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Phase Transitions Physics 313 Professor Lee Carkner Lecture 22
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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
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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
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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
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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 )
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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
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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
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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?
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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
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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)
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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
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Application: Mixing Chamber In general, the following holds for a mixing chamber: Mass conservation: Energy balance: Only if Q = W = pe = ke = 0
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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
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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:
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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:
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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
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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
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Equilibrium Consider the equilibrium case ( 1 /T 1 ) = ( 2 /T 2 ) Chemical potentials are equal in equilibrium
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