Phase Transitions Physics 313 Professor Lee Carkner Lecture 22
Exercise #21 Joule-Thomson Joule-Thomson coefficient for ideal gas = 1/c P [T( v/ T) P -v] ( v/ T) P = R/P = 1/c P [(TR/P)-v] = 1/c P [v-v] = 0 Can J-T cool an ideal gas T does not change How do you make liquid He? Use LN to cool H below max inversion temp Use liquid H to cool He below max inversion temp
First Order Phase Transitions Consider a phase transition where T and P remain constant If the molar entropy and volume change, then the process is a first order transition
Phase Change Consider a substance in the middle of a phase change from initial (i) to final (f) phases Can write equations for properties as the change progresses as: Where x is fraction that has changed
Clausius - Clapeyron Equation Consider the first T ds equation, integrated through a phase change T (s f - s i ) = T (dP/dT) (v f - v i ) This can be written: But H = VdP + T ds, so the isobaric change in molar entropy is T ds, yielding: dP/dT = (h f - h i )/T (v f -v i )
Phase Changes and the CC Eqn. The CC equation gives the slope of curves on the PT diagram Amount of energy that needs to be added to change phase
Changes in T and P For small changes in T and P, the CC equation can be written: or: T = [T (v f -v i )/ (h f - h i ) ] P
Control Volumes Often we consider the fluid only when it is within a container called a control volume What are the key relationships for control volumes?
Mass Conservation Rate of mass flow in equals rate of mass flow out (note italics means rate (1/s)) For single stream m 1 = m 2 where v is velocity, A is area and 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 Total energy per unit mass is: Since h = u +Pv = 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: For a steady flow situation: in [ Q + W + m ] = out [ Q + W + m ] In the special case where Q = W = ke = pe = 0
Application: Mixing Chamber In general, the following holds for a mixing chamber: Mass conservation: Energy balance: Only if Q = W = pe = ke = 0
Open Mixed Systems Consider an open system where the number of moles (n) can change dU = ( U/ V)dV + ( U/ S)dS + ( U/ n j )dn j
Chemical Potential We can simplify with and rewrite the dU equation as: dU = -PdV + TdS + j dn j The third term is the chemical potential or:
The Gibbs Function Other characteristic functions can be written in a similar form Gibbs function For phase transitions with no change in P or T:
Mass Flow Consider a divided chamber (sections 1 and 2 ) where a substance diffuses across a barrier dS = dU/T -( /T)dn dS = dU 1 /T 1 -( /T 1 )dn 1 + dU 2 /T 2 -( /T 2 )dn 2
Conservation Sum of dn’s must be zero: Sum of internal energies must be zero: Substituting into the above dS equation: dS = [(1/T 1 )-(1/T 2 )]dU 1 - [( 1 /T 1 )-( 2 /T 2 )]dn 1
Equilibrium Consider the equilibrium case ( 1 /T 1 ) = ( 2 /T 2 ) Chemical potentials are equal in equilibrium