Section 6: Phase Changes

Question 1

an equation of state f(P,V,T)=0 yields

Answer 1

a 3D surface of equilibrium states

Question 2

projections onto 2D graphs

Answer 2

PV graph - isotherm
PT graph - isochores
VT graph - isobars

Question 3

moving between two states on the 2D plots still

Answer 3

involves changes in the 3rd parameter

Question 4

phase changes produce

Answer 4

abrupt gradient changes on the PVT-surface

Question 5

1st order phase changes

Answer 5

melting and boiling point

accompanied by a change in specific volume

Question 6

PVT-surfaces have regions of

Answer 6

single-phase and regions of phase coexistence bounded by saturation lines

Question 7

the saturation lines for liquid and vapour meet at

Answer 7

the critical point

Question 8

for T>Tc

Answer 8

no condensation into a liquid is possible with increasing pressure

Question 9

for T<Tc

Answer 9

an increase in pressure can cause a change of state

Question 10

for T>Tc, gases and liquids are

Answer 10

indistinct

Question 11

natural choice of thermodynamic potential to analyse phase equilibria

Answer 11

gibbs free energy

Question 12

for two phases Gtot=

Answer 12

M1g1+M2g2

Question 13

at equilibrium dGtot=

Answer 13

g1dM1+g2dM2=0 as dgi=0

both dG and di=0

mass conservation for a closed system gives dMtot=dm1+dm2=0

yields g1=g2

Question 14

at equilibrium, the specific gibbs free energies are

Answer 14

equal

can be extended for more than two phases

eg at triple point gsolid=gliquid=gvapour

Question 15

adding the restriction of the phase equilibrium leaves

Answer 15

one degree of freedom along the a line on the PT plot

gradient of this phase boundary is (dP/dT)v

Question 16

for specific quantities the differential form of the Gibbs free energy is given by

Answer 16

dg=dP/ρ -sdT

Question 17

the change in entropy at the phase transition can be related to

Answer 17

a latent heat

since delta Q = T delta S = mL

this leads to the Clausius-Clapeyron equation for first order phase changes

Question 18

clausius-clapeyron equation is also useful as a

Answer 18

directly measurable proof of the 2nd law

Question 19

cc eqn - the volumes V1 and V2 are the volumes

Answer 19

occupied by the same amount of matter in the two phases

Question 20

cc eqn - depending on the units of L, m could be in

Answer 20

kilogram, moles or number of molecules

Question 21

mL is measured in

Answer 21

joules

Question 22

for first order changes there is a volume change associated with

Answer 22

the latent heat, L

most materials expand upon melting (except from eg water)

Question 23

1st order - for an isobaric path from A to B across the melting line we have

Answer 23

(dg/dT)p=-s
d2g/dT2)p=-(ds/dT)p=-cp/T

but along this path s>0 and cp>0 so plot of g(T) has negative slope and curvature

Question 24

at equilibrium the specific gibbs functions for solid and liquid are equal so we expect

Answer 24

the separate curves for liquid and solid to intersect at the phase boundary

Question 25

the equilibrium phase has the lowest value for

Answer 25

the gibbs function the gradient of g therefore has to be discontinuous

Question 26

gradient of g being discontinuous menas that the entropy

Answer 26

must be discontinuous across the phase boundary

Question 27

all first order phase changes have a

Answer 27

discontinuous 1st derivative of the gibbs function g

Question 28

the kinks in the gibbs functions at the transition is associated with the

Answer 28

latent heat for the phase transition

Question 29

across second-order phase transitions, first order derivatives are

Answer 29

continuous S=-(dG/dT)p and V=(dG/dP)T so deltaS=deltaV=0 and no latent heat and no volume change

Question 30

second order derivatives for second order phase transitions

Answer 30

discontinuous eg discontinuities in the bulk moduli, compressibilities and heat capacities

Question 31

third order phase transitions also occur eg

Answer 31

in ferromagnetics or superfluids