Understanding First Order Phase Transitions and Open Mixed Systems in Physics
This lecture by Professor Lee Carkner explores crucial concepts of first order phase transitions, focusing on scenarios where temperature (T) and pressure (P) remain constant, such as boiling water. The presentation dives into the Clausius-Clapeyron equation, emphasizing how changes in molar entropy, volume, and enthalpy relate to phase changes. It further examines control volumes in fluid dynamics, mass conservation in steady flow systems, and energy balances in mixing chambers. The Gibbs function and chemical potentials are also discussed in the context of open mixed systems.
Understanding First Order Phase Transitions and Open Mixed Systems in Physics
E N D
Presentation Transcript
Open Systems -- Part 2 Physics 313 Professor Lee Carkner Lecture 24 1
First Order Phase Transitions • Consider a phase transition where T and P remain constant • e.g. boiling water • Entropy, volume and enthalpy will change • If the molar entropy and volume change, then the process is a first order transition 2
Phase Change • Consider a substance in the middle of a phase change from initial (i) to final (f) phases • n0 is total number of moles • Can write equations for properties as the change progresses as: S = n0(1-x)si + n0xsf • Where x is fraction that has changed 3
Clausius - Clapeyron Equation • Consider the first T ds equation, integrated through a phase change T ds = cV dT + T (dP/dT) dv T (sf - si) = T (dP/dT) (vf - vi) • This can be written: dP/dT = (sf -si)/(vf - vi) • But H = VdP + T ds so the isobaric change in molar entropy is T ds, yielding: dP/dT = (hf - hi)/T (vf -vi)
Phase Changes and the CC Eqn. • The CC equation gives the slope of curves on the PT diagram • e.g. fusion, vaporization and sublimation • The (hf - hi) quantity is the molar latent heat • Amount of energy that needs to be added to change phase 5
Changes in T and P • For small changes in T and P, the CC equation can be written: DP/DT = (hf - hi)/T (vf -vi) • or: DT = [T (vf -vi)/ (hf - hi) ] DP • The CC equation can be used to compute differences in melting T with P 6
Control Volumes • Many engineering applications involve moving fluids • Often we consider the fluid only when it is within a container called a control volume • e.g. car radiator • What are the key relationships for control volumes? 7
Mass Conservation • For a steady flow system, mass is conserved • Rate of mass flow in equals rate of mass flow out (note italics means rate (1/s)) Smin = Smout • For single stream m1 = m2 r1v1A1 = r2v2A2 • where v is velocity, A is area and r is density
Energy of a Moving Fluid • The energy of a moving fluid (per unit mass) is the sum of the internal, kinetic, and potential energies and the flow work wflow = Pv • work needed to move mass • Total energy per unit mass is: q = u + Pv + ke + pe • Since h = u +Pv q = h + ke +pe (per unit mass)
Energy Balance • Rate of energy transfer in is equal to rate of energy transfer out for a steady flow system: Ein = Eout • For a steady flow situation: Sin[Q + W + mq] = Sout [Q + W + mq] • In the special case where Q = W = ke = pe = 0 Sinmh = Soutmh 10
Application: Mixing Chamber • A mixing chamber is where several streams of fluids come together and leave as one • In general, the following holds for a mixing chamber: • Mass conservation: Smin = mout • Energy balance: Sinmh = mouthout • Only if Q = W = pe = ke = 0
Open Mixed Systems • Consider an open system where the number of moles (n) can change • The internal energy now depends on the various n’s (one for each substance) • dU = (dU/dV)dV + (dU/dS)dS + S(dU/dnj)dnj • Where nj is the number of moles of the jth substance 12
Chemical Potential • We can simplify with mj = (dU/dnj) • and rewrite the dU equation as: dU = -PdV + TdS + Smjdnj • The first term is the work • The second term is the heat • The third term is the chemical potential or: dWC = Smjdnj 13
The Gibbs Function • Other characteristic functions can be written in a similar form • Gibbs function dG = VdP - SdT + Smjdnj • For phase transitions with no change in P or T: mj = (dG/dnj) 14
Mass Flow • How does mj involve mass flow? • Consider a divided chamber (sections 1 and 2) where a substance diffuses across a barrier dU = TdS +mdn dS = dU/T -(m/T)dn • The total dS is the sum of dS for each section dS = dU1/T1 -(m1/T1)dn1 + dU2/T2 -(m2/T2)dn2 15
Conservation • For an isolated, insulated system • Sum of dn’s must be zero: dn1 = -dn2 • Sum of internal energies must be zero: dU1 = -dU2 • Substituting into the above dS equation: dS = [(1/T1)-(1/T2)]dU1 - [(m1/T1)-(m2/T2)]dn1 16
Equilibrium • Consider the equilibrium case • No difference in S or T: (m1/T1) = (m2/T2) m1 = m2 • Chemical potentials are equal in equilibrium • T drives heat • P drives work • m drives mass flow 17
