Introductory Chemical Engineering Thermodynamics By J.R. Elliott and C.T. Lira
The fundamental property relation Piston+Cylinder ⇒ Closed system ⇒ dU =( Q + W ) dt For the heat exchange: Qdt = TdS (macroscopic definition) For the work Wdt = -PdV (W = Fdx) ⇒ dU =TdS - PdV This is the fundamental property relationship. H = U + PV ⇒ dH = dU + d(PV) = dU + PdV + VdP Substitution of the fundamental property relationship: ⇒ dH = (TdS - PdV) + PdV + VdP ⇒ dH = TdS + VdP dA = - SdT - PdV dG = - SdT + VdP We can integrate these expressions to find that A ≡ U - TS Helmholtz free energy G ≡ H - TS Gibbs free energy Elliott and Lira: Chapter 5 – Classical Thermodynamics Slide 1
Example. Enthalpy of H2O above its saturation pressure Determine the enthalpy of H2O at 20 ° C and 100 MPa. Note that values of pressure this high do not appear in the steam tables. Solution The expression for a change in enthalpy should remind us that shaft work for an adiabatic, reversible process ( dS =0) is ∆ H = ∫ VdP . But, VL ≈ constant = 1.003 cm3/g ⇒ ∆ H =1.003(100-.00234)8.314/8.314=100.30 J/g. ⇒ H =83.96 + 100.30 = 184.25 J/g. NOTE: W = ∆ H = 100 J/g ~ P (MPa) Elliott and Lira: Chapter 5 – Classical Thermodynamics Slide 2
Maxwell's Relations By recognizing that any variable is a function of any two other variables, U = U ( S , V ) ⇒ dU = ( ∂ U / ∂ S )V dS + ( ∂ U / ∂ V )S dV But the fundamental property relationship says: dU = T dS - PdV ( ) ( ) ( ) ∂ ∂ ∂ 2 ∂ ∂ ∂ 2 U S V , U S V , U S V , U S V ( , ) = = = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ S V S V V S V S S V V S ( ) ∂ ∂ ∂ ∂ 2 = 2 U S V , U S V ( , ) ( ) ( ) = − = P V T ∂ ∂ ∂ ∂ ∂ ∂ S V S V S V S dH = ( ∂ H / ∂ S ) dS + ( ∂ H / ∂ P )S dP = TdS + VdP ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = 2 2 V T H H H H = = = ⇒ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ S P P S S P P S S P S p p S p S dU = TdS - PdV ⇒ -( ∂ P / ∂ S ) V = ( ∂ T / ∂ V ) S dH = TdS + VdP ⇒ ( ∂ V / ∂ S )p = ( ∂ T / ∂ P ) S MAXWELL'S dA = - SdT - PdV ⇒ ( ∂ P / ∂ T ) V = ( ∂ S / ∂ V ) T RELATIONS dG = - SdT + VdP ⇒ -( ∂ V / ∂ T ) p = ( ∂ S / ∂ P ) T Elliott and Lira: Chapter 5 – Classical Thermodynamics Slide 3
Example. Entropy change vs. T at constant P Evaluate ( ∂ S / ∂ T )p in terms of Cp , Cv , T , P , and V Solution: Recall, dH = TdS + VdP If P is constant, then dP =0. Dividing through by dT , ∂ ∂ ∂ ≡ H H S = But, the definition of Cp is: Cp T ∂ ∂ ∂ T T T p p p ∂ S Cp = Therefore, ∂ T T p Elliott and Lira: Chapter 5 – Classical Thermodynamics Slide 4
Example . Entropy change as a function of T and P Derive a general relation for entropy changes of any fluid with respect to temperature and pressure in terms of Cp, Cv, P, V, T and their derivatives. dS = ( ∂ S / ∂ T )p dT + ( ∂ S / ∂ P ) T dP but ( ∂ S / ∂ T )p = Cp / T as derived above, and Maxwell's Relations show that ( ∂ S / ∂ P ) T = -( ∂ V / ∂ T ) P . ⇒ dS ( T , P )p = Cp dT / T - ( ∂ V / ∂ T )p dP (eqn. 5.35, a useful, general equation) Similarly dS ( T , V ) = Cv dT/T + ( ∂ P / ∂ T ) V dV (eq 5.36) dH ( T , P ) = Cp dT + [ V - T ( ∂ V / ∂ T ) P ] dP (eq 5.37) dU ( T , V ) = Cv dT + [ T ( ∂ P / ∂ T ) V - P ] dV (eq 5.38) Elliott and Lira: Chapter 5 – Classical Thermodynamics Slide 5
II. Generalized Fluid Properties What is so important about the variables Cp, Cv, P, V, T ? As a sample application, we can assume the ideal gas equation of state. V = RT / P ⇒ ( ∂ V / ∂ T ) P = R/P . Substitution ⇒ ∆ S = Cp ln( T 2 / T 1 ) - R ln( P 2 / P 1 ) Applying ∆ S = 0 ⇒ ln( T 2 / T 1 ) = R/Cp ln( P 2 / P 1 ) ⇒ ( T 2 / T 1 ) = ( P 2 / P 1 ) (Cp-Cv)/Cp − γ − γ γ = 1 1 / = 1 / = 1 / T P V P V P P = ⇒ 2 2 2 2 2 2 1 T PV P V P P 1 1 1 1 1 1 2 Once again, we arrive at this familiar relation, but this time, it is easy to recognize the necessary changes for applications to real gases. That is, we must simply replace the PVT relation by a more realistic equation of state. e.g. V = RT /( P+C ) which we may refer to as the “bogus” equation of state. ⇒ ( ∂ V / ∂ T ) P = R/(P+C) Substitution ⇒ ∆ S = Cp ln( T 2 / T 1 ) - R ln[( P 2 + C )/( P 1 +C )] Applying ∆ S = 0 ⇒ ln( T 2 / T 1 ) = R/Cp ln[( P 2 + C )/( P 1 +C )] ⇒ ( T 2 / T 1 ) = [( P 2 + C )/( P 1 +C) ] R/Cp Elliott and Lira: Chapter 5 – Classical Thermodynamics Slide 6
II. Generalized Fluid Properties Triple Product Rule (a.k.a. Chain Rule) Suppose: F = F ( A,B ) then dF = ( ∂ F / ∂ A ) B dA + ( ∂ F / ∂ B ) A dB . Consider what happens when dF = O (ie. at constant F ). Then, ∂ ∂ F A − − ∂ ∂ ∂ ∂ ∂ ∂ F A F A B F = + ⇒ = = A B 0 ∂ ∂ ∂ ∂ ∂ ∂ F B A B B B B F A F ∂ ∂ A F B A But, because C is some fourth property and F is constant. ∂ A ∂ ∂ ∂ ∂ C A A C = = F ∂ ∂ ∂ ∂ B B C B F F F ∂ C F Now the negative sign has disappeared. Did you notice the difference? Elliott and Lira: Chapter 5 – Classical Thermodynamics Slide 7
II. Generalized Fluid Properties Example. Application of the triple product relation. Evaluate ( ∂ S / ∂ V ) in terms of Cp , Cv , T , P , and V . Your answer may include absolute values of S if it not associated with a derivative. Solution: This problem illustrates a typical situation where the triple product rule is necessary because the free energy is held constant. It is easiest to express changes in the free energies as changes in other variables. Applying the triple product rule: ( ∂ S / ∂ V ) A = -( ∂ A / ∂ V ) S /( ∂ A / ∂ S ) V dA = - PdV -SdT ⇒ ( ∂ A / ∂ V ) S = - P - S ( ∂ T / ∂ V ) S and ( ∂ A / ∂ S ) V = 0 - S ( ∂ T / ∂ S ) V ∂ ∂ T P T T − = and 5.35 ⇒ ( ∂ T / ∂ V )S = ∂ ∂ Cv T S Cv V V Substitution ∂ − ∂ S PCv P ⇒ = + ∂ ∂ V ST T A V Elliott and Lira: Chapter 5 – Classical Thermodynamics Slide 8
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