CHE 3161 (JUN 11) Solution for CHE3161_S1_2011 Question 1. (20 Marks) One mole of ideal gas with Cp = (5/2)R and Cv = (3/2)R undergoes the following two sequential steps: (i) Heating from 200 K to 600 K at constant pressure of 3 bar, followed by (ii) Cooling at constant volume. To achieve the same amount of Work produced by this two-step process, a single isothermal expansion of the same gas from 200 K and 3 bar to some final pressure, P can be performed. (1) Draw all the processes on a P-V diagram. [4 marks] (2) What is the final pressure, P of the isothermal expansion process assuming mechanical reversibility for both the processes? [14 marks] (3) Comment on the value of P of the isothermal expansion process assuming mechanical reversibility for the two-steps process while mechanical irreversibility for the isothermal expansion process. [2 marks] Solution: (1) P 1 Isobaric 2 Isochoric Isothermal 3 4 V Page 1 of 12 hence W12 = − P2V2 + P1V1 Applying Ideal Gas Law: W12 = − RT2 + RT1 = R(T1 − T2 ) W23 = − ∫ 32 PdV =0 Therefore.CHE 3161 (JUN 11) (2) For the two-steps process: ! !!" = − ! !" ! = −!∆! = −!! (!! − !! ) = −!! !! + !! !! Since P1 = P2. Page 2 of 12 . Total W13 = W12 + W23 = R(T1 − T2 ) For the isothermal expansion process: W14 = RT1 ln P4 P1 If the two works have to be the same: P RT1 ln 4 = R (T1 − T2 ) P1 ln (1) P4 (T1 − T2 ) = P1 T1 ⎡ (T − T2 ) ⎤ P4 = P1 exp⎢ 1 ⎥ ⎣ T1 ⎦ ⎡ (200 − 600) ⎤ = 3 exp⎢ ⎥ 200 ⎣ ⎦ = 0.406bar (3) P increases since the left hand term of Eqn (1) will be multiplied by a factor of less that 1 (1 =100% efficiency). 0320) Starting with an initial guess of Z = 1.15 15 = 0.0372 ! ! " Z + !$ The derivative is: Page 3 of 12 .3531 369.0320) = 1+ 0. (20 Marks) Calculate the compressibility (Z). [20 marks] Solution: For the given conditions: Tr = 80 + 273. residual entropy (SR).08664Tr We can now solve iteratively for Z using the equation: Z = 1+ ! ! q! (Z ! ! ) (Z ! 0. and iterating gives.8 42." ) = "#1+ (0. and ω = 0.48 bar.3403 "Tr 0.0320 Tr Tr 2 ! (Tr .574" ! 0.48 The dimensionless EOS parameters for the R/K EOS are: ! =! Pr P = 0.0328 q= !! (Tr ) 0. residual enthalpy (HR). Z = 0.4278! (Tr ) = = 5.176" 2 ) (1! Tr1/2 )$% = 1. The critical properties of propane are Tc = 369.CHE 3161 (JUN 11) Question 2.480 +1.8442 Then the integral I is: I= 1 Z + #$ ln = 0.08664 r = 0.9550 Pr = = 0.1709 Z(Z + ! ) Z(Z + 0.8 K. Pc = 42.152. and residual Gibbs energy (GR) of propane at 80oC and 15 bar using the Soave/Redlich/Kwong equation of state.0320 ! 0. 4915 RT # d lnTr & SR d ln " (Tr ) = ln(Z ! ! ) + qI = !0.3448 R d lnTr Therefore.K-1 Knowing these.mol-1.031 J.574" ! 0.CHE 3161 (JUN 11) 0.mol-1.867 J. H R = !1443.6877 d lnTr # ! (Tr ) & Next.mol-1 Page 4 of 12 .176" 2 ) $ r ' = !0.5 " T % d ln ! (Tr ) = !(0. G R = H R ! TS R = !430.572 J.480 +1. S R = !2. we can use these values to calculate the residual enthalpy and entropy from: " d ln ! (Tr ) % HR = Z !1+ $ !1' qI = !0. yi P = xi ! i Pi sat Page 5 of 12 . ln ! 2 = A x12 § At 144oC. Determine the equilibrium pressure P and vapour composition y1 from the following information: § ln !1 = A x22 . therefore P = P2sat + x1 (P1sat ! P2sat ) (3) Equation 3 predicts that P is linear in x1. we can first determine the value for A. (2) Based on the known information. no azeotrope is possible.294 Solution: (1) For a binary system obeying Raoult’s law. y1P + y2 P = x1 P1sat + x2 P2sat As y1 + y2 =1 and x1 + x2 = 1.CHE 3161 (JUN 11) Question 3. and then calculate equilibrium pressure and vapour composition. Thus no maximum or minimum can exist in this relation. y1P = x1 P1sat (1) y2 P = x2 P2sat (2) equations (1) + (2) give. From modified Raoult’s law. (20 Marks) (1) Prove: An equilibrium liquid/vapour system described by Raoult’s law cannot exhibit an azeotrope.66 KPa § The system forms an azeotrope at 144oC for which x1az = y1az = 0.6 is in equilibrium with its vapour at 144oC. (2) A liquid mixture of cyclohexanone(1)/phenol(2) for which x1 = 0. P1Sat = 75. Since such an extremum is required for the existence of an azeotrope.20 and P2Sat = 31. ln !1 = A x22 .CHE 3161 (JUN 11) At the azeotrope.4696 P = x1 !1P1sat + x2! 2 P2sat = 38. !1 P sat ln 2sat ! P A= 2 22 = 2 1 2 x2 ! x1 x2 ! x1 ln Putting in the known numbers for satuation pressures and compositions at the azeotrope: A = -2. yi= xi. !1 P2sat = ! 2 P1sat Given the conditions.6. x2 = 1-x1 = 0.1898 kPa The vapour composition y1 is: y1 = x1 !1P1sat = 0.4. !i = P Pi sat Therefore. !1 = exp(A x22 ) = 0. ln ! 2 = A x12 Then.7146 ! 2 = exp(A x12 ) = 0.0998 Next. at x1 = 0.8443 P Page 6 of 12 . ln !1 = A(x22 ! x12 ) !2 Therefore. then. Using the expressions obtained in (4a). (a) (b) Determine an expression as a function of x1 for (i) the partial molar volume of species 1. [8 marks] [12 marks] Solutions: (a) V = 120x1 + 70x2 + (15x1 + 8x2) x1x2 But x1 + x2 = 1 ∴x2 = 1 – x1 V = 120 x1 + 70(1 – x1) + [15x1 + 8(1 – x1)] x1(1 – x1) Reagreement and simplification of equation will lead to: V = -7x13 – x12 + 58x1 + 70 dV = −21x12 − 2 x1 + 58 dx1 (i) Using Eq.15). (ii) the partial molar volumes at infinite dilution V1! and V2! . V2 . (20 Marks) The molar volume (cm3 mol-1) of a binary liquid system of species 1 and 2 at fixed T and P is given by the equation V = 120x1 + 70x2 + (15x1 + 8x2) x1x2. (ii) the partial molar volume of species 2. (11. V1 = V + x2 dV dx1 V1 = !7x13 ! x12 + 58x1 + 70 + (1! x1 )(!21x12 ! 2x1 + 58) Reagreement and simplification of equation will lead to: Page 7 of 12 . V1 .CHE 3161 (JUN 11) Question 4. calculate the values for (i) the pure-species volumes V1 and V2 . V1 = 14(1)3 – 20(1)2 – 2(1) + 128 V1 = 120 cm3 mol-1 For pure species volume.CHE 3161 (JUN 11) V1 = 14x13 ! 20x12 ! 2x1 +128 (ii) Using Eq. V1! x1 = 0 Thus. V2 x2 = 1 or x1 = 0 Thus.16). V1 x1 = 1 Thus. V2 = 14(0) + 02 + 70 V2 = 70 cm3 mol-1 (i) For partial volume at infinite dilution. V1! = 14(0)3 – 20(0)2 – 2(0) + 128 V1! = 128 cm3 mol-1 For partial volume at infinite dilution. V2! x2 = 0 or x1 = 1 Page 8 of 12 . (11. V2 = V ! x1 dV dx1 V2 = !7x13 ! x12 + 58x1 + 70 ! x1 (!21x12 ! 2x1 + 58) Reagreement and simplification of equation will lead to: V2 = 14x13 + x12 + 70 (b) (i) For pure species volume. CHE 3161 (JUN 11) Thus. V2! = 14(1) + 12 + 70 V2! = 85 cm3 mol-1 Page 9 of 12 . Tc. calculate the compostions of the equilibrium mixture (yn-C4H10 and yiso-C4H10) by two procedures: (a) Assume an ideal-gas mixture.1 K.48 bar Solutions: Given T = 425 K.1= 37. [6 marks] [14 marks] For n-C4H10: ω1 = 0.CHE 3161 (JUN 11) Question 5. v = ∑ vi = 1 − 1 = 0 Assume species 1 ≡n-C4H10.1= 425. (20 Marks) Equilibrium at 425 K and 15 bar is established for the gas-phase isomerisation reaction: n-C4H10(g) →iso-C4H10(g) If there is initially 1 mol of reactant and K = 1.974 Page 10 of 12 . (b) Assume an ideal solution. K = 1.200.96 bar For iso-C4H10: ω2 = 0. y1 = y2 = (a) 1− ε = 1− ε 1 + 0(ε ) ε 1 + 0(ε ) =ε For an ideal-gas mixture: −v ⎛ P ⎞ ∏i ( yi ) = ⎜⎜ P ⎟⎟ K ⎝ o ⎠ v1 v2 y1 × y2 = K vi (1 − ε ) −1 × ε = K K= ε 1− ε = 1.181. no = 1.974. Tc. Pc.974.2= 36.2 = 408. Pc.1 K. species 2 ≡iso-C4H10. P = 15 bar. ε = 0.96 Tr . 2 = 425 = 1.1 Using Equation (3.083 − Using Equation (3. ε = 0. 2 = P Pr . Tc. Pc.395 {− 0.172 = −0.336 y2= ε = 0.395 37.200.1 K. 2 × Tc .1 = P Pr .48 Tr .041 408. 2 × Pc .1 × Pc .339 + 0.2 Using Equation (11.2 = 408.033 14. 2 = 15 = 0.1 = T 15 = 0.1 K.664 y1= 1 .1 Using Equation (3.1= 37.65) to determine Bo 0.1 = 425 =1 425.66) to determine B1 B11 = 0.1= 425.664 (b) For an ideal solution: −v ⎛ P ⎞ ∏i ( yiφi ) = ⎜⎝ P o ⎟⎠ K vi For species 1 ≡n-C4H10: ω1 = 0.181. ⎡ 0.48 bar Pr . Tc.68) to determine φ1.411 36.96 bar Pr .422 = −0.2= 36. Pc.2(−0. 2 = T Tr .6 B1o = 0.139 − 0.CHE 3161 (JUN 11) Thus.339 11.65) to determine Bo Page 11 of 12 .1 = Tr .1 × Tc .872 ⎣ 1 ⎦ φ1 = exp⎢ For species 2 ≡iso-C4H10: ω2 = 0.033)}⎤⎥ = 0. 6 Using Equation (3.ε = 0.883 ⎣ 1.974 Thus. ε = 0.661 Page 12 of 12 .661 y1= 1 .313 1.872)]−1[ε (0. ⎡ 0.29 × 10 −3 ) ⎥ = 0.29 × 10 −3 4.CHE 3161 (JUN 11) B2o = 0.974 vi [(1 − ε )(0.0411.139 − 0.422 = −0.2 1.041 ⎦ φ2 = exp⎢ { } −v ⎛ P ⎞ ∏i ( yiφi ) = ⎜⎝ P o ⎟⎠ K ( y1φ1 ) v1 × ( y2φ2 ) v2 = 1.883)]1 = 1.66) to determine B1 B21 = 0.041 Using Equation (11.172 = −6.083 − 0.68) to determine φ2.181(−6.313 + 0.411 ⎤ − 0.339 y2= ε = 0.