Thermodynamics of an Electron Gas: Entropy, Pressure, and Sound Speed, Study notes of Thermodynamics

Download Thermodynamics of an Electron Gas: Entropy, Pressure, and Sound Speed and more Study notes Thermodynamics in PDF only on Docsity! Example 1 The fundamental relation for the entropy of an electron gas can be approximated as S(U,V,N) = B N1/6 V1/3 U1/2, where (A) B = 23/2π4/3kBm1/2Navag1/6/(31/3hP). (B) Here, kB denotes the Boltzmann constant that has a value of R /NAvag = 1.3804×10–26 kJ K–1, hP is the Planck constant that has a value of 6.62517×10–37 kJ s, m denotes the electron mass of 9.1086×10–31 kg, N the number of κµολε of the gas, V its volume in m3, and U its energy in kJ. Determine s , T, and P when u = 4000 kJ k mole–1, and v = 1.2 m3 kmole–1. Solution The value of B = 5.21442 kg1/2 k mole1/6 s K–1. From Eq. (A), s = S/N = (B/N) N1/6( v N)1/3( u N)1/2 = B v 1/3 u 1/2, i.e., (C) s = 5.21442 (kg1/2 K–1 Kmole1/6 s)(1.2 m3 k mole–1) 1/3 (4000 kJ kmole–1) 2. Recalling that the units kg (m/s2) m ≡ J. s = 350 kg1/2 m kJ1/2 kmole–1 K–1. = 350.45 kJ kmole–1 K–1. From the entropy fundamental equation 1/T = (∂s /∂ u )v̄. Differentiating Eq. (C) with respect to u and using this relation, 1/T = (1/2) B v 1/3/ u 1/2 = 0.04381 or T = 22.8 K. (D) Similarly, since P/T = (∂s /∂ v )u, Upon differentiating Eq. (C) and using the above relation, P/T = (1/3) B u 1/2/ v 2/3 = 94.35 kPa K–1. (E) Using the value for T = 22.83 K, the pressure P = 2222.4 kPa.. The enthalpy h = u + P v = 4000 + 2222.4 × 1.2 = 6666.9 kJ kmole–1. Remarks Eq. (C) can be expressed in the form u (s , v ) = s 2/(B2 v 2/3). (F) Equation (F) is referred to as the energy representation of the fundamental equation (cf. Chapter 5). Rewriting Eq. (D) u (T, v ) = 1/4 B2 v 2/3T2. (G) Differentiating this relation with respect to T we obtain the result cv = (∂u/∂T) v = (1/2) B2/3 v 2/3T. (H) Dividing Eq. (E) by Eq. (D) we obtain the expression u (P, v ) = (3/2) P v . (I) Likewise, using the entropy fundamental state equation (Eq. (A)), we can also tabulate other nonmeasurable thermodynamic properties such as a (= u – Ts ) and g (= h – Ts ). Eliminating u in Eqs. (D) and (E) we obtain the state equation P = P(T, v ) for an electron gas in terms of measurable properties, i.e., P = (B/6) T2/ v 1/3. (J) If this state equation (in terms of P, T and v ) is known, it does not imply that s , u , h , a , and g can be subsequently determined. This is illustrated by considering the temperature and pressure relations T = ∂s /∂ u , and P/T = ∂s /∂ v (K) One can use Eq. (J) in (K). These expressions indicate that Eqs. (K) are differential equations in terms of s and, in order to integrate and obtains =s(T, v), an integration constant is required which is unknown. Therefore, a fundamental relation is that relation from which all other properties at equilibrium (e.g., T, P, v , s , u , h , a , g , cp, and cv) can be directly obtained by differentiation alone. While the Eq. (A) represents a fundamental relation, we can see that the relation Eq. (J) does not. Example 2 Obtain an expression for the entropy change in an RK gas when the gas is isothermally compressed. Determine the entropy change when superheated R–12 is isothermally compressed at 60ºC from 0.0194 m3 kg–1 (state 1) to 0.0126 m3 kg–1 (state 2). Compare the result with the tabulated value of s1 = 0.7259, s2 = 0.6881. Solution Consider the RK state equation P = RT/(v–b) – a/(T1/2v(v+b)) (A) From the third Maxwell’s relation Eq. (22) and Eq. (A), (∂s/∂v)T = (∂P/∂T)v= R/(v–b) + (1/2) a/(T3/2v(v+b)). (B) Integrating Eq. (B), s2(T,v2) –s1(T,v1) = Rln((v2–b)/(v1–b)) +(1/2)(a/(T3/2b)) ln(v2(v1+b)/(v1(v2+b))). (C) The critical conditions for R–12 are Tc = 385 K, and Pc = 41.2 bar. Therefore a =208.59 bar (m3 kmole–1) 2 K1/2, and b = 0.06731 m3 kmole–1. The molecular weight M = 120.92 kg kmole–1, and a = a /M = 208.59 bar (m3 kmole–1) 2K1/2÷120.92 kg kmole–1 = 1.427 k Pa (m3 kg–1) 2 K1/2, and b = 0.557×10–3 m3 kg–1. Since, R = 8.314 ÷ 120.92 = 0.06876 kJ kg–1 K–1, s2 – s1 = 0.06876 ln((0.0126 – 0.000557) ÷ (0.0194 – 0.000557)) + (1/2)(1.427 ÷ (3331.5 0.000557)) ln (0.0126 (0.0194 + 0.000557) + (0.0194 × (0.0126+ 0.000557))). = – 0.06876 × 0.448 – 0.211 × 0.01495 = –0.03396 kJ kg–1 K–1. Example 3 Show that both the isothermal expansivity βP = (1/v)(∂v/∂T)P and the isobaric compressibility coefficient βT = – (1/v)(∂v/∂P)T tend to zero as T → 0. Solution Example 1 shows that (∂s/∂T)v → 0 and (∂s/∂P)v → 0 as T → 0. From the fourth of the Maxwell’s relations, (∂v/∂T)P = –(∂s/∂P)T, so that (∂v/∂T)P → 0. Similarly, using the third Maxwell’s relation and the cyclic relations it may be shown that (∂P/∂T)v = –((∂v/∂T)/(∂v/∂P)) = (∂s/∂v)T → 0 as T → 0. Since ∂v/∂T → 0, it is apparent that ∂v/∂P → 0 as T → 0. Thereafter, combining Eqs. (1) and (K) c2 = k(T,v) v2 {RT/(v – b)2 + a/v3} (L) Remarks If, in the VW state relation, a = b= 0, the expression reduces to the sound speed for an ideal gas. In that case, k = k(T). High pressures often develop within the clearance space in turbine seals, and gas leaks are governed by the resulting choked flow conditions. The value of the sound speed through a real gas is required in order to evaluate this condition. In the case of liquids and solids, a very large pressure is required to cause a small change in the volume so that (∂P/∂v)T → ∞ and, consequently, c → ∞. Therefore, sound travels at faster speeds in liquids and solids. Applying Eq. (L) to the case of an ideal gas, co2 = k(T) RT, and dividing Eq.(I) by co2 , (c2/co 2) = – (k(TR, vR´)/(ko(TR) TR)) vR2 (∂PR/∂vR´)TR. b. Example 6 The state of a copper bar is initially at a pressure of 1 bar and temperature of 250 K. It is compressed so that the exerted pressure is 1000 bar. Assume that the compression is adiabatic and reversible (i.e., the material reverts to its original state once the load is removed) with, and that βP = 48×10–6 K–1, βT = 7.62×10–7 bar–1, v = 1.11×10–4 m3 kg–1, cp = 0.372 kJ kg–1 K–1, and cv = 0.364 kJ kg–1 K–1. Determine the change in the internal energy and the intermolecular potential energy of the solid. Solution ( )2 1 T 1 2 1v v v P P− ≈ −β − (A) With T T O V v 1       ∂ ∂−=β , and integrating ( )2 T 2 1 1 Vln P P V   = −β −    solving for v2 134 2 gm101109.1v −−×= . From Eq. (37 ) and integrating ( )2 P 2 1 1 T v Tln V V T C β= − − β or ( ) ( )12 V 1 T P 12 VVC T PT − β β −≈− Using results from (A) ( ) K375.0PP C TT 12 V P 12 =− β ≈ (B) c.) For a adiabatic reversible process, the first law yields, ∆u = – ∫P dv = –∫P (∂v/∂P) dP = ∫βT P v dP ( ) 2 2 1 2 2 vPP Tβ−= = ((1000 × 100)2÷2 – 1002÷2) × 1.11×10–4 × 7.62×10–9 = 0.00423 kJ kg–1 or 4.23 J kg–1. d.) Note that one may use following equation also, which is similar to the result in Example 7. du = cv dT + (T βP/βT – P) dv. Now, thermal part of change in “u” is given as, cv dT = 0.364 × 0.38 = 0.138 kJ kg–1, (D) The intermolecular potential energy of the solid ∆(ipe) = (TβP/βT – P) dv where second term is same as the answer in part (c) = (250 K × 48×10–6 K–1 ÷ 7.62x10–7 bar–1) * 8.8x10-8 m3kg-1*100 kPa bar-1 -0.00423 ∆ (ipe) = 0.1398-0.00423 = -0.13kJkg-1 (E) Therefore, the net change in the internal energy of the solid is du = 0.138– 0.1356 = 0.00239 kJ kg–1, At the minimum intermolecular potential energy, compression should cause the "ipe" to increase. Since the ipe decreases with compression, this indicates that the solid is not at that minimum value. In this example, the temperature increases by 0.38 K, increasing the thermal portion of the internal energy by 0.138 kJ, but the IPE decreases by 0.13 kJ. ( )121T12 PPvvv −β−≈− Recall that (∂T/∂v)s = – TβP/(βTcv) so that dvs = –dTsβTcv/(T βP). Likewise, since (∂T/∂P)s = TvβP/cP, dTs = dPs(TvβP/cP). Hence, ∆T = 0.36 K. Consequently, the temperature following compression is 250.36 K. The internal energy change dus = –(P dvs) = (P βT cv/(T βP)) dTs. An approximate solution for the internal energy change is as follows: dus = – Pdvs = (P βT cv/(T βP)) dTs = 1000 bar× 7.62×10–7 bar–1× 0.364 kJ–1 kg–1 K–1× 0.36 K÷(250 K× 48×10–6 K–1) = 0.00832 kJ kg–1. A more accurate evaluation is as follows dv = –7.62×10–7 × 1000 × 1.11 × 10–4 = –8.46×10–8 m3 kg–1. Therefore, ∆u = – ∫P dv = –∫P (∂v/∂P) dP = ∫βT P v dP = ((1000 × 100)2÷2 – 1002÷2) × 1.11×10–4 × 7.62×10–9 = 0.00423 kJ kg–1 or 4.23 J kg–1. Note that du = cv dT + (T βP/βT – P) dv. Now, cv dT = 0.364 × 0.36 = 0.131 kJ kg–1, and the intermolecular potential energy of the solid ipe = (TβP/βT – P) dv = (250 K × 48×10–6 K–1 ÷ 7.62x10–7 bar–1 – 1000 bar) × 100 bar kPa–1 × (–8.46×10–8 m3 kg–1). = – 0.125 kJ kg–1. Therefore, the net change in the internal energy of the solid is du = 0.131– 0.125 = 0.006 kJ kg–1, At the minimum intermolecular potential energy, compression should cause the IPE to increase. Since the IPE decreases, this indicates that the solid is not at that minimum value. In this example, the temperature increases by 0.36 K, increasing the thermal portion of the internal energy by 0.131 kJ, but the IPE decreases by 0.125 kJ. c. Example 7 Obtain an expression for dh and du for a liquid in terms of cp, βP, βT, cv, dT and dP. Simplify the relations for an incompressible liquid. Solution Rewriting Eq. (43), dh = cp dT + (v – T βp v) dP, (A) where βp = (∂v/∂T)p. Therefore, du = dh – d(Pv) = cp dT –Pdv – T βp v dP However, v = v(T,P), so that dv = (∂v/∂T)P dT + (∂v/∂P)T dP = v (βp dT – βT dP). (B) Hence, du = (cp – Pvβp) dT + v (PβT – TβP) dP (C) For incompressible liquids βP = βT = 0, and Eqs. (A) can be expressed in the form dh = cp dT + v dP., i.e., h = h(T,P), and (D) Further, (dh/dT)P = cp. (E) For an incompressible fluid Eq. (C) assumes the form du/dT = cp, i.e., u = u (T) alone. (F) Upon comparing Eqs. (E) and (F), (∂h/∂T)p = du/dT = cp. (G) Note that du/dT = (∂u/∂T)v, since v is constant. Therefore, (∂h/∂T)p = du/dT = cp = cv = c (H) d. Example 8 Determine the entropy of H2O assuming the relation s= 0 for the saturated liquid at the triple point, and hfg = 2503 kJ kg–1. If the ideal gas specific heat of water is reproduced by the relation c p,o (kJ kmole–1 K–1) = 28.85 + 0.01206 T+100,600/T2, determine its entropy at a pressure of 250 bar and a temperature of 873 K. Solution Typically, properties are tabulated with respect to arbitrary reference conditions, e.g., Tref and Pref. For saturated liquid water, it is customary to set that reference condition at the triple point, i.e., Tref = TTP = 273 K and PTP = 0.00611 bar at which the entropy is assumed to have a value of zero. Since Tds + vdP = dh, during vaporization at a specified pressure, ds = dh/T, i.e., sg – sf = (hg – hf)/T = hfg/T. Therefore, sg(TTP,PTP) – 0 = hfg/T = 2503÷273 = 9.17 kJ kg–1 K–1. Applying the RK equation at this state, 0.00611 = 0.08314×273÷( v –0.0211) – 142.64÷(2732× v ( v +0.0211)), i.e., v = 3715 m3 kmole–1. The compressibility factor based on this value of the specific volume Z(TTP,PTP) = P v /( R T) = 0.00611×3715÷(0.08314×273) = 1. Furthermore, employing Eq. (53), s(273, 3715) – so(273,3715) ≈ 0, and using Eq. (57) s(273, 0.00611) – so(273,0.00611) ≈ 0. This result is expected, since the pressure is low so that the vapor behavior is like that of an ideal gas. Hence, s(273,0.00611) = so (273, 0.006 bar) = 9.17 kJ kg–1. (A) Since.