Download Open Systems - Thermodynamics and Statistical Mechanics - Lecture Slides and more Slides Thermodynamics in PDF only on Docsity! Thermodynamics and Statistical Mechanics Open Systems and Chemical Potential Docsity.com Diffusive Interaction •If particles are added to a system, the energy of the system can change, because of the chemical potential of the added particles in their new environment. A term is needed to account for this effect. •dU = TdS – PdV + µdn Docsity.com Chemical Potential J/kmole 10138.3 kmoles 102.0 J 627.6 J 627.6 J/kcal) kcal)(4184 (0.15 K) )(0.15Kkgkcal kg)(1.0 0.1( 7 5- 1-1- ×−= × −= ∆ −= == ⋅⋅= ∆= µ µ n Q TmcQ P Docsity.com Chemical Potential eeV/molecul 326.0 J/eV 106.1 J/molecule 10213.5 kmolemolecules/ 1002.6 J/kmole 10138.3 19 20 26 7 −= × ×− = × ×− = − − µ µ µ Docsity.com Chemical Potential VSnSnV VSnSnV n U V UP S UT dn n UdV V UdS S UdU dnPdVTdSdU ,,, ,,, ∂ ∂ = ∂ ∂ −= ∂ ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ = +−= µ µ Docsity.com Equilibrium Conditions Consider two systems, A1 and A2, that can interact thermally, mechanically, and diffusively. For either system, n T V T PU T S nVPSTU ∆−∆+∆=∆ ∆+∆−∆=∆ µ µ 1 or Docsity.com Equilibrium Conditions The change in entropy for the combined system is given by, ∆S0 = ∆S1 + ∆S2, where ∆S1 and ∆S2 are given by the expression on the previous slide. Then, 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 0 1 1 n T V T PU T n T V T PU T S ∆−∆+∆+ ∆−∆+∆=∆ µ µ Docsity.com Equilibrium Conditions Since the two systems are interacting only with each other, we have, ∆U2 = – ∆U1 ∆V2 = – ∆V1 ∆n2 = – ∆n1 Docsity.com Equilibrium Conditions 21 2 2 1 1 21 2 2 1 1 21 21 1 2 2 1 1 1 2 2 1 1 1 21 so ,0 so ,0 so ,011 011 µµµµ µµ ==− ==− ==− =∆ −−∆ −+∆ − TT PP T P T P TT TT n TT V T P T PU TT Docsity.com Approach to Equilibrium •To examine the approach to equilibrium, we shall replace ∆U1 by Q1. To do so, use •∆U1 = Q1 – P1∆V1 + µ1∆n1. Then, 1 2 21 1 2 21 1 21 1 2 2 1 1 1 2 2 1 1 1 21 0 )()(11 11 n T V T PPQ TT n TT V T P T PU TT S ∆ − −∆ − + −= ∆ −−∆ −+∆ −=∆ µµ µµ Docsity.com Approach to Equilibrium •∆S0 > 0, so each term must be positive. 1 2 21 1 2 21 1 21 0 )()(11 n T V T PPQ TT S ∆−−∆−+ −=∆ µµ If T1 > T2 , ∆Q1 < 0 If P1 > P2 , ∆V1 > 0 If µ1 > µ2 , ∆n1 < 0 Docsity.com
