The constitutive coordinates of a mole of a one-component fluid can be taken to be the set (T,V).

Transcription

1 MSE 20A hermodynamics and Phase ransformations Fall, 2008 Problem Set No. 6 Problem : he constitutive coordinates of a mole of a one-component fluid can be taken to be the set (,). (a) Show that the isometric specific heat, C, the isothermal compressibility,, and the coefficient of thermal expansion, å, are a complete set of first-order thermodynamic properties for the one-component fluid. A complete set of thermodynamic properties is provided by the complete set of second derivatives of any form of the fundamental equation. For one mole of a onecomponent fluid, the Helmholtz free energy, F(,), is a suitable form of the fundamental equation. he properties are: F = - C F = F = F = - P = - å..2.3 (b) Express the first-order thermodynamic properties that can be derived from the energy function in terms of C, å and. he three independent properties in this case are E SS = S = S - = C.4 E = - P = S S = + å2 C v.5 where we have used the expressions and S = C v C P.6

2 C P = C + å2.7 derived in the notes. E S = E S = = (,S) S (,S) = (,S) (,) (,S) (,) - = - S S - = C P = å C.8 (c) Express the first-order thermodynamic properties that can be derived from the Gibbs free energy in terms of C, å and. G = - S = - C P P = - C - å2.9 where we have used the relation C P = C + å2.7 which is derived in the notes. G PP = P = -.0 G P = G P = - S P = P = å. Problem 2: (a) Find the specific heat, C, the isothermal compressibility,, and the coefficient of thermal expansion, å, for a multi-component ideal gas. From the previous problem set, for a multicomponent ideal gas, F = F(,,{N}) = E - S = NC - NC ln(c) - N ln() + k From eq..-.3, N k ln(n k ) - Nsº 2. Page 2

3 F = - C = - NC 2.2 C = NC 2.3 F = = N 2 = P 2.4 = P 2.5 F = F = - å = - N = - P 2.6 å = 2.7 (b) Find the properties C P, s and å s for the multicomponent ideal gas. From eq..7, C P = C + å2 = N(C+) 2.8 From eq..6, S = C v C = P C C+ = C C+ P 2.9 From eq..8, å å = s C = P NC 2.0 å s = C 2. Problem 3: Find the restrictions on the first-order thermodynamic properties, e j, of a mole of a one-component fluid that are imposed by the conditions of stability. You need consider only the properties, e j, that are determined by the energy function. he first-order properties of a one-component fluid are E SS, E, and E S = E S, whose values are given above. he molar forms are: Page 3

4 e = e ss = c v 3. e 22 = e vv = v s = c P vc v 3.2 e 2 = e 2 = e vs = e sv = å c v 3.3 which form the 2x2 matrix e ij. he matrix e ij has two eigenvalues, which are the solutions to the equation det[e ij - ij ] = where ij is the Kronecker. Equation.4 yields the quadratic equation 2 - (e + e 22 ) + (e e 22 - e 2 2 ) = which has the two solutions = (e +e 22 ) 2 ± 2 (e -e 22 ) 2 + 4e he greater eigenvalue, 2, is found by taking the positive sign in equation 3.6. he lesser,, is found by taking the negative sign. or he condition, 0, requires that which is precisely the condition (e + e 22 ) 2 (e - e 22 ) 2 + 4e e e 22 - e det(e ij ) Substituting the values of e = e ss, e 22 = e vv and e 2 = e sv into equation 3.9 yields the condition c P c v vc - å v c v 2 = vc v2 c P - vå2 Page 4

5 = vc v Since both and v are positive and finite, it is sufficient that both c v and be positive and finite. o show that it is also necessary that c v and be positive and finite, consider the additional conditions of stability, that the diagonal terms of the property matrix be positive definite. Since e ss = c v 0 3. the specific heat, c v, is positive and finite, unless the zero of e ss coincides with the zero of. It then follows from 3.0 that the same condition pertains to. Alternatively, stability with respect to an isentropic change in volume yields the condition e vv = v s = c P vc v Since 0 by eqs. 3.0 and 3., it follows that c p 0. o explore behavior at the stability limit, note from eq. 3.0 that either c v,, or both must becomes singlar there. In fact, both types of behavior are observed. Singularities in c v are often found at instabilities, such as those observed in mutations such as ferromagnetic, ferroelectric and ordering transitions. Other transitions, such as the superconducting transitions, have only steps in c v, which is well-behaved as the instability is approached from either side. o see that this is possible note that = - (,) (P,) = - (,) (,S) (P,) (,S) - e ss = e ij 3.3 where e ij is the determinant of the property matrix. Since e ij 0 at the limit of stability, then at the stability limit unless e ss also vanishes there (c v ). It may be shown by straightforward Jacobian transformation that c P = - S = e vv P e ij å = v v = e sv P v e ij Page 5

6 It follows from eqs. 3.3, that all elements of the property matrix e ij can be finite and well-behaved at the stability limit while the properties, c P and å are all singular at the stability limit. Problem 4: At low temperature ( less than about one-half of the Debye temperature, Œ D ) the Helmholtz free energy of a simple one-component solid is given, to a good approximation, by the relation F(,,N) = Eº(,N) - π4 5 Nk Œ D 3 = Eº(,N) - E D /3 4. where Eº is the binding energy at 0 K, and the E D is the thermal energy due to lattice vibrations. he Debye temperature is a function of the molar volume, v = /N only. (a) Find the entropy as a function of, and from it find the isometric specific heat. Show that C v vanishes in the limit 0. From eq. 4. the entropy is S(,,N) = - F From which the specific heat is = 4π4 5 Nk Œ D C = S which vanishes in the limit 0. = 3Nk 4π Œ D 4.3 (b) Show that the internal energy is just E = Eº + E D 4.4 By definition, E = F + S = Eº - E D /3 + 4E D /3 = Eº + E D 4.5 (c) Show that the pressure is given by Page 6

7 P = - Eº + E D 4.6 where is the Grªuneisen parameter = - dln(œ D) dln(v) = - Œ D Œ D 4.7 he pressure of the solid is given as a function of (,,N) by the partial derivative P = - F = - Eº - 3π4 5 Nk4 Œ 4 dœ D D d = - Eº - 3π4 5 Nk If we define the Grªuneisen parameter Œ D 3 dln(œ D ) dln() 4.8 = - dln(œ D) dln(v) = - Œ D Œ D 4.7 which specifies the volume dependence of the Debye temperature, the pressure is given by where E D is the vibrational energy. P = P(N,,) = - Eº + E D 4.9 (d) his example is curious in that one can apparently derive the thermal equation of state, P(N,,), from the caloric equation of state, E(N,,), that is, the caloric equation of state is equivalent to the fundamental equation. o understand why this happens, show that E = - 2 F 4.0 Using this relation, show that the Helmholtz free energy is determined by the caloric equation of state to within an additive function of the form F' = f(,n) 4. hen show that the function f(,n) must vanish if the fundamental equation of the phase is to satisfy the hird Law. [he fundamental equations of gases and solutions do not satisfy the 3rd Law, so the procedure does not work for these phases.] Page 7

8 Eq. 4.0 is simply another way of writing E = F + F = E + S 4.2 Given eq. 4.0, F = - E 2 d + f(,n) 4.3 where f(,n) is an undetermined function that is independent of. Hence, F(,,N) = - E 2 d + f(,n) 4.4 he entropy, S(,,N), is S = - F = - E 2 d - E + f(,n) 4.5 According to the hird Law, the entropy of an unconstrained solid vanishes at = 0. It follows that S(,,N) can contain no term that is independent of. Hence, f(,n) must vanish unless the first two terms contain a part (g(,n) that is independent of, and must equal - g(,n) if it exists. Since f(,n) is determined by E(,,N), the caloric equation of state is sufficient to fix the fundamental equation. Problem 5: Prove that, in the limit 0, å = 0 5. C P = C = P = S = 5.4 (a) he coefficient of thermal expansion å = = P{N} 2 G P = - S P 5.5 {N} Page 8

9 In the limit, 0, the hird Law requires that S P = {N} which establishes 5.2. (b) Given that S(,) = 0 C d 5.7 and S(,P) = 0 CP d 5.8 S diverges as 0 unless lim 0 [C ] = A n 5.9 lim 0 [C P ] = A m 5.0 where m and n are greater than zero. Equation 5.2 follows. (c) From the definition of the isothermal compressibility, it can be shown that P = - å + (å) P {N} 5. which vanishes by eq (d) From eq of the notes: S = C C 5.2 P Since C P = C + å2 5.3 hen in the limit 0, C P = C, which establishes 5.4. Page 9

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