SOLUTIONS to Problem Set I - EEE 2010

# 718 hn hn 38301 kjmol hv hv 35645 kjmol q

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Unformatted text preview: nd Watson correction (4.13) Trn = Tn/Tc; Trn = 0.659; Trsat = Tsat/Tc; Trsat = 0.718 . . ΔHn = ; ΔHn = 38.301 kJ/mol . ΔHv =∆ . ΔHv = 35.645 kJ/mol ; ∆ Q = n.ΔH; ∆ n = 100 kmol/hr; ; ΔH = 49.38 kJ/mol Q = 1.372 ×103 kW Ans. Similarly, For Benzene ΔHv = 28.273 kJ/mol; ΔH = 55.296kJ / mol; Q = 1.536×103kW For Toluene ΔHv = 30.625 kJ/mol; ΔH = 65.586kJ / mol; Q = 1.822×103kW d. Solution to Problem 4.15 (SVNA) For Benzene Tc = 562.2K; Pc = 48.98bar; Tn = 353.2K T2sat = 358.7K; Cp = 162J / mol⋅K1 T1sat = 451.7K; Estimate ΔHv using Riedel equation (4.12) and Watson correction (4.13) Trn = Tn/Tc; Trn = 0.628; Tr2sat = T2sat/Tc; Tr2sat = 0.638 ΔHn = ΔHv =∆ . . . . ; ; ΔHn = 30.588 kJ/mol ΔHv = 30.28 kJ/mol Assume the throttling process is adiabatic and isenthalpic Guess vapor fraction (x): x=0.5 x = Find(x); x=0.498 Ans. Given Cp (T1sat – T2sat) = x.ΔHv; e. Solution to Problem 4.49 (SVNA) C. HEAT EFFECTS OF MIXING a. Solution to problem 12.28 (SVNA) LiCl.2H2O ---> Li + 1/2 Cl2 + 2 H2 + O2 (1) Li + 1/2 Cl2 + 10 H2O ---> LiCl(10 H2O) (2) 2(H2 + 1/2 O2 ---> H2O) (3) -------------------------------------------------------------------LiCl.2H2O + 8 H2O (l) ---> LiCl(10 H2O) ΔH1 = − (−1012650) J (Table C.4) ΔH2 = −441579 J (Pg. 457) ΔH3= 2 (−285830 J) (Table C.4) ΔH = ΔH1 + ΔH2 + ΔH3 ΔH = −589 J (On the basis of 1 mol of solute) Since there are 11 moles of solution per mole of solute, the result on the basis of 1 mol of solution is ΔH/11 = −53.55 J b. Solution to problem 12.29 (SVNA) 2(HCl + 2.25 H2O -----> HCl(2.25 H2O)) (1) HCl(4.5 H2O) -----> HCl + 4.5 H2O (2) ---------------------------------------------HCl (4.5 H2O) + HCl -----> 2 HCl(2.25 H2O) 12.29 ΔH1 = 2 (−50.6⋅kJ) (Fig. 12.14 @ n=2.25) ΔH2 = 62 kJ (Fig. 12.14 @ n=4.5 with sign change) ΔH = ΔH1 + ΔH2 ΔH = −39.2 kJ Ans. c. Solution to problem 12.30 (SVNA) Moles of LiCl added: n'LiCl = 0.472 kmol Mole ratio, original solution: nH2O/nLiCl = 21.18 Mole ratio, final solution = nH2O/(nLiCl+n’LiCl) = 8.15 0.2949(LiCl(21.18 H2O) ---> LiCl + 21.18 H2O) (1) 0.7667(LiCl + 8.15 H2O ---> LiCl(8.15 H2O)) (2) --------------------------------------------------------------------------------------0.2949 LiCL(21.18 H2O) + 0.4718 LiCl ---> 0.7667 LiCl(8.145 H2O) ΔH1=nLiCl (35kJ/mol) (Fig. 12.14, n=21.18) ΔH2= (nLiCl+n’LiCl)(-32 kJ/mol) [Fig. 12.14, n=8.15) Q = -14213kJ Q = ΔH1+ ΔH2; d. Solution to problem 12.34 (SVNA) BASIS: 1 second, during which the following are mixed: (1) 12 kg hydrated (6 H2O) copper nitrate (2) 15 kg H2O n2=15/18.015 kmol/sec n1= 12/295.61 kmol/sec; n1=0.041 kmol/sec; n2 = 0.833 kmol/sec Mole ratio, final solution: (6n1+n2)/n1 = 26.51 6(H2 + 1/2 O2 ---> H2O (l)) (1) Cu + N2 + 3 O2 ---> Cu (NO3)2 (2) Cu (NO3)2.6H2O ---> Cu + N2 + 6 O2 + 6 H2 (3) Cu (NO3)2 + 20.51 H2O ---> Cu (NO3)2(20.51 H2O) (4) -----------------------------------------------------------------------------------------Cu (NO3)2.6H2O + 14.51 H2O (l) ---> Cu (NO...
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## This homework help was uploaded on 04/03/2014 for the course CHME 2020 taught by Professor None during the Spring '08 term at Rensselaer Polytechnic Institute.

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