Introductory Chemical Engineering Thermodynamics

By J. Richard Elliott and Carl T. Lira

Chapter 8 Equilibrium in a Pure Fluid

Chapter 8 - Phase Equilibrium in a Pure Fluid 1 Example. in a Variable Volume Cylinder A piston-cylinder contains some and some at -12°C. The piston is forced down a specified distance and the pressure is increased. The vessel is allowed to cool back to its original . What is the final pressure in this vessel? This is a trick question. If there is vapor left, then the pressure will go back to where it started. If not, then we are compressing a liquid and the pressure could become quite high. We would need to compute how high based on an . First, let's examine the case of no change in pressure. Since the temperature and pressure from beginning to end are constant, the change in Gibbs energy of the system from beginning to end must be zero. ⇒ For the whole system G = MLGL+ MV GV dG = MLdGL +MV dGV + GL dML +GV dMV ⇒ Since T and p are constant, dGL =dGV =0. But dML = -dMV 0 = GL -GV or GL =GV . This is a very significant result! If the pressure comes back to the original pressure then GL = GV . If GL≠GV , then the pressure does not come back to the original pressure. But we know that when the pressure does not come back it must be because there is only liquid. In other words,

GL= GV is a constraint for phase equilibrium. If we specify one more constraint (eg. T), then all of our other state properties can be calculated.

Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 2 The Clausius-Clapeyron Equation: V If starting from VLE, and maintaining VLE while changing T, dG =dGL V sat V L sat L Fundamental property relation ⇒ V dP - S dT = V dP - S dT V V V L L L By definition of G: G = H - TS = H - TS = G V L V L ⇒S - S = (H - H )/T vap ()V − L ()()V − L dp = H H dT 1/ RT = H H dT vap ()vap V vap L ()2 V()L Clausius-Clapeyron Equation p T p V − p V 1/ RT RT Z − Z

sat dividing by P ⇒ dP sat ()H V − H L dT ()1/ RT ()H V − H L dT = ()()= () P sat T P satV V − P satV L 1/ RT RT 2 Z V − Z L dP sat  1  − dT − ()H V − H L = dlnP sat ; d  = ⇒ dln P sat = ()d()1/ T P sat  T  T 2 R Z V − Z L

Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 3 Example. Clausius-Clapeyron near or below the point What is the value of the pressure in a piston-cylinder at -12°C with vapor and liquid V L V L propane present? Recall that Z ≈1 and Z ≈0 and (H - H )≈constant. Integrate  P sat  ()H V − H L  1 1  ln  = −  −  1atm  R T Tb  ω ⇒ (P sat ) [-(1+0.152)] For propane, =0.152 Tr =0.7 = Pc *10 = 3.034 atm. at Tr=0.7⇒ T = 259K The CRC lists the boiling temperature of propane as -42°C=231K. ln(3.034/1) = - ∆H/R(1/259 - 1/231) ) -∆H/R = -2372 sat Therefore, P (261K) = exp[-2372(1/261 - 1/231)] = 3.277 atm.

Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 4 Short-cut Estimation of Saturation Properties If ∆Hvap/∆Zvap was constant, then we could recover the simple low-P form. To find sat out, plot ln(P ) vs. 1/T. A straight line means the slope is constant.  Pvap  ∆Η  1 1  0 0 ln   = −  −   P  R∆Ζ  T T  1 1.2 1.4 1.6 1.8 2 -2 123 ref   ref  -1 -4 -2 -6 log(Pr) log(Pr) 1/Tr 1/Tr

Setting Pref = Pc and Tref =Tc , ∆H  Tc Tc  ∆H  1  sat 1 ∆H  1   1  ln P sat = −  −  = 1−  ⇒ log()P = 1−  ≡ A1−  r ∆ ∆ r ∆ R ZTc  T Tc  R ZTc  Tr  2.303 R ZT  Tr   Tr   1  7 @T = 0.7 log P sat ≡ −()ω + 1 = A − 1 = A()− 3 / 7 ⇒ A = ()1+ω r 10 r  .7  3 sat 7 ∴ log (P ) = ()1+ ω ()1− 1 SHORT-CUT EQUATION 10 r 3 Tr sat 7 e.g. For propane P (-12°C) = P *10** ()1+ω ()1− 1 = 3.26 atm c 3 0.7057

Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 5 Example. General Application of the Clausius-Clapeyron equation. Liquid is pumped to a vaporizer as a saturated liquid under a pressure of 1.88 MPa. The butane leaves the exchanger as a wet vapor of 90 percent quality and at practically the same pressure as it entered. From the following information, estimate the load on the vaporizer per gram of butane entering.

Tc =425.2 K; Pc =3.797 MPa; ω=0.193; Solution: Use VP eqn. to find the T at which this takes place ⇒ Tsat = 383 K  v − ig  L ig  H H   H − H  ⇒     PREOS  nRT  = -0.9952;  nRT  = -5.2567; Therefore, Hv = (-0.9952+5.2567)8.314*383 = 13568 J/mol 90% quality→ Q = 0.9*13568 = 12211 J/mol * 1mol/58g = 211 J/g

Note: Alternatively, we could have used the short-cut Clausius-Clapeyron another way sat Clausius-Clapeyron ⇒ ln(P ) = - ∆ Hvap /RT(ZV-ZL ) + B ∆H 7 vap = 2.3025 ()1+ ω Tc = 2726 R∆Z 3 V L ∆Hvap = 2726*R*(Z -Z ) = 2726*8.314*(0.67434-0.07854) = 13503 J/mol

Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 6 II. Generalized Fluid Properties Overview of UNIT II.

1. Calculus of State Functions a) FPR b) Maxwell’s Relations c) P=P(T,V) ⇒ dU(T,V), dS(T,V) ⇒ dH(T,P), dS(T,P) 2. Equations of state: e.g. P = RT/(V-b) - a/V2RT 3. Departure Functions: e.g. (U-Uig)/RT = ∫-T(∂Z/∂T)dρ/ρ e.g. (U-Uig)/RT = -aρ/RT = -aP/ZR2T2 4. Phase Equilibrium in a Pure Fluid: sat 7 e.g. VLE ⇒ log (P /P) = /3(1+ω)(1-1/Tr)

Chapter 8 - Phase Equilibrium in a Pure Fluid Slide 7