GROUP (22) 1 SHEET (3) 2nd Law of Thermodynamics 2 Question No. (1) The initial state of one mole of an ideal gas is P = 10 atm and T = 300K. Calculate the entropy change in the gas for (a) An isothermal decrease in the pressure to 1 atm; (b) A reversible adiabatic decrease in the pressure to 1 atm; (c) A constant volume decrease in the pressure to 1 atm. The Answer of Question No. (1) a) Isothermal ∆S = NRln V2 P1 = NRln V1 P2 10 = 1 ∗ 8.314 ∗ ln 1 ∆S = 19.14 J/K b) Reversible Adiabatic ∆S = 0 c) Constant Volume T2 P2 ∆S = NCv ln = NCv ln T1 P1 3 3 1 = 1 ∗ ∗ 8.314 ∗ ln 2 10 = − 28.71 J/K 4 Question No. (2) One mole of an ideal gas is subjected to the following sequence of steps: a.Starting at 25 ˚C and 1 atm, the gas expands freely into a vacuum to double its volume. b.The gas is next heated to 125 ˚C at constant volume. c. The gas is reversibly expanded at constant temperature until its volume is again doubled. d.The gas is finally reversibly cooled to 25 ˚C at constant pressure. Calculate ∆U, ∆H, q, W, and ∆S in the gas. The Answer of Question No. (2) a) Expands freely into a vacuum to double its volume RT1 0.082 ∗ 298 V1 = = = 24.436 L P1 1 V2 = 2 V1 = 48.872 L V2 P1 = V1 P2 1 =2 P2 5 P2 = 0.5 atm T2 = T1 = 298 K ∆U = W = q = ∆H = 0 V2 ∆S = NR ln = 1 ∗ 8.314 ln 2 V1 = 5.763 J/K b) Heated to 125 ˚C at constant volume T3 = 398 K W=0 ∆U = NCv ∆T 3 = 1 ∗ ∗ 8.314 ∗ (398 − 298) 2 = 1247.1 J q = ∆U = 1247.1 J ∆H = NCp ∆T 5 = 1 ∗ ∗ 8.314 ∗ (398 − 298) 2 = 2078.5 J 6 ∆S = NCv ln T2 T1 3 398 = 1 ∗ ∗ 8.314 ∗ ln 2 298 = 3.6086 J/K c) Reversibly expanded at constant temperature until its volume is again doubled V4 = 2V3 = 2 ∗ 48.872 = 97.744 L ∆U = 0 W = NRT ln V3 V4 1 = 1 ∗ 8.314 ∗ 398 ∗ ln 2 = − 2293.6 J q = − W = 2293.6 J ∆H = 0 V4 ∆S = NR ln = 1 ∗ 8.314 ∗ ln 2 V3 = 5.763 J/K d) Reversibly cooled to 25 ˚C at constant pressure W = P(V4 − V5 ) = NR(T4 − T5 ) = 1 ∗ 8.314 ∗ (398 − 298) = 831.4 J 7 ∆U = NCv ∆T 3 = 1 ∗ ∗ 8.314 ∗ (298 − 398) 2 = − 1247.1 J q = ∆U − W = −1247.1 − 831.4 = − 2078.5 J ∆H = qp = − 2078.5 J T5 ∆S = NCp ln T4 5 298 = 1 ∗ ∗ 8.314 ∗ ln 2 398 = − 6.014 J/K Process ∆U (J) W (J) q (J) ∆H (J) ∆S (J/K) a) 0 0 0 0 5.763 b) 1247.1 0 1247.1 2078.5 3.6086 c) 0 0 5.763 d) -1247.1 Total 0 -2293.6 2293.6 831.4 -2078.5 -2078.5 -6.014 -1462.2 1462.2 8 0 9.1206 Question No. (3) Calculate the final temperature and the entropy produced when 1500 grams of lead at 100 ˚C is placed in 100 grams of adiabatically contained water, the initial temperature of which is 25 ˚C. Given that: Cp(H2O) =18 cal/(deg.mole) and Cp(Pb) =6.38 cal/(deg.g.atom). The molecular and atomic weights of H2O and Pb are 18 and 207 grams, respectively. The Answer of Question No. (3) ∑ dq = 0 NCp ∆T)water + NCp ∆T)lead = 0 Nw Cp(w) (Tf − Tw ) = NPb Cp(Pb) (TPb − Tf ) 100 1500 ∗ 18.03 ∗ (Tf − 25) = ∗ 6.38 ∗ (100 − Tf ) 18 207 ∴ Tf = 48.6846 ℃ ∆Stotal = ∆Swater + ∆Slead ∆Swater = Nw Cp(w) ln 9 Tf Tw = 100 321.6846 ∗ 4.18 ∗ 18.03 ∗ ln 18 298 = 32.02 J/K ∆Slead Tf = NPb Cp(Pb) ln TPb 1500 321.6846 = ∗ 4.18 ∗ 6.38 ∗ ln 207 373 = − 28.6 J/K ∆Stotal = 32.02 − 28.6 = 3.42 J/K 10 Question No. (4) From 298 K up to its melting temperature of 1048 K, the constant pressure molar heat capacity of RbF is given as: Cp = 7.97+ 9.2 x 10-3 T + 1.21 x 105 T-2 cal/ (deg.mole) and from the melting temperature to 1200 K, the constant pressure molar heat capacity of liquid RbF is given as: Cp = -11.3 + 0.833 x 10-3 T + 350.7 x 105 T-2 cal/ (deg.mole) At its melting temperature the molar heat of fusion of RbF is 6300 cal/mole. Calculate the increase in the entropy of 1 mole of RbF when it is heated from 300 K to 1200 K. The Answer of Question No. (4) ∆S1 : Heating solid RbF from 300 K to 1048 K. ∆S2 : Melting RbF at 1048 K. ∆S3 : Heating liquid RbF from 1048 K to 1200 K. Tm ∆S1 = ∫ Ti Tm NC δq p(s) =∫ dT T T Ti 1048 =∫ (7.97T −1 + 9.2 ∗ 10−3 + 1.21 ∗ 105 T −3 ) dT 300 11 1048 5 1.21 ∗ 10 = [7.97 ln T + 9.2 ∗ 10−3 T + T −2 ] −2 300 ∆S1 = 17.468 J/K qm ∆S2 = T 6300 = = 6.01145 J/K 1048 Tf Tf NC δq p(l) ∆S3 = ∫ =∫ dT T T Tm Tm 1200 =∫ 1048 ( −11.3 + 0.833 ∗ 10−3 + 350.7 ∗ 105 T −3 ) dT T 1200 5 350.7 ∗ 10 = [−11.3 ln T + 0.833 ∗ 10−3 T + T −2 ] −2 1048 = 2.3846 J/K ∆Stotal = ∆S1 + ∆S2 + ∆S3 = 17.468 + 6.01145 + 2.3846 = 25.864 J/K 12 Question No. (5) Two moles of an ideal gas are contained adiabatically at 30 atm pressure and 298 K. The pressure is suddenly released to 10 atm, and the gas undergoes in an irreversible adiabatic expansion as a result of which 500 cal of work are performed. Show that the final temperature of the gas after the irreversible expansion is greater than that which the gas would attain if the expansion from 30 to 10 atm had been conducted reversible. Calculate the entropy produced as a result of the irreversible expansion. Cv for the gas equals to 1.5R. The Answer of Question No. (5) If the process is reversible from 1 2 P2 𝛾−1 T2 𝛾 ( ) =( ) P1 T1 5 −1 3 10 ( ) 30 5 3 T2 =( ) 298 T2 = 192 K 13 For the irreversible process from 1 3 ∆U = W NCv (T3 − T1 ) = W 3 2 ∗ ∗ 8.314 ∗ (T3 − 298) = −4.18 ∗ 500 2 T3 = 214.2 K ∴ T3 > T2 To calculate ∆Sirr , choose any reversible path from state 1 to state 3. Consider the reversible path 1 2 3 From 1 2 adiabatic ∆S = 0 From 2 3 constant pressure ∆S = NCp ln T3 5 214.2 = 2 ∗ ∗ 8.314 ∗ ln T2 2 192 = 4.548 J/K ∴ ∆Sirr = 4.548 J/K 14 Question No. (6) At a pressure of 1 atm the equilibrium melting temperature of lead is 600 K, and at this temperature the latent heat of fusion of lead is 1150 cal/mole. If 1 mole of supercooled liquid lead spontaneously freezes at 590 K and 1 atm pressure, calculate the entropy produced. The constant pressure molar heat capacity of liquid lead, as a function of temperature at 1 atm pressure, is given as Cp (Pb)l = 7.75 – 0.74 x 10-3 T cal/deg And the corresponding expansion for solid lead is given as Cp (Pb)s = 5.63 + 2.33 x 10-3 T cal/deg The Answer of Question No. (6) The freezing process happened spontaneously (irreversibly), so to calculate ∆Sirr consider the following reversible steps: 1. Heating supercooled liquid lead from 590 K to 600 K. 2. Solidifying the liquid lead at 600 K. 3. Cooling solid lead from 600 K to 590 K. 600 ∆S1 = ∫ 590 600 NC δq p(Pb(l)) =∫ dT T T 590 15 600 (7.75T −1 − 0.74 ∗ 10−3 ) dT =∫ 590 = [7.75 ln T − 0.74 ∗ 10−3 T]600 590 = 0.12285 cal/K −qm ∆S2 = T −1150 = 600 = − 1.91667 cal/K 590 ∆S3 = ∫ 600 590 NC δq p(Pb(s)) =∫ dT T T 600 590 =∫ (5.63T −1 + 2.33 ∗ 10−3 ) dT 600 = [5.63 ln T + 2.33 ∗ 10−3 T]590 600 = − 0.1179 cal/K ∆Sirr = ∆S1 + ∆S2 + ∆S3 = 0.12285 − 1.91667 − 0.1179 = − 1.9117 cal/K 16 Question No. (7) A sample of 1 mole Ar is expanded isothermally at 0 ˚C from 22.4 lit to 44.8 lit; if the process is carried out: a) reversibly, b) against constant external pressure equal to the final pressure of the gas, or C) freely. Calculate ∆S (in J/(deg.mole)) for each case. The Answer of Question No. (7) a) Isothermal V2 ∆S = NR ln V1 44.8 = 1 ∗ 8.314 ∗ ln 22.4 = 5.763 J/K b) Constant pressure T2 V2 ∆S = NCp ln = NCp ln T1 V1 5 44.8 = 1 ∗ ∗ 8.314 ∗ ln 2 22.4 = 14.407 J/K 17 c) Freely: it is irreversible process. Consider the reversible isothermal path (during free expansion temperature does not change (ideal gas)) V2 ∆S = NR ln V1 44.8 = 1 ∗ 8.314 ∗ ln 22.4 = 5.763 J/K 18 Question No. (8) A sample of 1 mole H2O(g) is condensed isothermally and reversibly to liquid water at 100 ˚C. The standard enthalpy of vaporization of water at 100 ˚C is 40.656 KJ/mole. Find ∆S in (KJ/(deg.mole)) for this process. The Answer of Question No. (8) −∆H ∆S = T = −40.656 373 = − 0.109 KJ/K 19 Question No. (9) The heat capacity of one mole sample of a perfect gas was found to vary with temperature according to the expression Cp(J/(deg.mole) = 20.17+0.3665 T. Calculate ∆S in (KJ/(deg.mole)) when the temperature is raised from 25 ˚C to 200 ˚C if the process is carried out: a) at constant pressure, or b) at constant volume. The Answer of Question No. (9) a) Constant pressure T2 NC p ∆S = ∫ dT T T1 473 =∫ (20.17T −1 + 0.3665) dT 298 = [20.17 ln T + 0.3665T]473 298 = 73.456 J/K = 0.073456 KJ/K b) Constant volume Cp − Cv = R Cv = Cp − R 20 Cv = 20.17 + 0.3665T − 8.314 = 11.856 + 0.3665T T2 NCv ∆S = ∫ dT T T1 473 =∫ (11.856T −1 + 0.3665) dT 298 = [11.856 ln T + 0.3665T]473 298 = 69.615 J/K = 0.069615 KJ/K 21 Question No. (10) Two moles of N2 expands from 10 lit, 25 ˚C to 20 lit final state. Find ∆S obtained if the process is isothermal, assuming that Cv= (5/2) R, and Cp= (7/2) R, and the gas behaves ideally. The Answer of Question No. (10) V2 ∆S = NR ln V1 = 2 ∗ 8.314 ∗ ln = 11.525 J/K 22 20 10 Question No. (11) For each step, and for the net cycle shown in the next page, if 1 mole of a perfect gas performed the cycle and assuming that all processes to be reversible; find: a) P, V, T at points 1, 2, 3, and 4 and b) ∆S for every step and for the net cycle. P (atm.) 1 10 atm ↗ Isothermal 2 → adiabatic 4 0 0.5 3 1 1.5 2 2.5 V (lit) The Answer of Question No. (11) At point 1 V1 = 0.5 L P1 = 10 atm T1 = P1 V1 0.5 ∗ 10 = = 60.976 K NR 0.082 23 At point 2 V2 = 2 L T2 = T1 = 60.976 K NRT2 P2 = V2 1 ∗ 0.082 ∗ 60.976 = = 2.5 atm 2 At point 4 V4 = 1 L V4 𝛾 P1 ( ) =( ) V1 P4 5 3 1 10 ( ) =( ) 0.5 P4 P4 = 3.15 atm P4 V4 1 ∗ 3.15 T4 = = = 38.41 K NR 0.082 At point 3 V3 = 2 L P3 = P4 = 3.15 atm P3 V3 2 ∗ 3.15 T3 = = = 76.82 K NR 0.082 24 Process from 1 2 (Isothermal) ∆S = NR ln V2 V1 2 = 1 ∗ 8.314 ∗ ln 0.5 = 11.525 J/K Process from 2 3 (Constant volume) T3 ∆S = NCv ln T2 3 76.82 = 1 ∗ ∗ 8.314 ∗ ln 2 60.967 = 2.88 J/K Process from 3 4 (Constant pressure) T4 ∆S = NCp ln T3 5 38.42 = 1 ∗ ∗ 8.314 ∗ ln 2 76.82 = − 14.4 J/K Process from 4 1 (Adiabatic) ∆S = 0 25 For the net cycle ∆Scycle = 0 Point P (atm) V (L) T (K) 1 10 0.5 60.976 2 2.5 2 60.976 3 3.15 2 76.82 4 3.15 1 38.41 Process 1 2 2 3 3 4 4 1 Total ∆S (J/K) 11.525 2.88 -14.4 0 0 26