.kg/m3)(0.003m3 ) 19.0kJ. (3.40) This value is smaller, because the ﬁnal volume is much larger than that for the isothermal compression. 2 The following example analyzes the maximum theoretical eﬃciency of a heat engine, which illustrates the importance of enthalpy in combustion. Example 3.2.3 A heat engine exploits heat ﬂowing from a hot object to a colder one. Typically, the energy is provided by burning fuel, and the environment is the colder object. An example of such an engine is the steam cycle discussed in detail in §5.2.1. Find the maximum theoretical work available from one mole of hydrogen gas as a function of temperature. Solution: A simpliﬁed schematic of a heat engine is shown in Figure 3.4.
The process is broken into two subsystems as drawn. We begin with an energy balance on the furnace, assuming that the work on the environment is purely mechanical and reversible dU = dQrev − P dV ¯ d(U + P V ) = dH = → Qf = −∆Hrxn (T ). dQrev ¯ dQrev = − dQf ¯ ¯ (3.41) 76 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS
The second line follows because the system is operated isobarically. The third follows from the deﬁnition of H , and noting that the sign convention for Qf is diﬀerent from the usual, as is shown in the ﬁgure. The last line comes from integrating the diﬀerential from the initial state to the ﬁnal state. The diﬀerence in enthalpy between the products and the reactants, both at temperature T and pressure P , is called the heat of reaction, indicated by ∆Hrxn (T ). Figure 3.4: A simpliﬁed thermodynamic schematic of a heat engine. We divide the overall system into two subsystems. The top subsystem contains the furnace, reactants and products. It operates isothermally and isobarically, so that it might exchange mechanical work with the environment Wenviron through expansion (or contraction). It passes heat to the heat engine, which is the second subsystem. The heat engine involves cyclical behavior, so that its ﬁnal thermodynamic state must equal its initial state. Aside from taking heat from the furnace, it also expels heat to the environment, while producing usable mechanical work, Wout . To accomplish this work production, it is necessary that the reaction take place at a higher temperature than the environment: T > Tatm .
Next, performing an energy balance on the heat engine yields (being careful with the proper signs for heat and work ﬂows, as shown in the drawing) dU = dQf − dQc − dWout ¯ ¯ ¯ (3.42) We can integrate this equation from the initial to the ﬁnal state. Since the engine operates as a cycle, its ﬁnal internal energy must be the same as its ﬁnal, making the left side zero. Hence, Wout = Qf − Qc = −∆Hrxn (T ) − Qc , (3.43) 3.3. MAXWELL RELATIONS
where we have used the result from the energy balance on the furnace, Eqn. (3.42). To ﬁnish, we need to ﬁnd the heat ﬂow Qc . Clearly, the smaller Qc , the more work will be available. The limitation in Qc comes from an overall entropy balance. Here we perform an entropy balance over just the heat engine, assuming reversible heat transfer with its environment. The entropy balance is then dStotal = dSenviron + dSengine , (3.44) 77 where the subscripts are ‘total’ for the universe, ‘environ’ for the environment, and ‘engine’ for the engine. Still assuming reversibility, we know the change in entropy for the environment from heat ﬂow is dQc /Tatm − dQf /Trxn . Note ¯ ¯ that the arrows in the sketch determine our signs here, and that we assume the heat delivered from the furnace is at temperature Trxn . We thus obtain 0=− dQf ¯ dQc ¯ + + dSengine , Trxn Tatm (3.45) assuming reversibility. We integrate this equation noting that the temperatures are constant Qf Qc 0= − , (3.46) Tatm Trxn where we have again exploited the fact that the engine operates as a cycle. We can now ﬁnd the amount of work available for the heat engine Wout = (−∆Hrxn (T )) 1 − Tatm Trxn (3.47) While we assumed that the reaction was exothermic above, this expression actually holds for an endothermic reaction as well. In that case, each term in the parentheses is negative, and work is still available. Also note that the greater the temperature diﬀerences, the larger the available work. These results require that the entire system operates reversibly. Hence, entropy may not be generated from heat ﬂow across ﬁnite temperature diﬀerences, all processes must be quasistatic, there is no friction, etc. Of course, real processes always operate irreversibly. So these results allow one to estimate how close a real system is to ideal performance, or to get rough estimates when designing. Also, many cyclical engines do not operate in this manner. An automobile engine, for example, exploits the expansion of the reaction of fuel to produce work, rather than exploiting heat ﬂow. Hence, the analysis done here does not apply to typical combustion engines. 2 3.3 Maxwell Relations In the last section, we saw that the enthalpy, H , the Helmholtz potential, F , and the Gibbs free energy, G, contain complete thermodynamic information when expressed in their canonical independent variables. In Chapter 2, from the deﬁnitions 78 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS V
s d d F
d d d d d d T U G S
(3.59). H P Figure 3.5: Thermodynamic square for remembering Eqs. (3.48) through (3.51), and (3.56) through of temperature, pressure and chemical potential, we have already found the total diﬀerentials for both internal energy and entropy, Eqs. (2.27) and (2.29). By simple manipulation we can also ﬁnd the total diﬀerential for the Helmholtz potential in its canonical independent variables dF = dU − T dS − SdT = −SdT − P dV +
m µi dNi ,
i where we used the diﬀerential for internal energy Eqn. (2.27) to obtain the second line. We can perform similar manipulations on the other potentials to ﬁnd their total diﬀerentials. The results are dU dH dF = = T dS − P dV + T dS + V dP +
i µi dNi ,
i (3.48) (3.49) (3.50) (3.51) µi dNi , µi dNi ,
i = −SdT − P dV + dG = −SdT + V dP + µi dNi .
i From these diﬀerential expressions we can also see how to ﬁnd equations of state from the generalized potential. In addition to those shown in the second column of 3.3. MAXWELL RELATIONS Table 3.1, we can write T= ∂H ∂S ,
P ,N 79 P =− ∂F ∂V .
T ,N (3.52) Therefore, all thermodynamic information can be retrieved from the generalized potentials as functions of their canonical variables, just as we have already done from the fundamental energy relation. Since H, F and G are constructed from simple manipulations of the analytic function U , they are also analytic (§A.1). As analytic functions, their secondorder derivatives must be independent of the order of diﬀerentiation. Hence, we can write ∂2H ∂2H = . ∂S∂P ∂P ∂S (3.53) Recall from the appendix that we must be careful to remember what independent variables are being held constant in secondorder partial derivatives. Writing the term on the left side more explicitly and using Eqn. (3.49) we can write ∂2H ∂S∂P = = Similarly, ∂2H ∂P ∂S = = ∂ ∂P ∂T ∂P ∂H ∂S .
S,N ∂ ∂S ∂V ∂S ∂H ∂P .
P ,N S,N P ,N (3.54) P ,N S,N (3.55) Putting Eqs. (3.54) and (3.55) into Eqn. (3.53) yields our ﬁrst Maxwell relation ∂V ∂S =
P ,N ∂T ∂P .
S,N Of course, we can use the same procedure for U, F and G to obtain three more relations. When we do, we ﬁnd the four Maxwell relations 80 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS ∂T ∂V ∂V ∂S ∂P ∂T ∂S ∂P S,N =− = ∂P ∂S ∂T ∂P ∂S ∂V ∂V ∂T (3.56)
V ,N (3.57)
S,N P ,N =
V ,N (3.58)
T ,N T ,N =− .
P ,N (3.59) It is not necessary to memorize these equations. They are straightforward to derive from the diﬀerentials, or generated from the thermodynamic square in the next section, when needed. Also, any generalized potential—including Massieu functions—can be used to generate such a relation. It is already clear how the Maxwell relations might be useful. For example, it might be extremely diﬃcult to measure a change in entropy with pressure for a system at constant temperature. However, the last Maxwell relation allows us to measure the change in volume with temperature at constant pressure to ﬁnd the same quantity. When the Maxwell relations are used in combination with the techniques outlined in Chapter 4, we will be able to ﬁnd any conceivable thermodynamic quantity in terms of such measurable quantities. 3.4 The Thermodynamic Square In this section we outline how we can reconstruct all of the results of §§3.1–3.3 in a straightforward manner using a diagram constructed from a mnemonic device. The phrase to remember is the unpleasant sentence Very Few Thermodynamics UnderGraduate Students Have Passed.2 First we draw a square, and place all of the boldfaced letters at the corners and sides of the square beginning at the upper lefthand corner. Then, we draw two arrows from the corners of the square, both going upwards. The result is shown in Figure 3.5. We are ﬁnished constructing the diagram. The diagram is used to construct the total diﬀerentials by the following. Note that the energies are at the sides of the square, with the canonical independent
2 One may also choose among the phrases Va Falloir Trimer Grave Pour Harmoniser Son Univers , or Una Verdadera Funci´n Thermodinamica Guarda Premisas Hondamente Simples . o 3.4. THE THERMODYNAMIC SQUARE 81 variables at the adjacent corners. The independent variables make up the diﬀerentials on the righthandside of the equation. The corresponding coeﬃcients are connected by the arrows. Pick out the energy whose diﬀerential you would like to ﬁnd. Right down the diﬀerential of that energy on the left side of your equation. If an arrow points from the coeﬃcient to the independent variable (in other words towards the side with the energy of interest), then the sign of that diﬀerential on the righthandside is negative. Otherwise, it is positive. Example 3.4.1 Find the diﬀerential for internal energy using the thermodynamic square. Solution: The square shows that the canonical independent variables for U
are (S, V ), hence the diﬀerentials dS and dV must show up on the right side. Their coeﬃcients according to the square are T and −P , respectively. Hence, we can immediately write down Eqn. (2.27) dU = −P dV + T dS. Note that the square does not deal with changes in mole number, so one must remember always to add + i µi dNi . 2 The Maxwell relations may also be found from the square. First, note that all Maxwell relations involve derivatives of a variable on one corner with respect to the variable on an adjacent corner, while the variable on the opposite corner is held constant. The corresponding derivative in the Maxwell relation involves swapping the independent variables. The new dependent variable is opposite the new ﬁxed variable. The sign of the relation is again given by the arrows; if the arrows both point to (or both point away from) the dependent variables, the sign is positive. If one arrow points towards one of the dependent variables, but the other arrow points away from the other dependent variables, the sign is negative. Example 3.4.2 Find a Maxwell relation for
∂S ∂V T . Solution: We note that the ﬁxed variable T lies opposite to the dependent
variable S , and that the independent variable V lies adjacent to S , so there exists a Maxwell relation for the derivative. The diﬀerential related to this one by a Maxwell relation must have V as the ﬁxed variable, and T as the changing variable. The intensive variable opposite V must be the dependent variable, or P . Hence, our second diﬀerential is ∂P ∂T .
V Both arrows point away from the dependent variables S and P , so the sign must be the same on each. Hence, we have just found Eqn. (3.58) using the thermodynamic square. 2 82 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS It is highly recommended that the reader attempt to reconstruct all of the differentials and the Maxwell relations with a blank sheet of paper, a pencil, and remembering that very few thermodynamics undergraduate students have passed. 3.5 Coeﬃcient of Thermal Expansion, Isothermal Compressibility and ConstantPressure Heat Capacity Everyday experience shows that many materials are functions of the ﬁrstorder derivatives T, P, and µi . For example, when objects become warmer, their volume increases. Gases and liquids can shrink considerably under an increase of pressure, and ice freezes at a lower temperature when the chemical potential of salt in the ice is increased. These eﬀects are all examples of secondorder derivatives .3 The three most frequently tabulated secondorder derivatives for pure substances are α := κT cP 1 v 1 := − v := T ∂v ∂T ∂v ∂P ∂s ∂T , “coeﬃcient of thermal expansion”
P (3.60) (3.61) (3.62) , “isothermal compressibility”
T , “constantpressure heat capacity”
P These deﬁnitions should be memorized. Using the thermodynamic transformations introduced in §3.6, it is possible to relate all other secondorder derivatives to these three. These quantities can be found from measurement, tables, or equations of state like those given in §§B and D.2. It is straightforward to imagine experiments to measure α or κT . To measure CP , we use Eqn. (2.46) to write CP := T ∂S ∂T =
P ∂ Qqs ∂T .
P (3.63) Therefore, if one has information on how much heat is required to raise the temperature of a substance by a small amount at constant pressure, the ratio of heat divided by temperature change is the heat capacity. Such measurements are routinely performed in a diﬀerential scanning calorimeter, or DSC. These devices raise the temperature of a substance at a controlled rate, and measure the rate of
3 Since pressure and temperature are ﬁrstorder derivatives of the fundamental variable U (S, V, N ), we call quantities that are derivatives of these secondorder derivatives. 3.5. SECONDORDER COEFFICIENTS 83 heat required for the change. If these rates are suﬃciently slow that the process is d d quasistatic, then the ratio of dt Q to dt T yields the heat capacity as a function of temperature.4 As an aside, we point out that intuitively we expect each of these quantities to be positive. In §4.1 we will see that the postulates indeed show that κT and cP are required to be positive. Example 3.5.1 Find the isothermal compressibility for the simple van der Waals ﬂuid. Solution: From Examples 2.5.1 and 2.6.1, we have seen that the proposed
fundamental entropy relation given in Eqn. (2.21) leads to the van der Waals mechanical equation of state P= RT aN 2 − 2, V /N − b V (2.35) and the simple van der Waals thermal equation of state Eqn. (2.34). We begin with the deﬁnition of κT Eqn. (3.61) κT := = = − − 1 v v ∂v ∂P 1
∂P ∂v T 2 T RT v 3 − 2a (v − b) v ( v − b) 2 2. (3.64) To obtain the second line we applied the relation Eqn. (A.18). Then, in order to ﬁnd the derivative in the denominator, we take the derivative of each side of the van der Waals mechanical equation of state (2.35) with respect to v , and insert the result into the second line above. After some algebraic manipulation, the third line is the result. Note that the denominator can pass through zero for some values of v . This means that the van der Waals ﬂuid’s isothermal compressibility can sometimes be inﬁnity! Is this result physical? The surprising answer to this question must wait until the next chapter. 2 Example 3.5.2 Find the constantpressure and constantvolume heat capacities for the general ideal gas given in Example 3.1.2.
Glassy liquids are ones that relax extremely slowly, and can fall out of equilibrium upon cooling. Hence, most DSCs change the temperature too fast to be quasistatic for these, and spurious results can be obtained.
4 84 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS Solution: The deﬁnition for the constantpressure heat capacity Eqn. (3.62)
requires that we have s(T, P ). However, Eqn. (3.17) gives entropy as a function of temperature and volume. If we use the ideal gas law found in Example 3.1.3 to eliminate volume from Eqn. (3.17), we obtain S = Ns0 + N C (T ′ ) ′ RT dT + NR log , T′ P v0 T0 general ideal gas.
T (3.65) Inserting this expression into the deﬁnition Eqn. (3.62) gives us the constantpressure heat capacity for the general ideal gas described by Eqn. (3.16) cP (T ) = R + C (T ), general ideal gas. Note that it was necessary to use the Leibniz rule Eqn. (A.25) to obtain this result. The constantvolume heat capacity has the logical deﬁnition cV := T ∂s ∂T ,
v (3.66) so we may use Eqn. (3.16) directly to ﬁnd cV ≡ C (T ). 2 (3.67) We now see the physical signiﬁcance of the function C (T ) introduced in the earlier example. C (T ) is actually called the “ideal constantvolume heat capacity.” Polynomial expressions for several purecomponent substances are given in Table D.2. The secondorder derivatives are extremely useful for solving problems. In the following example, we show how to extract the work available from a hydrogen fuel cell, which operates isothermally and isobarically using the heat capacity, and a result from §3.2. Example 3.5.3 Prove that the amount of electrical work available from a fuel cell is equal to the change in the system’s Gibbs potential, if the process is done isothermally, isobarically, and reversibly. Use these results to ﬁnd the maximum work available from a hydrogen fuel cell utilizing the reaction 1 H2 + O2 → H2 O(g), 2 for 300K < T < 1000K . 3.5. SECONDORDER COEFFICIENTS 85 Solution: A fuel cell takes reactants (e.g., hydrogen and oxygen) and reacts
them to make electricity. Figure 3.6 shows a picture of a fuel cell, and Figure 3.7 shows a schematic of its operation. The oxidation reaction and the reduction reaction take place at diﬀerent electrodes, and the transport of the electrons takes place not through the liquid phase, but rather through an electrical circuit. The circuit can then extract the work available from the reaction. Figure 3.6: Photograph of an experimental fuel cell using methanol as a fuel. Visible in front are the
fuel ports, and in back, the connection to a copper electrode. Photo by William Mustain. These systems are open , since reactants and products ﬂow through the fuel cell. However, we will simplify the system by looking at the ﬁnal state (the products + fuel cell), the initial state (the reactants + fuel cell), the work done, and the heat ﬂow. The system is still open, however, since electrons ﬂow across the system boundary, providing the electrical work (Figure 3.8). For our system, we begin with the conservationofenergy equation dU = dQ + dW. ¯ ¯ (2.1) For our system, there are two possible kinds of work: mechanical (from compression or expansion), and electrical work. We are assuming here that everything happens reversibly. It is important to note here that we have written the 86 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS Figure 3.7: Sketch of a fuel cell. This sketch is taken without permission from the Fuel Cell Handbook
[3]. electrical work as positive when it provides work to the surroundings. We begin with the diﬀerential for internal energy dU = = = T dS − P dV + T dS − P dV + T dS − P dV + µi dNi i in out µin dNe + µout dNe , e e in µin − µout dNe , e e (3.68) since only the electrons cross the boundary, and the electrons entering one electrode must balance those exiting the other. Clearly, the ﬁrst term on the right side is the reversible heat exchange with the environment, the second term is the reversible work, and the third term is the electrical work dQrev ¯ dWvolume ¯ dWelec ¯ = T dS, = −P dV, in out = −µin dNe − µout dNe . e e (3.69) Rearranging the energy balance yields dU + P dV − T dS = − dWelec . ¯ (3.70) We recognize that the left side of this equation is just the diﬀerential for G when isothermal and isobaric, since d(U − T S + P V ) = dG dG = − dWelec , isothermal, isobaric, reversible. ¯ (3.71) Therefore, if we wish to know how much electrical work is possible to extract from our reactants, we need only know the change in Gibbs free energy. Remember that work obtained from the fuel cell would be positive, so we require a reaction that has negative ∆Grxn . 3.5. SECONDORDER COEFFICIENTS 87 Figure 3.8: To analyze the maximum theoretical eﬃciency of a fuel cell, we can lump the reactants, products, and fuel cell in our system. Here we allow no mass to pass across the boundary of our system, which operates at constant pressure and temperature. Because the system is isobaric, it may exchange work with the environment, which is assumed reversible. It also produces useful electrical work. Because it is isothermal, it may exchange heat with the environment. We may use experimental data available from JANAF [15], or from the NIST Chemistry Webbook to plot the speciﬁc Gibbs potential, which is the maximum allowed work per mole of hydrogen available from the fuel cell. The relevant available data are the heats of formation of each compound ∆Hf◦,i [Tref ] at a reference temperature Tref and pressure, and the ideal heat capacities ideal CP,i . We will assume here that the pressure is suﬃciently low that the gases are ideal. The heat of formation of any compound at temperature T is the diﬀerence in enthalpy between the compound and it constituent elements. For example, for water 1 ∆Hf ,H2 O [T ] := HH2 O [T ] − HH2 [T ] − HO2 (g) [T ]. 2 (3.72) From this deﬁnition of heat of formation, the heat of reaction for the hydrogen reaction is the same as the heat of formation for water. However, we now need to correct for the change in temperature. The temperature correction is achieved through the heat capacity. Beginning 88 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS
with the deﬁnition for H , we write dH = = = = d(U + P V ) dU + P dV + V dP T dS + V dP + µdN T dS, isobaric, isomolar (3.73) where we used the chain rule of diﬀerentiation to obtain the second line, the diﬀerential for U, Eqn. (2.27), to obtain the third, and the shown restrictions to obtain the last line. Dividing by dT , and holding P and N constant yields ∂H ∂T =T
P,N ∂S ∂T P,N = CP ∼ C ideal = P (3.74) where we used the deﬁnition of heat capacity, and our assumption of ideality. We then integrate each side over T from Tref to T , and use the fundamental theorem of calculus (A.24) to obtain H [T ] ∼ H [Tref ] + =
T Tref ideal CP [T ′ ]dT ′ . (3.75) Using Eqn. (3.75) for each of the substances in Eqn. (3.72), the heat of formation at a reference temperature, and the heat capacities, we can ﬁnd the heat of formation for water at any temperature. The result is shown in Figure 3.9. Our analysis of the fuel cell operation is not yet complete, because some fuel cells also require an input of heat to keep them operating isothermally. This heat input also costs money, so it either must be taken from the available work, or arise from irreversibility, during which some of the fuel must be burned. We can ﬁnd the heat ﬂow necessary for our example from the conservation of energy equation dQrev ¯ We use this result in the next example Q = T ∆Srxn = ∆Hrxn − ∆Grxn . (3.77) = T dS. (3.76) Figure 3.9 shows that ∆Grxn > ∆Hrxn . Therefore, Q < 0, and heat ﬂows out of the hydrogen fuel cell. Since, at these temperatures, the fuel cell is operating above room temperature, there is no additional cost of heating or refrigerating. In Figure 3.9 we also show the maximum theoretical work possible for a heat engine that burns hydrogen, assuming Tatm = 298K . We see that the heat engine has a higher theoretical eﬃciency, but only if there is an extremely cold reservoir to expel the heat. 2 3.5. SECONDORDER COEFFICIENTS
250 240 230 Energy, kJ/mol 220 210 200 89 −∆hrxn −∆grxn −∆hrxn + Tatm ∆srxn
300 400 500 600 700 Temperature, K 800 900 1000 190 200 Figure 3.9: Diﬀerence in enthalpy and Gibbs free energies of the reactants and products for the
hydrogen reaction considered in Example 3.5.3. In that example we show how the maximum work available for a fuel cell per mole of hydrogen is −∆grxn . In Example 3.2.3 we show that the maximum work available from a heat engine burning hydrogen is −∆hrxn + Tatm ∆srxn . These data are taken from the NIST web page. In the previous example, we found the maximum electrical work available from a hydrogen fuel cell operating isothermally and isobarically at a range of temperatures. Fuel cells are sometimes touted as having greater theoretical eﬃciency than heat engines. Applying the results of the last examples to Example 3.2.3, we can then ﬁnd the conditions where work available from a fuel cell is greater than work available from a heat engine for the same amount of fuel. This comparison is made in the next example. Example 3.5.4 When does a fuel cell have a higher theoretical eﬃciency than a heat engine for a given reaction? Solution: In Example 3.5.3, we analyzed a hydrogen fuel cell. From that
analysis we discovered that the fuel cell expelled heat to its environment. Since the fuel cell was operating at temperatures well above ambient, this incurred no more cost. However, if the fuel cell required energy to maintain its constant temperature, or if it operated at lower temperatures, the answer would be diﬀerent. Therefore, several possible outcomes are possible for the comparison. First of all, however, the fuel cell and heat engine both require that ∆Grxn < 0, or the reaction will not take place (see Table 3.2). On the other hand, ∆Srxn 90 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS
can be positive or negative. Also, knowledge of the operating temperature relative to ambient temperature Tatm is important to determine whether we must incur the costs of heating or cooling the fuel cell. We consider the resulting four cases one at a time. 1. ∆Srxn > 0 and T > Tatm . Since ∆Srxn > 0, the heat into the fuel cell Q = T ∆Srxn > 0 is positive. Since the temperature of the cell is above ambient, it requires heat to maintain its temperature, Eqn. (3.76). To continue the comparison, we make a simplifying assumption that some amount of the fuel is burned outside of the fuel cell to supply the heat. Let us ﬁnd the amount of fuel that would be burned. For every mole of fuel, x reacted in the cell, 1 − x is burned to keep the cell warm. The amount of heat necessary to warm the cell per mole of fuel is Q/N = xT ∆srxn from Eqn. (3.77). The heat available from burning the 1 − x moles of fuel is Q = −∆hrxn (1 − x). Setting these two expressions for Q equal tells us what fraction of fuel must be burned to keep the cell operating x= ∆Hrxn . ∆Grxn (3.78) Therefore, the amount of work available is Welec = −x∆Grxn = − ∆Hrxn ∆Grxn = −∆Hrxn . ∆Grxn (3.79) Hence, the fuel cell has the same theoretical maximum of work as a heat engine, but only if there exists a heat reservoir at zero temperature. In reality, the ineﬃciency of a fuel cell probably generates heat, reducing x in our analysis, but simultaneously reducing W . At any rate, the analysis here is the theoretical maximum. 2. ∆Srxn > 0 and T < Tatm . Here, the fuel cell still requires heat. However, since it operates below ambient temperature, no fuel must be burned, and heat can be absorbed from the environment. Also, since ∆Srxn > 0, Welec = −∆Grxn > −∆Hrxn , (3.80) and the fuel cell has a higher maximum theoretical eﬃciency than a heat engine. In practice, existing fuel cells all operate at elevated temperatures, so this advantage is not exploited. The elevated temperature is required for a practical reason: faster kinetics. In theory, a fuel cell could be designed to be more eﬃcient than a heat engine at low temperatures. 3. ∆Srxn < 0 and T > Tatm . For this case, Q < 0, and heat is dissipated to the environment. Since the fuel cell is hotter than the environment, no additional cost is incurred for 3.6. THERMODYNAMIC MANIPULATIONS Fuel Cell Welec = −∆Hrxn Welec = −∆Grxn = −∆Hrxn + T ∆Srxn Welec = −∆Grxn > −∆Hrxn Not Possible Without Refrigeration Heat Engine Wengine = (−∆Hrxn )(1 − (Tatm /T )) Wengine = (−∆Hrxn )(1 − (Tatm /T )) Wengine = (−∆Hrxn )(1 − (Tatm /T )) Wengine = (−∆Hrxn )(1 − (Tatm /T )) 91 ∆Srxn (T ) > 0, T > Tatm ∆Srxn (T ) < 0, T > Tatm ∆Srxn (T ) > 0, T < Tatm ∆Srxn (T ) < 0, T < Tatm Table 3.3: This analysis assumes isothermal, isobaric, and reversible operation of both machines. The fuel cell and furnace operate at temperature T , in an environment of temperature Tatm . The entropy of reaction may be either positive or negative at this temperature, as indicated. cooling. Therefore, the available work is Welec = −∆Grxn < −∆Hrxn , (3.81) and the fuel cell has less available work than a heat engine. A real comparison requires knowing the reaction and ambient temperatures, however. 4. ∆Srxn < 0 and T < Tatm . The fuel cell still expels heat, but now it will not ﬂow to the environment, except perhaps using a refrigerator, which costs money. This is the worst case of all. Not only is −∆Grxn < −∆Hrxn , we do not even get all of the work from −∆Grxn , because we have to pay for a refrigerator. A summary of these results is shown in Table 3.3. Currently, all fuel cells operate at temperatures well above ambient, so only the ﬁrst two lines are typical. The hydrogen fuel cell considered in Example 3.5.3 is a special case of the second line of Table 3.3. In Problem 3.3.F. we consider a methanol fuel cell, which is described by the ﬁrst line of the table. 2 In the examples here, we calculated the maximum possible work, using a closed system operating isobarically and isothermally. In Chapter 5 we consider the analogous open systems. There we consider systems not operating reversibly, and calculate the eﬃciency. Still, the analyses here form the basis for such comparisons. 3.6 Thermodynamic Manipulations The application of thermodynamics requires estimation of many sorts of derivatives. For example, in the last section we found that analysis of a fuel cell required estimates of changes in Gibbs free energy, enthalpy and entropy. These estimates 92 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS were calculated using just the constantpressure heat capacity. Since the heat capacity is straightforward to measure experimentally, it was possible to ﬁnd such data. However, sometimes we require quantities that are not so easy to measure, and are, therefore, typically not tabulated. For example, in turbine engine design we assume that part of the cycle is done isentropically, and we need to know how the temperature changes with pressure. Hence, we need to ﬁnd Pf T [Pf ] = T [P0 ] +
P0 ∂T ∂P dP.
S,N But it is very unlikely that you will ﬁnd an equation of state for temperature as a function of pressure and entropy for many substances. On the other hand, we are aware from the structure of thermodynamics that all equations of state are derivable from a single equation (any of the fundamental relations). Hence, there must be only a ﬁnite number of equations of state that are independent, and the equation of state T = T (S, P ) must be expressible in terms of this independent set. Up to this point, we have derived equations of state from fundamental relations. However, it turns out that we need integrate only two equations of state to ﬁx the fundamental relations for a pure substance [13, pp.59–65]. Therefore, if we have either two equations of state, or equivalent tabulated data for our system, the derivative of interest is ﬁxed. We can also show that, for a singlecomponent system, only three secondorder derivatives are independent—all others can be derived from these . This is very important, so we repeat it: Complete thermodynamic information is contained in either two equations of state, or in the dependence of three secondorder derivatives on temperature and pressure. This means that sometimes we will not need to have a fundamental relation to make predictions—some measurement of secondorder derivatives might be suﬃcient. This fact will be exploited in the following chapter. In this section, we show how to reduce all derivatives (except those involving N ) to the three secondorder derivatives introduced in §3.5. The procedure suggested here may not be the most direct method for all cases, but is guaranteed to work. We use the method of Jacobians reviewed in the appendix and the thermodynamic square introduced in §3.4, and one Maxwell relation. The procedure is as follows: 3.6. THERMODYNAMIC MANIPULATIONS 93 1. Using the Jacobian manipulations, reduce all derivatives to those using T and P as the independent variables. This reduction may also be done by the three relations derived in the appendix, but is more easily done with Jacobians. 2. All derivatives of potentials (U, H, F, G) may be reduced by use of their diﬀerentials, Eqs. (3.48)–(3.51). Recall that these are easily found from the thermodynamic square, Fig.(3.5). 3. You should now be left only with derivatives of S or V with respect to T or P . The derivatives of V , and (∂S/∂T )P are the secondorder derivatives, Eqs. (3.60)–(3.62). The other derivative (∂S/∂P )T can be found from the fourth Maxwell relation, Eqn. (3.59). Note that it is sometimes more convenient to exchange the ﬁrst and second steps. The general procedure is used in the following illustrative examples. Example 3.6.1 Find
∂T ∂P S in terms of tabulated quantities. Solution: The ﬁrst step of the procedure is to make T and P the independent variables. We use the ﬁrst property of the Jacobian Eqn. (A.14) to write ∂ (T, S ) ∂T = ∂P S,N ∂ (P, S )
∂ (T, S ) ∂ (P, T ) ∂ (P, T ) ∂ (P, S ) ∂ (S, T ) ∂ (S, P ) =− ∂ (P, T ) ∂ (T, P ) ∂S ∂S =− , ∂P T ∂T P = which can also be found from Eqn. (A.20). We are ﬁnished with step 1. We have no derivatives of potentials, so we go on to step 3. We note that the denominator is 1/T times the constantpressure heat capacity. The numerator can be eliminated using the fourth Maxwell relation − ∂S ∂P =
T ∂V ∂T = V α,
P where α is the coeﬃcient of thermal expansion, deﬁned by Eqn. (3.60). Hence, we are ﬁnished ∂T αV T = . (3.82) ∂P S CP 94
2 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS Example 3.6.2 Find ∂U ∂T P in terms of tabulated quantities. Solution: The independent variables are already T and P , so we go on to step 2. We have a derivative of the potential U , so we use its diﬀerential from the thermodynamic square. Dividing each side of the diﬀerential for U Eqn. (2.27) by dT and holding P and N constant we obtain ∂U ∂T =T
P ∂S ∂T P −P ∂V ∂T .
P (3.83) At step 3 we recognize that we have only secondorder derivatives left. Therefore, using their deﬁnitions, Eqs. (3.60) and (3.62), we ﬁnd ∂U ∂T 2 = CP − P V α. (3.84) P Example 3.6.3 Find the constantvolume heat capacity in terms of tabulated quantities. Solution: The constantvolume heat capacity is deﬁned as
CV := T ∂S ∂T .
V (3.85) We use the Jacobian manipulations to ﬁnd CV = = = T ∂ (S, V ) ∂ (T, V ) ∂ (S, V ) T ∂ (T, P )
∂S ∂T P ∂V ∂T ∂ (T, V ) ∂ (T, P )
∂S ∂P T ∂V ∂P T T ∂V ∂P .
T P To obtain the last line we used the deﬁnition of the Jacobian. Since there are no derivatives of potentials, we go to step 3 of the procedure. Note that all of the terms are secondorder derivatives, except the upper righthand term of the determinant; however, we can use the Maxwell relation to reduce it. Using ﬁrst the Maxwell relation in expanding the determinant, and then the deﬁnitions of the secondorder derivatives we ﬁnd CV = = T
∂V ∂P T 2 ∂S ∂T P ∂V ∂P +
T ∂V ∂T 2 P CP − Vα T . κT (3.86) 3.6. THERMODYNAMIC MANIPULATIONS
2 95 The last result simpliﬁes to a wellknown expression for an ideal gas. For the ideal gas, it is straightforward to prove that κT = NRT /(P 2 V ) and α = NR/(P V ). ideal = C ideal − R. Hence, we ﬁnd CV P Example 3.6.4 Using the data from the steam tables §D.4, ﬁnd the diﬀerence in chemical potential of steam at 2MPa and 5MPa, both at 300◦ C. Solution: We can write the diﬀerence of the chemical potential as the integral over a derivative
P1 µ[P1 ] − µ[P0 ] = P0 ∂µ ∂P dP.
T ,N (3.87) The fact that temperature and pressure are the independent variables suggests using the Gibbs potential. In fact, from the Euler relation, Eqn. (3.20), we can write µ(T, P ) = g (T, P ). (3.88) Taking the derivative with respect to P of each side ∂µ ∂P =
T ,N ∂g ∂P v T = (3.89) where we used the diﬀerential for g to obtain the second line. Eqn. (3.87) becomes
P1 µ[P1 ] − µ[P0 ] = v (T, P )dP.
P0 (3.90) The steam tables provide numerical values for the speciﬁc volume for many values of temperature and pressure. These must be integrated numerically. Many simple methods exist for estimating the area under the curve. Here we choose to integrate a polynomial ﬁt to the data. For other methods of integration, see [67]. Figure 3.10 shows the data taken from the steam tables in Appendix D.4 at the temperature and range of pressures of interest. Also shown is a polynomial ﬁt to the data. We ﬁnd that the expression v ≈ A0 + A1 P + A2 /P, (3.91) describes the data extremely well. Insertion of this expression into our result Eqn. (3.90) yields the approximation 1 2 2 µ[P1 ] − µ[P0 ] ≈ A0 ∗ (P1 − P0 ) + A1 ∗ (P1 − P0 ) + A2 ∗ log(P1 /P0 ). (3.92) 2 96
speciﬁc volume, cm3 /g CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS × 120 × 100 80 60 2e+06 data × × polynomial ﬁt × × × × × × × × × × ×× ×× ×× ×× ×× ×× ×× ××× × 3e+06 4e+06 Pressure, Pa 5e+06 Figure 3.10: Speciﬁc volume for supersaturated steam at 300◦ C as a function of pressure. Also shown
is a polynomial ﬁt to these data, Eqn. (3.91), with parameters given by Eqn. (3.93). Using a LevenbergMarquardt method5 for ﬁtting a nonlinear function to data, we obtain estimates for the parameters A0 A1 A2 ≈ −5.9024cm3/g, ≈ ≈ −3.2222 × 10−7 cm3 /g·Pa, 2.609 × 108 cm3 ·Pa/g. (3.93) Using the fact that 1MPa·cm3 = 1J, we ﬁnd that the chemical potential diﬀerence is µ[P1 ] − µ[P0 ] ≈ 217.5J/g. 2 (3.94) 3.7 One and TwoDimensional Systems In §2.9 we considered the fundamental entropy relation of a simple, ideal rubber band. In this section, we generalize the structure of thermodynamics a little bit more to incorporate the generalized thermodynamic potentials introduced in this chapter. In particular, we revisit the rubber band to allow nonsimple and nonideal behavior. We also consider Langmuir adsorption isotherms to describe the attachment of molecules to adsorption sites on a solid surface. In the following subsection we derive our ﬁrst fundamental relation from two equations of state that are found experimentally.
5 See, for example [67, §15.5]. Note that a powerlaw expression v = A0 P a may work even better, and is easier to ﬁt. 3.7. ONE AND TWODIMENSIONAL SYSTEMS 97 3.7.1 Nonideal Rubber Band Experiments show that the tension in a rubber band increases linearly with temperature at a ﬁxed length, for stretches greater than approximately 10% [54]. We can use this experimental fact to show that stretching a rubber band isothermally decreases the entropy of the rubber band, but does not change its internal energy. We write the experimental observation as T = T ψ (L), (3.95) where ψ (L) is some arbitrary, but measurable function of length, such that the tension is zero at the rest length L0 , ψ (L0 ) = 0, but positive for L > L0 . Since T and L are the independent variables (the variables easily manipulated in experiment), it is natural to use F as our fundamental relation. Using the relationship between pressure and Helmholtz potential implied by its diﬀerential, we can integrate Eqn. (3.95) from L0 to arbitrary L at constant temperature to obtain
L F (T, L) − F [L0 ] = = L0 L L0 ∂F ∂L dL
T T (T, L)dL
L =T
L0 ψ (L′ )dL′ . (3.96) If we take the derivative with respect to T of each side of this equation, we can obtain an equation of state for entropy
L S (T, L) − S [L0 ] = − ψ (L′ )dL′ . (3.97) L0 Note that the entropy decreases with stretching. We can ﬁnd S [L0 ] from an integration over temperature
T S [L0 ] = S [T0 , L0 ] +
T0 T ∂S ∂T dT
L L=L0 = S [T0 , L0 ] +
T0 CL [T ′ , L0 ] ′ dT , T′ (3.98) 98 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS where CL := T (∂S/∂T )L is the constantlength heat capacity for the rubber band.6 Similarly, we can ﬁnd F [L0 ] from an integration
T F [L0 ] = F [T0 , L0 ] +
T0 T ∂F ∂T dT
L = F [T0 , L0 ] − S [T ′ , L0 ]dT ′ . (3.99) T0 We can insert Eqn. (3.98) into Eqn. (3.96) to ﬁnd
L F (T, L) = T L0 T T′ T0 T0 ψ (L′ )dL′ + F0 − (T − T0 )S0 − CL [T ′′ , L0 ] ′′ ′ dT dT . T ′′ (3.100) The double integral can be simpliﬁed by changing the order of integration to arrive at
L F (T, L) = F0 − (T − T0 )S0 + T
T T0 L0 ψ (L′ )dL′ − (3.101) T − T′ T′ CL [T ′ , L0 ]dT ′ . Note that we have derived a fundamental relation from the two measurable quantities T (T, L) and CL (T, L0 ). We can use these results, and the deﬁnition for F to ﬁnd an equation of state for U U = F + TS
T = U0 +
T0 CL [T ′ , L0 ]dT ′ , (3.102) where U0 := F [L0 , T0 ]+ T0 S [T0 , L0 ]. Hence, we have proven that the internal energy does not depend on length at constant temperature for a rubber band whose tension changes linearly with temperature. This result may seem surprising, given the fact that a stretched rubber band stores the ability to do work. However, the apparent paradox can be understood in the following way.7 When a rubber band is stretched isothermally, work is done
Actually, it is not straightforward to measure the constantlength heat capacity, since the rest length tends to change with temperature. A constant tension heat capacity is easier to measure. 7 Note that this observation appears to be a paradox only if one deﬁnes energy as ‘the ability to do work’.
6 3.7. ONE AND TWODIMENSIONAL SYSTEMS 99 on it, and, hence, to remain isothermal, heat is dissipated to the surroundings ( dQ = − dW ). At the same time, the entropy decreases from the stretching, despite ¯ ¯ the heat ﬂow out, as we see from Eqn. (3.97). The rubber band can do work by pulling a weight, say. However, to remain isothermal, the internal energy remains constant, and heat must be drawn in from the surroundings ( dQ = − dW ), while the ¯ ¯ weight is being pulled. Therefore, the rubber band uses heat from the surroundings to perform work. It simply uses its lowered level of entropy to drive the heat ﬂow. Although we have allowed arbitrary dependence of the tension on length through the function ψ (L), we have still idealized the situation somewhat by assuming that the tension is strictly linear with T . In Problem 3.7.C. we consider how to determine the amounts of energetic and entropic contributions to the elasticity from experimental data. If one accounts for thermal expansion, linearity in T is an excellent approximation. 3.7.2 Unzipping DNA In order to replicate, DNA must ﬁrst separate its two strands from the helix. This process is called denaturation. In most aqueous solutions DNA and proteins denature only at elevated temperature—well above body temperature. However, in the body these processes occur all the time. Therefore, there is an interest in understanding how external forces on DNA can cause the strands to denature at lower temperatures. In a recent experiment [19], workers unzipped the two strands of a DNA double helix by pulling them apart. A schematic of the process is shown in Figure 3.11. One strand of the DNA is attached to a magnetic bead, and the other is attached to a solid surface (via a spacer). A desired tension can be applied to the strand by turning on a magnetic ﬁeld of known strength. At constant temperature, the tension is gradually increased. As the tension increases, more of the chain unzips, until the chain completely denatures. In this way, they were able to construct a ‘temperaturetension phase diagram.’ The data for this phase diagram are shown in Table 3.4. These data show that the strand can be denatured at body temperature if a force is exerted to pull them apart. In the following example, we use a statistical mechanical model derived in a later chapter to predict these data. Example 3.7.1 A simple statistical mechanical model may be constructed for the partially peeled DNA strand shown in Figure 3.11. If the ends of the peeled strand are pulled distance L apart with tension T at temperature T , the Gibbs potential is 100 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS Figure 3.11: Schematic of the apparatus used to unzip the doublestranded DNA segment. Figure taken without permission from [19]. derived to be sinh(x) x
˜ 2(M +1) nb 2 nb ˜ ˜ G(T, T , M ) = Gden (T, M ) − kB T log − sinh(x) x − λ(T ) ˜ λ(T )M +1 , (3.103) ˜ where M is the number of base pairs in the DNA, x := T aK /kB T , aK is the Kuhn ˜ step length of a denatured strand, kB = R/NA is Boltzmann’s constant, nb is the ∆b number of base pairs per Kuhn step, and λ(T ) := exp kBgT . We have also used the diﬀerence in Gibbs potential between a single attached and unattached base pair ∆gb (T ) := gu − ga . Eqn. (3.103) assumes that the free end of the coiled chain has the two strand ends covalently bonded. Therefore, when all of the base pairs are ˜ detached, it is assumed that the DNA is one long single strand of 2M /nb Kuhn steps. The real DNA strand has no such covalent bonding at its ends, of course. 3.7. ONE AND TWODIMENSIONAL SYSTEMS Tension, T [pN] 3.4353 3.6235 4.4706 5.4118 6.6353 8.8941 9.9294 9.9294 9.9294 10.4 10.4 11.059 11.624 12.0 15.482 16.518 16.894 17.741 35.906 Melt Temperature, Tm [◦ C] 49.944 45.033 42.034 40.128 40.405 36.047 30.048 34.413 36.323 27.321 31.413 30.051 24.051 27.053 21.881 19.974 20.249 18.069 15.129 101 Table 3.4: For a given force (in picoNewtons), the temperature (in degrees Celsius) is found at which
lambda phage DNA can be pulled apart in an aqueous solution, by using the apparatus sketched in Figure 3.11. At zero tension, the ‘melting temperature’ (the temperature at which the DNA spontaneously denatures, or separates) is given. These data are taken from [19]. Hence, the model predicts complete peeling when all the base pairs are detached, and one long strand is made. Using the above fundamental relation, ﬁnd the distance L between the two ends of the unpeeled strands as a function of tension and temperature. For a given temperature, at a critical tension, this separation goes to inﬁnity, which is the soughtafter ‘temperaturetension phase diagram’. Compare your result with the experimental results given in Table 3.4. Assume that ∆gb (T ) = ∆hb − T ∆sb , where ∆hb and ∆sb are temperature independent. Assume that aK = 15˚, nb = 5 (These estimates A are from [87]). What values of ∆s and ∆h give the best ﬁt? How does your estimate of ∆s compare with the reported value (estimated using a diﬀerent method, and in a diﬀerent solvent) of ∆s = 85J/K.mol? What do you predict for the melting temperature at zero tension? Solution: Note that the fundamental relation given here is in terms of Kuhn 102 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS
steps of size aK , whereas Exercise 2.9.E used the persistence length lp . The persistence length is equal to half the Kuhn step length, aK = 2lp . The contour length of the two models should remain the same, so Np = 2NK . ˜ First we ﬁnd the separation length L as a function of (T, T , M ). Recall the diﬀerential for the Gibbs potential, dG = −SdT + V dP + µdN , which, for the ˜ 1D case becomes dG = −SdT − LdT + µdM . Hence, we ﬁnd the length from L= = = − − ∂G ∂T ∂x ∂T
˜ T ,M T ∂G ∂x ˜ T ,M coth(x) − 1 2 aK x nb ˜ M +1 1 − , (3.104) ˜ ˆ (T, T )M +1 ˆ (T, T ) 1−λ 1−λ
2/nb where we have introduced ˆ λ(T, T ) := λ(T ) = exp x sinh(x) ∆gb kB T T aK /kB T sinh(T aK /kB T ) 2/nb . (3.105) We used the chain rule to obtain the second line of Eqn. (3.104), and we skipped a step of slightly messy algebra to obtain the last line. Eqn. (3.104) gives the average distance between the two free strand ends at a given tension, and a given temperature. This length has two coupled contributions, however: the tension pulls the detached segment of strands straight, and the temperature and tension detach additional base pairs. To ﬁnd the relative magnitude of these contributions, we can complete detach all the base pairs theoretically by letting the free energy of attachment become large and ˆ negative: ∆gb → −∞. In that case, λ = 0, and the length becomes LIL = coth(x) − ˜ 1 2 M aK . x nb (3.106) This is the length the chain would have for a given tension if all the base pairs were detached, and is called the Langevin force law [see Exercise 3.7.B]. If we take the ratio of the two lengths, we obtain the expression L LIL = ˜ ˜ 1 + 1/M 1/M − , ˜ ˆ ˆ 1 − λ(T, T )M +1 1 − λ(T, T ) (3.107) This ratio now includes only the eﬀect of detaching base pairs. When it goes to 1, all the base pairs must be detached. We can make a plot of this ratio as ˆ ˜ a function of λ, as shown in Figure 3.12 for M = 1000. From this ﬁgure we see 3.7. ONE AND TWODIMENSIONAL SYSTEMS
ˆ that the chain “melts” when λ approaches 1. The value never becomes exactly ˆ , which means very negative values for ∆gb , or large 1, except for very small λ ˆ tensions. However, whenever λ is near one, it becomes very easy for the number of attached base pairs to ﬂuctuate to zero, at which point the two strands will be irreversibly separated. It makes sense, then, to assume that chain becomes ˆ completely unpeeled when λ = 1.
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 103 L LIL 0.6 0.8 1 ˆ λ 1.2 1.4 Figure 3.12: The endtoend distance of a partially unzipped DNA strand as ˜ a function of the tensionmodiﬁed basepair partition function, λ. The distance is made dimensionless by the length the chain would have if it were completely unzipped. See Eqn. (3.107).
Hence, the ‘modiﬁed melting temperature’ Tm = Tm (T ), is determined by ˆ setting λ(Tm , T ) = 1. Or, taking the logarithm of each side, we can write − ∆sb 2 T aK /kB Tm ∆hb + = log , kB Tm kB nb sinh(T aK /kB Tm ) (3.108) where we replace the change in Gibbs potential per base pair upon melting, with the corresponding change in enthalpy and entropy, ∆gb = ∆hb − T ∆sb . If we neglect the temperature dependence of these two quantities, then we have four parameters ∆hb , ∆sb , nb , and aK , three of which are already estimated x from other experiments. To ﬁt the remaining parameter, we plot log sinh(x) vs. 1/Tm . For this plot it is useful to convert Boltzmann’s constant kB = = (1.38066 × 10−23 J/K)(1N·m/J)(1012 pN/N)(1010 ˚/m) A ˚ 0.138066 pN·A/K. (3.109) Figure 3.13 shows the ﬁt of our equation to the data (excluding the last point). The slope and intercept of the line are, then, −nb ∆hb /(2kB ) ∼ −11, 600 ± 900 = 104 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS
K, and nb ∆sb /(2kB ) ∼ 36 ± 3, respectively. Or, ∆hb ∼ 6.4 × 10−20J and ∆sb ∼ = = = 2.0 × 10−22J/K. This latter value compares favorably with other estimates. The melting temperature at zero tension is, according to Eqn. (3.108), Tm (T = 0) = ∆hb /∆sb ∼ 320K, which is too low. = The last data point in the table does not ﬁt well with the others. This point may not ﬁt because (1) ﬂuctuations are not accounted for in thermodynamics, and a large ﬂuctuation at high tensions could lead to irreversible peeling8 , (2) the free energy per base pair may be temperature dependent, (3) the model is too simple, in that it does not account for socalled stacking of the adjacent nucleotides, or (4) there may be experimental error at that high tension. 2 10 T aK /kB T / sinh(T aK /kB T ) 1 0.1 0.01 0.0031 0.0032 1/Tm 0.0033 0.0034 Figure 3.13: DNA unzipping data of Table 3.4 along with ﬁt of statistical mechanical model, Eqn. (3.103). The ﬁt to the data yields the change in enthalpy and entropy per base pair for denaturation. Note that the last data point in the table has been omitted from both the ﬁt and the plot here. 3.7.3 Langmuir Adsorption Here we consider a solid surface that has Ms moles of sites available for binding N ≤ Ms moles of species A. Species A exists in an ideal gas phase above the solid surface, but the molecules may attach to these adsorption sites, detach, or hop between adjacent sites on the surface. See Figure 3.14. Such a system is important in catalysis, and some separation processes called pressureswing adsorption (PSA). Adsorption measurements are also commonly used to calculate available surface area (See Problem 3.7.D).
8 Fluctuations are covered in §6.7. 3.7. ONE AND TWODIMENSIONAL SYSTEMS 105 Figure 3.14: Sketch of a crystalline surface that can adsorb molecules from a gas phase. A discrete number of adsorption sites exist on the solid surface, where the molecules (drawn here as circles) can sit. Pressureswing adsorption is used to separate oxygen from air for asthma patients, methane from waste decomposition to produce fuel, and to remove moisture from gas streams [75]. A schematic of one such PSA process is shown in Figure 3.15. A tank containing porous solid substrate is fed a mixture of gases from the top. One of the species, say Oxygen is preferentially adsorbed to the solid. At the bottom end of the tank, the Nitrogenenriched gas exits. Before the solid adsorption sites are ﬁlled with Oxygen, the ﬂow is stopped, and the pressure in the tank is lowered, which allows the Oxygen to desorb. An inert gas is then fed at the bottom of the tank to ﬂush the Oxygenenriched gas out the top. In order to design the operation and size of such a process, it is necessary to be able to predict the amount of gas that can be adsorbed on the solid substrate. In order to model the adsorption on the solid surface, we replace the volume with the more natural thermodynamic variable Ms . Alternatively, we could use the area, and the density, or number of sites per area, but the development is the same. Therefore, we require a fundamental relation of the type F = F (T, Ms , N ) F (T, MS , N ) = Ms RT log 1 − N Ms NRT log [qsite (T )] . + NRT log N Ms − N − (3.110) In the language of statistical mechanics the function qsite (T ) is called the single site partition function. Its derivation is outside the scope of this textbook, since knowledge of quantum mechanics is required.9 However, it contains information about the possible energies of interaction between one molecule and one adsorption site. The sites are assumed to have a simple potential energy interaction with a molecule. Namely, the molecules are assumed not to interact energetically with one another. An explicit expression for q can be found if a few assumptions are made: there is an energy diﬀerence/mole ǫ0 (< 0) between the minimum of the energy well for
9 The details can be found in [41, Chapter 7]. 106 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS Figure 3.15: Sketch of a simple pressureswing adsorption process to separate two components of a gas. The tank contains solid, porous substrate which preferentially adsorbs one of the gas species on its surface. The process has three steps. The ﬁrst step pressurizes the tank with the mixture to be separated. The gas, now rich in the nonadsorbing species, is purged from the tank. When the pressure in the tank is lowered, the other species now desorbs, creating a gas phase rich in adsorbing species. an adsorption site, and a molecule inﬁnitely far away from the surface; a molecule sitting in this well will have only two vibration frequencies from thermal motion, one for vibrations perpendicular to the surface ω⊥ , and one for vibrations parallel ω ; associated with these vibrations are two characteristic vibration energies/mole, ˜ ˜ ǫ⊥ = 1 hω⊥ NA and ǫ = 1 hω NA . Planck’s constant is h. Finally, there is an energy 2 2 barrier V0 between adjacent sites. In the Langmuir adsorption model, we also neglect the ability of the molecules to hop between adsorption sites, and characterize the system with only the three fundamental energies ǫ⊥ , ǫ and ǫ0 . This assumption will work at lower energies when kB T ≪ V0 , or when T is less than approximately 3.7. ONE AND TWODIMENSIONAL SYSTEMS 250 K for most systems. With these approximations it is possible to derive qsite (T ) = exp 8 sinh2
ǫ RT −ǫ 0 RT 107 sinh ǫ⊥ RT . (3.111) From this relation, all thermodynamic properties of the lattice gas can be found. In particular, we can ﬁnd the percentage of occupied adsorption sites as a function of the pressure in the gas phase above the surface. In Problem 3.7.D., we use the results of the following example to analyze adsorption of N2 and O2 . Example 3.7.2 Find the socalled Langmuir adsorption isotherm—the fraction of ﬁlled sites as a function of pressure at constant temperature, for an adsorption surface in equilibrium with an ideal gas. Solution: Since the gas molecules are free to be either adsorbed or in the
gas phase, their chemical potential and temperature must be the same in the two phases. Note that the surface has no volume, so there is no pressure for the molecules adsorbed.10 From the diﬀerential for the Helmholtz potential, we can ﬁnd the chemical potential of the adsorbed phase from the fundamental Langmuir relation, Eqn. (3.110) µ= ∂F ∂N T ,Ms = RT log θ 1−θ 1 , qsite (T ) (3.112) where θ := N/Ms is the fraction of adsorption sites occupied. To ﬁnd the chemical potential in the ideal gas phase, we multiply both sides of Eqn. (B.2) by N , and take the derivative of each side with respect to N µIG = = where µ◦ (T ) :=
T ∂F ∂N
◦ T ,V µ (T ) + RT log P, (3.113) f0 − s0 (T − T0 ) + RT − RT log RT N v0 T0 T′ − T T′ . cideal (T ′ )dT ′ + v (3.114) The surface does have an area, however. Hence, one can ﬁnd the derivative of the internal energy with respect to area at constant entropy and mole number—analogous to pressure. The resulting quantity is called the surface tension, and is covered in greater detail in Chapter 12. 10 108 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS
Setting the two chemical potentials equal, and solving for θ yields θ= where χ(T ) := exp χ(T )P , 1 + χ(T )P µ◦ (T ) qsite (T ) RT (3.115) (3.116) Figure 3.16 shows some representative adsorption isotherms predicted by the Langmuir model. 2 1 0.8 0.6 s θ
0.4 0.2 0 increasing χ Pressure, P Figure 3.16: Fraction of ﬁlled sites as a function of pressure for several values of χ, as predicted by the Langmuir model, Eqn. (3.115). Table 3.7 shows adsorption data for nitrogen and oxygen on zeolite. Problem 3.7.D. asks the student to ﬁt these data to the Langmuir adsorption isotherm equation. The simple statistical mechanical model used to derive the Langmuir adsorption model (given in §6.4) does not require that the adsorption sites be on a solid surface. Primarily, it assumes that adsorption sites do not interact energetically with one another. Hence, the result is much more general than just molecules adsorbing on solid surfaces. In fact, the equation is often used to model other adsorption phenomenon, such as ligands adsorbing to proteins, or polymer chains to colloidal particles. These examples are considered in the exercises. In the following example, we use the Langmuir model to describe the binding of proteins to sites on DNA. However, an additional subtlety must be taken into account. For classical Langmuir adsorption, the chemical potential in the gas phase is not aﬀected by the degree of adsorption θ . However, in solution, every molecule adsorbed decreases the concentration of unadsorbed molecules. 3.7. ONE AND TWODIMENSIONAL SYSTEMS 109 Example 3.7.3 Proteins bind to speciﬁc sites of DNA. It is believed that the presence of other species can aﬀect this adsorption, and we wish to examine that eﬀect here. Recently Heyduk and Lee [39] synthesized a 32basepairlong fragment of DNA, which they tagged with a ﬂuorescent dye. The dye makes it possible to measure optically the fraction of DNA fragments that are bound by a speciﬁc protein. We won’t be concerned here with how this measurement works, but rather just assume that it does. Table 3.5 shows optical data for solutions of the E. Coli cyclic AMP receptor protein (total concentration cT ) and a 32basepair fragment DNA ligand of top tal concentration cT = 0.0111µM. Also present is cyclic adenosine monophosphate D (cAMP), in two diﬀerent concentrations, which is believed to play a role in binding, by forming complexes with the protein. We wish to study this role by using the Langmuir model, and seeing how the binding parameter χ is aﬀected by the concentration of cAMP. The anisotropy A is ﬂuorescence data giving a measure of the amount of binding, and is assumed to be linearly proportional to the concentration of bound complex cB , and unbound DNA fragment cD A = AD cD + AB cB , (3.117) where AD and AB are the (constant) anisotropies of the free and bound DNA, respectively. Use the data in the table to estimate the binding parameter χ for each concentration of ligand. Does its value depend upon cAMP concentration? Can anything be inferred about the presence of cAMP on DNA/protein binding? Solution: Assuming constant volume, we note that the total concentration
of protein cT is constant, and the sum of free protein cp and bound protein p cB concentrations: cT = cp + cB . Similarly, one can balance the total concenp tration of DNA, cT = cD + cB . The generalization of Langmuir adsorption to D concentration is cB χcD = , (3.118) cT 1 + χcD p assuming that the DNA is in excess, and equating the chemical potentials of bound and unbound DNA. From these three equations, it is possible to relate the concentration of bound complex cB to the total concentration of DNA ligand and protein. Inserting the balances into Eqn. (3.118), and solving for cB yields cB = 1 1 + χ cT + cT + p D 2χ 1 + χ cT + cT p D
2 − 4 χ2 c T c T , Dp (3.119) 110 cT p [nM] 0 3.92 7.97 11.76 15.68 19.6 25.35 33.06 40.77 59.84 78.79 97.47 116.2 134.6 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS Anisotropy, A 0.170 0.174 0.177 0.179 0.180 0.182 0.183 0.185 0.185 0.189 0.190 0.191 0.192 0.192 cT p [nM] 0.131 2.091 3.92 5.88 7.84 9.80 11.76 13.59 15.81 17.38 21.43 25.22 30.97 36.72 Anisotropy, A 0.169 0.175 0.179 0.182 0.185 0.188 0.189 0.191 0.191 0.192 0.193 0.194 0.194 0.196 Table 3.5: Anisotropy data A for DNA ligand binding to an E. Coli cAMP receptor protein, as measured by ﬂuorescence. In this table, cT is the total concentration p of receptor protein (bound and unbound). The ﬁrst two columns are for a concentration of cAMP equal to 0.5µM, and the right two columns are 500µM cAMP concentration. Data are taken from [39].
Inserting this equation into Eqn. (3.117) and replacing cD with cT gives the D anisotropy A as a function of χ, AD , AB , and the total concentrations cT and p cT . D A = AD cT + D (AB − AD ) 1 + χ cT + cT + p D 2χ 1 + χ cT + cT p D
2 − 4 χ2 c T c T . Dp (3.120) The concentrations are known (from the table), and we seek an estimate for χ. We can ﬁnd AD from the ﬁrst entry of the table where cT = 0, giving us p AD = 0.16954/(0.0111µM ) = 15.3µM −1 . Hence, we have two parameters to ﬁt to the data: AB and χ. Using a LevenbergMarquardt ﬁt11 we are able to ﬁt Eqn. (3.120) to the data, and ﬁnd estimates for these remaining two parameters, AB and χ. We ﬁt Eqn. (3.120) to the ﬁrst two columns where cAMP concentration is 0.5µM. The result is shown in Figure 3.17. Our ﬁt gave χ = 0.0216 ±
Such a ﬁt is straightforward with a software package called gnuplot, using the command “ﬁt”. Gnuplot is freeware under the GNU license, and is available for most computer operating systems. Spreadsheets, Mathematica, Matlab, or similar software can also accomplish this ﬁt.
11 3.8. SUMMARY 111 0.205 0.2 0.195 0.19 0.185 0.18 0.175 0.17 0.165 Anisotropy, A 0 cAMP 0.5µM cAMP 500µM 20 40 60 80 100 120 protein concentration, cT [µM] p 140 Figure 3.17: Fit of Eqn. (3.120) to the data given in Table 3.5. The lower curve is cAMP concentration of 0.5µM, where our nonlinear ﬁt gave values of χ = 0.0216 ± 0.0014µM −1 , and AB = 20.04 ± 0.1µM −1 . The upper curve is cAMP concentration of 500µM, where the parameters were found to be χ = 0.055 ± 0.005µM −1 , and AB = 21.3 ± 0.2µM −1 .
0.0014µM −1, and AB = 20.04 ± 0.1µM −1 . For the higher concentration of cAMP, we obtained χ = 0.055 ± 0.005µM −1, and AB = 21.3 ± 0.2µM −1 . The value of AB is not strongly dependent on cAMP concentration, as we expect. In fact, if we assume that AB = 20.6, the average of these two values, then we also obtain reasonable ﬁts for χ = 0.0162 ± 0.0007, and 0.076 ± 0.004, respectively. On the other hand, χ does depend strongly on the concentration of cAMP, indicating that cAMP does play a role in the binding of the protein to the DNA fragment. Apparently, cAMP enhances the ability of the protein to bind. Exercise 3.7.I uses a reaction model for the same binding process. 2 3.8 Summary In the previous chapter, we covered internal energy U , entropy S , temperature T , pressure P , chemical potential µi , heat ﬂow Q, work W , and the ﬁve postulates of thermodynamics. We proved that these postulates and deﬁnitions ﬁt our intuition 112 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS about temperature and heat ﬂow, pressure and expansion, and about mass ﬂow. In this chapter, we used Legendre Transforms to deﬁne new thermodynamic quantities with complete thermodynamic information: • Enthalpy H := U + P V , • Helmholtz potential F := U − T S , • Gibbs free energy G := U − T S + P V , (3.9) (3.10) (3.11) We proved that an isolated system at equilibrium has minimum internal energy at constant entropy, volume and mole number. The other subsystems are summarized in Table 3.2. Each of the generalized potentials has a canonical diﬀerential, which can be used to derive equations of state, or Maxwell relations. These diﬀerentials are dU = T dS − P dV + T dS + V dP +
i µi dNi ,
i (3.48) dH = µi dNi , µi dNi ,
i (3.49) (3.50) (3.51) dF = −SdT − P dV + dG = −SdT + V dP + µi dNi .
i The Maxwell relations are derived from these diﬀerentials by using the analytic property of the potentials, Eqn. (A.5). The diﬀerentials, and the Maxwell relations are conveniently memorized through the thermodynamic square, Figure 3.5. These are not the only such generalized potentials, but are the most common. For example, in Problem 3.3.D. we introduce another such potential common in statistical mechanics. Along the way, we derived the Euler Relation
r U = TS − PV + and the GibbsDuhem Relation
m µi Ni ,
i=1 Euler Relation, (3.20) 0 = SdT − V dP + Ni dµi ,
i=1 GibbsDuhem Relation. (3.22) 3.8. SUMMARY 113 The Euler relation shows that the speciﬁc Gibbs potential is identical to the chemical potential for pure substances. The GibbsDuhem relation is useful for mixtures, considered later in this book. We also proved that the generalized potentials are useful for calculating work in isothermal and/or isobaric conditions. We deﬁned three (of many possible) secondorder derivatives • The coeﬃcient of thermal expansion α :=
1 • The isothermal compressibility κT := − v 1 v ∂v ∂T P . ∂v ∂P T . ∂s ∂T P . • The constantpressure heat capacity cP := T All other secondorder derivatives can be found in relation to these, for a singlecomponent system. In fact, any thermodynamic quantity for a singlecomponent system can be found from these quantities (§3.6). Complete thermodynamic information for a pure substance is also contained in two equations of state (e.g., P = P (T, v ), u = u(T, v )), or in three relations for secondorder derivatives (e.g., α = α(T, P ), κT = κT (T, P ), cP = cP (T, P )). As example models, we considered the general ideal gas, Eqn. (3.16), and a nonlinear elastic strand, Eqn. (3.101). We also analyzed the following problems • Adiabatic and isothermal compression of ammonia, Example 3.2.2. • A heat engine, Example 3.2.3. • A fuel cell, Example 3.5.4. • Deriving a fundamental relation for a nonideal, elastic strand, §3.7.1. • Unzipping a single strand of DNA by using tension and heat, Example 3.7.1. • Langmuir adsorption on a solid surface, Example 3.7.2. • Binding of a protein to a DNA fragment, Example 3.7.3. In §3.7.1, we showed how to generate a fundamental relation from two equations of state for a general elastic string. We also predicted the fractional adsorption of molecules on a solid surface as a function of pressure, the socalled “Langmuir isotherm.” 114 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS 3.9 Exercises 3.1.A. Verify that the two fundamental energy relations: U1 (S ) = AS c and U2 (S ) = A(S + C0 )c (A and C0 may be functions of V but not S ; c is a constant) give the same equation of state for U (T ), but that they do give diﬀerent fundamental Helmholtz relations. 3.1.B. Show that the fundamental enthalpy relation H = AN P γ exp Sγ , NR (3.121) is a simple ideal gas. How are the constants here related to those in Eqn. (2.39)? 3.1.C. The fundamental Helmholtz potential for a ﬂuid is given as F (T, V, N ) =N f0 − N s0 (T − T0 ) − NRT log V /N − b v0 aN 2 − V
T T0 (B.12) T′ − T T′ cideal (T ′ )dT ′ . v +N Find the thermal and mechanical equations of state for this ﬂuid, and explain why it might be called a general, van der Waals ﬂuid. 3.1.D. Show that the fundamental Helmholtz relation f =f0 − s0 (T − T0 ) − RT log
T T0 v−b v0 a + √ log bT v v+b + (B.28) T′ T T′ cideal (T ′ )dT ′ . v is a general RedlichKwong ﬂuid. Find the mechanical and thermal equations of state for this ﬂuid. 3.1.E. The fundamental Helmholtz potential for a ﬂuid is given as f =f0 − s0 (T − T0 ) − a v − RT log − v v0 bRT (v0 − v ) 3(v + v0 )b − 4vv0 − 2b2 + 2 (v − b)2 v0 (B.22)
T T0 T′ − T T′ cideal (T ′ )dT ′ . v Find the thermal and mechanical equations of state for this ﬂuid. The P V T relation is called the CarnahanStarling generalization to the van der Waals equation of state. 3.9. EXERCISES 3.1.F. The fundamental Helmholtz potential for a ﬂuid is given as f =f0 − s0 (T − T0 ) − RT log v v0 + B0 RT − A0 − C0 1 + T2 v 115 (B.58) aα 1 bRT − a 1 + − 2 2 v 5 v5 C 1 1 γ + 2 exp − 2 − T2 γ 2v v
T T0 1 γ exp − γ 2 v0 + T′ − T T′ cideal (T ′ )dT ′ . v Find the thermal and mechanical equations of state for this ﬂuid. The P V T relation is called the BenedictWebbRubin equation of state . 3.1.G. The fundamental Helmholtz potential for a ﬂuid is given as f =f0 − s0 (T − T0 ) − RT log a(T ) log γ−β v+β−b v+γ−b v−b v0
T − T′ − T T′ cideal (T ′ )dT ′ . v (B.53) +
T0 Find the thermal and mechanical equations of state for this ﬂuid. The P V T relation is called Martin’s Generalized Cubic equation of state . 3.1.H. The fundamental Helmholtz potential for a ﬂuid is given as f =f0 − s0 (T − T0 ) − RT log a(T ) √ log 2 2b v−b v0 +
T0 +
T (B.43) T′ − T T′ cideal (T ′ )dT ′ . v √ v + b(1 + 2) √ v + b(1 − 2) Find the thermal and mechanical equations of state for this ﬂuid. The P V T relation is called the PengRobinson equation of state . Since v0 is the speciﬁc volume where the ﬂuid is ideal, it is much larger than b, and the second ratio inside the logarithm may be safely approximated as 1. 3.1.I. Show that the fundamental Gibbs relation G = Ng0 + [(c + 1)NR − Ns0 ] (T − T0 ) + NRT log T0 T
c+1 P P0 is an ideal gas, where c, R, s0 , T0 , and P0 are constants. Describe what each of these constants are. 116 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS Figure 3.18: A three step process to use a real gas as a refrigerator. the gas is at temperature T1 = 50◦ C, and pressure P1 = 10atm. In gas is compressed adiabatically to temperature T2 = 100◦ C. In step b, isobarically to T3 = 50◦ C; and in step c, it is expanded isothermally original state. In state 1, step a, the it is cooled back to its ˜ 3.1.J Find the generalized potential U (S, P, µ) that has S , P , and µ as canonical independent variables. Write its diﬀerential. 3.1.K. Similar to Example 2.6.6, we wish to use a gas as a refrigerator, except this time the gas is real instead of ideal. We use the van der Waals equation of state for oxygen as our gas. The gas is in a container that has a piston so that it can exchange work with its environment. All the walls are adiabatic except one, which is isothermal. However, we also have some insulation so that we can make that wall adiabatic at will. The gas begins with temperature T1 = 50◦ C, and pressure P1 = 10atm. In the ﬁrst step, step a, we compress the gas adiabatically (i.e., with the insulation covering the diathermal wall) to temperature T2 = 100◦ C. Next, during step b, we isobarically cool the gas to T3 = 50◦ C. Step c then isothermally expands the gas back to the ﬁrst state. See Figure 3.18. We wish to ﬁnd the amount of cooling performed per step, per mole of gas. To aid in the calculation, you might perform the following steps. • Using the expressions in Appendix B for the van der Waals ﬂuid, ﬁnd the constants for the P vT EOS, a and b. From the fundamental relation, ﬁnd the equations of state, s = s(T, v ) and u = u(T, v ) for the general van der Waals ﬂuid. • Make two tables to be ﬁlled in. The ﬁrst table should have boxes for the 3.9. EXERCISES 117 pressure, temperature and speciﬁc volume at each of the points 1, 2 and 3. The second table should have boxes for ∆u, Q and W for each of the steps a, b and c. Fill in the boxes from the problem statement. • Now begin ﬁlling in each of the boxes from the necessary calculations. For example, you could use the P vT EOS to ﬁnd the speciﬁc volume v1 . Then, step a is assumed adiabatic and reversible, so it is isentropic. Use your entropy equation of state above to ﬁnd the speciﬁc volume v2 . From T2 and v2 , you should be able to ﬁnd the pressure P2 . From your u EOS, you can also ﬁnd the change in internal energy in step a, ∆ua . From an energy balance, you can ﬁnd Qa . • Continue ﬁlling out the tables in this way. When ﬁnished, calculate the total work necessary, and amount of cooling that can be accomplished. Which steps require work, and which steps provide cooling? As a bonus, you could also run the machine backwards and use heat to generate work. In that case, how much work is generated per amount of heat added to the system? 3.1.L. Use the thermodynamics diagram for ammonia on p.205 to estimate the change in chemical potential for ammonia at constant T = 140◦ C, but pressure changing from 20 to 1MPa. 3.2.A. Prove that a system in contact with both a pressure and a temperature reservoir reaches equilibrium when dGsys = 0. 3.2.B. Use the result of Example 3.2.1 to ﬁnd the quasistatic work required to compress one mole of oxygen isothermally at 170 K from 2 to 0.8 liter using a PengRobinson equation of state. Can you think of another way to ﬁnd this result? 3.2.C. Using the results of Example 3.2.1, ﬁnd the work necessary to compress reversibly and isothermally one mole of ethane at 330K from 8 to 1 liters. Use the BenedictWebRubin equation of state. 3.2.D. Using the ﬁgure on p.205, ﬁnd the work necessary to compress ammonia isothermally from a density of 2kg/m3 to 250 kg/m3 at 340◦ C. 3.2.E. Using the results of Example 3.2.1, ﬁnd the work necessary to compress reversibly and isothermally one mole of DuPont refrigerant HCFC123 at 330K from 200 to 105 liters. Apparently, this ﬂuid is well described by a modiﬁed BenedictWebbRubin equation of state. Details of the appropriate equation of state are found at: http://www.dupont.com/suva/na/usa/literature/pdf/h47753.pdf. 118 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS This document also contains tables of thermodynamic properties, which may be used to avoid explicit use of the equation of state altogether. 3.2.F. Using the ﬁgure on p.205, ﬁnd the work necessary to compress 5 l of ammonia adiabatically from a density of 2kg/m3 and pressure of 250 kPa to a density of 60 kg/m3 . 3.3.A. Derive the diﬀerentials for the generalized potentials H and G. 3.3.B. Derive the remaining Maxwell relations, Eqs. (3.56), (3.58), and (3.59). 3.3.C. In Example 3.1.2, we found S (T, V ) in Eqn. (3.17) for a general ideal gas. Verify that the third Maxwell relation is satisﬁed for the general ideal gas. Verify the other three relations. 3.3.D. In this chapter we performed a Legendre transform on the internal energy in the variables S and V . One may also perform transforms in the variable N , or on the entropy. By performing a Legendre transform on U in the mole number, ﬁnd the deﬁnition of the generalized potential Ψ with canonical independent variables (T, V, µ). Find the diﬀerential for this variable, and show that ∂Ψ ∂µ = −N. (3.122) T ,V This generalized potential frequently arises in statistical mechanics, but is often the negative of the one deﬁned here. 3.3.E. Prove the Maxwelllike relation ∂ µi ∂P =
T ,{Nj } ∂V ∂Ni (3.123)
T ,P,{Nj =i } 3.3.F. A methanol fuel cell relies on the reaction 3 CH3 OH + O2 → CO2 + 2H2 O. 2 Find the maximum available work as a function of temperature for this fuel cell, assuming that it works reversibly. Also ﬁnd the maximum work available from a heat engine. Do your results agree with Table 3.3? 3.5.A. Find the coeﬃcient of thermal expansion and the heat capacity for the simple van der Waals ﬂuid. 3.5.B. Find the isothermal compressibility and heat capacity predicted by the RedlichKwong equation of state in §B.5. 3.9. EXERCISES 119 3.5.C. Find the isothermal compressibility and heat capacity predicted by the PengRobinson equation of state in §B.6. 3.5.D. It is reported that the isothermal compressibility at the boiling point (which is reported as 252.77◦ C) for hydrogen is 50.3MPa−1 . How does this value compare with the prediction by the van der Waals equation of state? Note that the constants for the van der Waals ﬂuid may be found from critical data by Eqn. (4.22). 3.5.E. Find α as predicted by the PengRobinson Equation of State. 3.5.F. Find α as predicted by the RedlichKwong model. 3.5.G. Use the SoaveRedlichKwong model (§B.5) to estimate the constantpressure heat capacity of water at 25◦ C and 2bar. 3.6.A. What is the constantvolume heat capacity for a simple ideal gas? 3.6.B. What is
∂U ∂T P for a general ideal gas? 3.6.C. Reduce the following derivatives in terms of standard quantities. ∂S (1) ∂V P ; (2) ∂ V S ; (3) ∂ H T ; (4) ∂ V H ; (5) ∂ V H ∂T ∂V ∂P ∂S 3.6.D. Find an expression that relates the isentropic compressibility κS to the three secondorder derivatives α, κT , and cP : κS := − What is κS for a simple ideal gas? 3.6.E. Find the change in chemical potential for steam at 400◦ C which is compressed from 0.5 to 2MPa. Explain why one cannot use the steam tables to ﬁnd the change in chemical potential for steam undergoing temperature changes. Hint: look at the entry for the entropy of saturated liquid at 0◦ C. Does it agree with the Nernst postulate? How could this problem be ﬁxed, so that one could use the steam tables? 3.6.F. Express the JouleThomson coeﬃcient ∂T ∂P ,
H,N 1 v ∂v ∂P .
S,N (3.124) JouleThomson coeﬃcient (3.125) in terms of measurable quantities. 3.6.G. The JouleThompson coeﬃcient deﬁned in the previous problem is useful in ﬂow through valves. As is shown in Chapter 5.2, if one assumes that ﬂow across a valve is adiabatic, it is also isenthalpic. Hence, the change in temperature as the pressure drops across the valve can be found from the JouleThompson coeﬃcient. Find the change in temperature when the pressure of ammonia drops from 200MPa and 180◦ C, to a pressure of 50MPa in two ways: (1) using the thermodynamic diagram on p.205, and (2) using the PengRobinson model. 120 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS
∂T ∂P S,N 3.6.H. 1. Find ideal gas. as a function of temperature and volume for a general ∂T 2. Make a plot of ∂P S,N for oxygen at room temperature as a function of pressure between 1 and 3 atm. 3. Integrate your expression from (1) to ﬁnd an expression that relates temperature to pressure. Note that the expression is not explicit in temperature (i.e., it is not possible to write an equation of the form T = T (P )). However, it can be made explicit in pressure: P = P (T ). Can you ﬁgure out a way to plot the temperature as a function of pressure during the reversible, adiabatic compression of part (2)? 3.6.I. Using the result of Problem 3.5.G. and thermodynamic manipulations, ﬁnd the constantpressure heat capacity of liquid water at the same temperature and pressure. 3.7.A. Find the mechanical and thermal equations of state for the rubber band with the fundamental Helmholtz relation F 1 = F0 − S0 (T − T0 ) − Nc RT log 1 − 2
T T0 T′ − T T′ L − L0 L1 − L0 2 + (3.126) CL (T ′ )dT ′ . The quantities Nc , L0 , and L1 are constants. CL is some known function of T . 3.7.B. Another statistical mechanical model for a single strand of DNA (see Problem 2.9.E.) is called the inverse Langevin force law with Gibbs potential
T G(T, T ) = G0 − (T − T0 )S0 − T0 kB T NK kB T log sinh T aK T − T′ 0 CT (T ′ )dT ′ − T′ T aK , kB T (3.127) where, as before, NK is the number of segments, aK is the length of a segment, kB is Boltzmann’s constant, and we have introduced the zerotension heat capacity for 0 the chain CT (T ′ ). How well does this model describe the data given in Table 2.1 on page 59? Be sure to read the comments about persistence length versus Kuhn step length in the solution to Example 3.7.1. 3.9. EXERCISES 121 3.7.C. Here we derive a method to measure the contributions of entropy and internal energy to the elasticity T . For isothermal stretching, we may write T = = ∂F ∂L ∂U ∂L
T T −T ∂S ∂L (3.128)
T where the second line is obtained from the deﬁnition for F . We then seek a way to estimate the contribution from each of the two terms on the right side of this expression. First, derive a generalized Maxwell relation to relate (∂S/∂L)T to measurable quantities (T, L, T ). Then, use the generalized diﬀerential for U to obtain (∂U/∂L)T in terms of measurables. By ‘generalized’, we mean here where we have replaced the usual variables (S, V, N ) with the new set appropriate for a rubber band (S, L). Finally, explain how measurements of tension at diﬀerent values of T but ﬁxed L can be used to estimate the contributions of entropy and internal energy to the elasticity at a given L. Use the data in Table 3.6, which contains tension versus extension length data ∂S for a crosslinked rubber, to make a plot of ∂ U T and −T ∂L T (divided by area) ∂L versus (L − L0 )/L0 for rubber. To a ﬁrst approximation, you may assume that the volume is constant during stretch, so that the area A = A0 L0 /L2 , where A0 is the crosssectional area in the unstretched state. T = 10◦ C Ext. Stress [%] [kg/cm2 ] 3.20 0.484 6.17 0.800 13.5 1.35 22.2 2.01 38.2 3.00 72.9 4.52 T = 30◦ C Ext. Stress [%] [kg/cm2 ] 2.52 0.419 5.49 0.781 12.3 1.38 21.3 2.07 37.5 3.11 71.8 4.77 T = 50◦ C Ext. Stress [%] [kg/cm2 ] 2.29 0.428 5.49 0.809 11.7 1.40 21.0 2.14 37.5 3.25 71.3 5.01 T = 70◦ C Ext. Stress [%] [kg/cm2 ] 1.83 0.372 5.03 0.772 11.2 1.41 19.9 2.22 37.5 3.43 70.2 5.27 Table 3.6: Tension divided by crosssectional area versus % Extension (100 × (L − L0 )/L0 ) for several temperatures as measured by Anthony, Caston and Guth. These data are for reversible, isothermal extensions. 3.7.D. Table 3.7 shows the amount of Oxygen and Nitrogen that adsorb to solid zeolite as a function of pressure at 293K. In order to design a pressure swing adsorber, 122 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS it is necessary to ﬁnd the Langmuir adsorption function χ(T ) at this temperature for each species. Using these data, ﬁnd the maximum amount of each species that can be adsorbed per gram of zeolite ρmax , and χ(T = 293K ). Find an estimate for the size of an oxygen or nitrogen molecule, and use this size to approximate the amount of surface area per mass of zeolite. Nitrogen ρ [mmol adsorbed/g solid] 0.0710 0.146 0.310 0.396 0.487 0.567 0.657 0.752 0.838 0.909 0.976 1.05 1.121 1.18 Oxygen ρ [mmol adsorbed/g solid] 0.0410 0.0626 0.0993 0.127 0.162 0.201 0.240 0.279 0.317 0.356 0.410 0.456 0.518 0.575 0.646 0.711 0.771 P [bar] 0.130 0.391 0.957 1.348 1.696 2.174 2.674 3.304 3.935 4.5 5.174 5.891 6.717 7.565  P [bar] 0.217 0.5 0.783 1.044 1.413 1.761 2.174 2.609 3.022 3.457 4.022 4.609 5.261 6 7 7.87 8.67 Table 3.7: Isothermal adsorption amount versus pressure for oxygen and nitrogen on a surface of 5˚ Zeolite at 293K. The amount adsorbed ρ is given in mmol of A adsorbent per gram of bulk zeolite. Data are taken from [75]. 3.7.E. The B.E.T.(BrunauerEmmettTeller) model for multilayer chemical adsorption. If we generalize the statistical mechanical Langmuir model for adsorption to allow for more than one molecule to adsorb on a single site, we ﬁnd the following expression for the generalized potential Ψ = −Ms RT log c + (c − 1)qsite (T ) exp µ c − qsite (T ) exp RT
µ RT , (3.129) where c is a dimensionless constant of order 100, and qsite (T ) is the singlesite partition function as described in §3.7. The other constants have the same physical 3.9. EXERCISES 123 meaning as in the Langmuir model. The generalized potential Ψ is introduced in Problem 3.3.D, and one might ﬁnd the result of that problem useful here. The constant c is related to the diﬀerence in energy between adding the ﬁrst molecule to the site, and the second, or higher molecule. Hence, in the limit that c → ∞, your answer should reduce to the Langmuir adsorption isotherm. You can use this limit to check your result. Make a plot of θ := N/Ms vs. χ(T )P for c = 160 for this model. Can you interpret the shape of the plot? 3.7.F. Colloidal systems consist of particles in a liquid that do not dissolve, but stay suspended [40]. Milk is such a system with proteins and fats suspended in water. Many colloidal systems appear ‘milky’. Paints are colloidal suspensions of clay particles, and many foods are also colloids. It is typically desirable to keep the particles suspended as long as possible. However, some particles prefer to aggregate into bigger particles, which then fall out of solution. To avoid this problem, some colloids are polymerically stabilized , by dissolving polymer chains in the liquid, which adsorb to the colloidal particles (see Figure 3.19). These chains typically adsorb only partially, and leave tails and loops sticking out into the solvent. These polymer brushes ‘push’ the particles apart and prevent agglomeration.12 Figure 3.19: Sketch of a spherical particle in a solvent with polymer chains adsorbed. The loops
and tails of the polymer can prevent the particle from agglomerating with other particles, making the suspension more stable. A simple way to understand the phenomenon of adsorption, is to model the process by a Langmuir adsorption isotherm. However, the polymer is no longer in the ideal gas phase, but rather in solution. Here we assume that the concentration of the polymer in the bulk phase Ceq plays the role of pressure for the ideal gas, so that µ ∼ µ◦ (T ) + RT log Ceq . = An additional complication that arises for polymer adsorption is that the area occupied by a polymer chain when adsorbed can change with temperature, depend12 The polymer chains have more entropy when their loops and tails can move around. Pushing two particles together decreases the amount of room the loops have. 124 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS ing on how much of the chain adsorbs, and how much of the chain forms loops. The temperature dependence of adsorption can exhibit rich behavior; however, it can be understood by a simple free energy argument: to adsorb, the chain minimizes free energy. Adsorption gives the chain more order, hence less entropy, so ∆Sads of adsorption is negative. Adsorption requires, then, that ∆Hads is also negative, so that ∆Gads = ∆Hads − T ∆Sads of adsorption is negative and adsorption is favorable. As temperature is changed, the quality of solvent can change, and the entropy of adsorption can be sensitive to temperature. Hence, the temperature dependence can be rather complicated. Table 3.8 gives the mass of polymer adsorbed per area for ethyl(hydroxyethyl)cellulose on dehydroxylated silica particles, reported in [47]. Fit these data to the Langmuir adsorption equation to ﬁnd the maximum adsorption and adsorption parameter χ as functions of temperature. Can you explain the trends that you obtain? Can you say anything about the change in entropy of adsorption if you assume that the energy of adsorption is nearly constant? T = 18◦ C Ceq Ads. Amt. (ppm) (mg/m2 ) 1. 0.616 2. 0.811 7.41 1.14 13.7 1.1 17.3 1.05 53.7 1.17 58.6 1.12 97 1.22 104 1.17 109 1.12 154 1.17 T = 25◦ C Ceq Ads. Amt. (ppm) (mg/m2 ) 1. 0.790 6.05 1.14 10.7 1.11 31.7 1.34 40.6 1.28 75.4 1.39 88.9 1.29 124 1.41 T = 37◦ C Ceq Ads. Amt. (ppm) (mg/m2 ) 2.03 1.19 10.1 1.51 34.1 1.71 40.3 1.66 56.5 1.94 62.3 1.88 101 1.99 147 2.01 158 1.95  Table 3.8: Adsorption data for ethyl(hydroxyethyl)cellulose on dehydroxylated silica particles in ultrahigh purity water, taken from [47]. 3.7.G. Pressureswing adsorption uses the fact that some molecules preferentially adsorb to the surface. If one begins with a gas phase that has equal moles of two species, and the pressure is raised, the preferentially adsorbed species will ﬁrst ﬁll the adsorption sites, and be more strongly depleted from the gas phase. If this procedure is repeated, on the depleted gas phase, the two species can be separated. If the gas above the surface is ideal, the chemical potential of each species in the 3.9. EXERCISES 125 mixture is just the chemical potential of pure species at the same temperature and partial pressure µi (T, P, xi ) = µ◦ (T ) + RT log(xi P ), i ideal gas. (3.130) However, since a site can hold only one molecule each, the free energy for two types of adsorbed species is slightly diﬀerent [33, p.426] F (T, MS , N1 , N2 ) RT = MS log 1 − N2 log N1 + N2 N1 + N1 log + MS MS − N1 − N2 N2 − N1 log [q1 (T )] − N2 log [q2 (T )] , MS − N1 − N2 (3.131) where qi (T ), i = 1, 2 describes the energy of interaction between the adsorption site and molecule i. Using this fundamental relation, and assuming that the gas phase is ideal, ﬁnd the fractions of species 1, θ1 := N1 /MS , and species 2, θ2 := N2 /MS , adsorbed as functions of temperature, pressure and mole fraction in the gas phase. 3.7.H. Construct a fundamental relation for a rubber band, assuming (1) that the volume is constant as a function of stretch, (2) that the tension is linear in temperature, (3) that the tension is given by the data in Table 3.9, and (4) that the constantlength heat capacity is welldescribed by a cubic function of temperature: CL = A0 + A1 T + A2 T 2 + A3 T 3 . 3.7.I. The binding of protein to DNA could also be modeled using a pseudoreaction expression cB K= . (3.132) cp cD Using a similar species balance as was done in Example 3.7.3, ﬁnd values for K at the two cAMP concentrations. How good are the ﬁts to the data? Does K show similar trends to χ with cAMP concentration? 126 CHAPTER 3. GENERALIZED THERMODYNAMIC POTENTIALS Extension, (L − L0 )/L0 0.0544 0.1096 0.169 0.2687 0.432 0.7717 1.282 1.928 2.560 3.210 3.704 Stress, [kgf /cm2 ] 0.4329 0.928 1.383 2.054 3.040 4.691 6.748 9.0542 11.56 14.78 18.76 Table 3.9: Elongational stress data for uniaxial stretch of crosslinked rubber. A rubber band with rest
length L0 is stretched, and the stress is measured as a function of the length L. Note that the stress is the tension divided by the cross sectional area of the rubber. Data are taken from Treloar, [93]. ...
View Full
Document
 Spring '08
 LIZ

Click to edit the document details