Chemistry 163B. One-Component. Phase Diagram Basics

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1 Chemistry 163B One-Component Phase Diagram Basics 1

2 qualitative factors in phase changes solid melting liquid freezing/fusion vaporization ENDOTHERMIC liquid gas condensation EXOTHERMIC sublimation solid gas deposition 2

3 vapor pressure over pure liquid (notation) P P * P 0 Gene E&R Raff (many others) 3

4 gas liquid as pressure increases (vary P, const T) 4

5 liquid vapor as heat added (vary T, const P) P T 1 5

6 dg for phase change at constant T,P A () phase A () phase same T,P for each phase dg SdT VdP dn one phases and constant T, P dg dn dn T P ( ) ( ) ( ) ( ), A A A A i i i dn dg dn ( ) ( ) A A ( ) ( ) ( ) T, P A A A dn 6

7 at equilibrium () = () ; is ESCAPING TENDENCY dg ( ) ( ) ( ) T, P A A A ( ) ( ) A A T,P dn at equilibrium dg = 0 dg for spontaneity dg < 0 T,P ( ) ( ) ( ) dn 0 T, P A A A dn 0 dn 0 ( ) ( ) ( ) A A A ( ) ( ) ( ) A A A molecules lost from phase molecules gained by phase ( ) A high low hyper mellow 7

8 Question: can my pressure cooker heat water to 200C without exploding? T T=298 K P =0.032 bar T=473 K P =15.5 =???. bar 8

9 phase equilibrium one-component systems (i.e pure substances) A() A() (T,P) 0 ( ) ( ) A A How can P and T covary to maintain equilibrium? T, P T, P ( ) ( ) T and P covary d ( ) d ( ) T and P covary T, P T, P ( ) ( ) dt, dp T, P T, P T, P T, P ( ) ( ) ( ) ( ) before after 9

10 conditions for remaining at phase equilibrium (one-component), covary T and P dt, dp T, P T, P T, P T, P ( ) ( ) ( ) ( ) d S dt V dp S dt V dp d ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) with ( ) ( ) T T T ( ) ( ) dt dt dt ( ) ( ) P P P ( ) ( ) dp dp dp ( ) ( ) ( ) ( ) S dt V dp S dt V dp ( ) ( ) ( ) ( ) S S dt V V dp 10

11 phase equilibrium (one-component) ( ) ( ) ( ) ( ) S S dt V V dp dp dt phase equilibrium S V = phase change eqn E&R since phase change is an equilibrium (reversible) process dp dt S phase equilibrium H T H TV 11

12 I. application to liquid gas (vapor) or solid gas dp dt phase equilibrium vaporization condensation or sublimation deposition H TV liquid gas (vapor) solid gas V and V are small compared to V V V solid liquid vapor vapor assume ideal gas V V dp dt d ln P dt or s g vap or sub 2 H RT T P H RT PH RT 2 RT P vapor 12

13 I. application to liquid gas (vapor) or solid gas dp dt or s g H PH 2 RT RT T P d ln P dt equilibrium H vap or sub RT 2 Clausius-Clapeyron eqn E&R 13

14 I. application to liquid gas (vapor) or solid gas d ln P dt equilibrium H vap or sub RT 2 for s or gas K P d ln K dt P equilibrium P gas H RT vap or sub 2 P P T H d ln P dt (assume H independent of T 2 RT 2 2 T 1 1 P H ln P R T T application to problems: E&R eqn 8.20 where is vaporization similar for sublimation to get vapor pressure given Tº boiling and H vap : at T = T P P = 1atm o 1 1 bp vapor Pvapor T H vap 1 1 H vap 1 1 ln 1atm R T T R T bp bp T 14

15 I. application to liquid gas (vapor) or solid gas application to problems: atm vap 1 1 to get T boiling when P atm 1atm : P H ln 1atm R T bp T Patm H T vap bp T ln 1 bp 1atm R T bp P H H T atm vap vap bp T ln bp 1atm R R T bp bp H Patm vap R T ln bp 1atm R H vap T bp 1 T RT bp bp P atm 1 ln Hvap 1atm T bp Tbp 15

16 I. application to liquid gas (vapor) or solid gas application to problems: to get T boiling when P atm 1atm : T bp 1 T RT bp bp P atm 1 ln Hvap 1atm Denver: elev=1610m P=0.822 atm T bp P 1atm 1 Tbp T T bp bp ouch Death Valley: elev =-82.5 m, P=1.010atm T bp P 1atm 1 Tbp T T bp bp 16

17 II. application to solid liquid dp dt phase equilibrium H TV T for phase equilibrium at P = 1 atm o melting what is T at other pressures? melting dt V T melting V dp ln P 1atm T H melting T melting H melting T V V ln T melting H melting liquid solid melting P 1atm will increased pressure raise or lower T melting? H melting > 0 (usual) V liquid > V solid (when??) V liquid < V solid T melting increases T melting decreases 17

18 phase rule one-component system (save proof for later) f f= degrees of freedom p=phases simultaneously present 2 variables : T, P (same for each phase) p-1 restrictions: () = () = () =. f: degrees of freedom = (variables-restrictions) f = 2-(p-1) = 3-p f = 3-p 1 phase: T,P vary independently 2 phases present: T and P covary 3 phases present: fixed T and P 18

19 phase diagrams one component: phase vs (P,T) P PHASE T BE[A]WARE: when we study multicomponent phase diagrams the axis variables may not be P,T 19

20 phase diagrams (f=3-p) state or phase as a function of P, T CO 2 high T gas 1 phase, f=2 vary both T,P 20

21 phase diagrams (f=3-p) state or phase as a function of P, T CO 2 lower T liquid 1 phase, f=2 vary both T,P 21

22 phase diagrams (f=3-p) state or phase as a function of P, T CO 2 lower T solid more 22

23 phase diagrams (f=3-p) state or phase as a function of P, T high T gas CO 2 lower T liquid lower T more solid 23

24 phase diagrams (f=3-p) state or phase as a function of P, T low P gas CO 2 raise P liquid raise P more solid 24

25 two-phase equilibrium (p=2) f= f=3-p =1 dp dt equilib H TV 25

26 phase diagrams (f=3-p) liquid solid equilibrium line (melting, freezing or fusion) T s and P s CO 2 for liquidsolid equilibrium 2 phases, f=3-2=1 T and P covary dp dt s H TV fus fus 26

27 phase diagrams (f=3-p) liquid gas equilibrium line (vaporization, condensation) T s and P s CO 2 for liquidgas equilibrium 2 phases, f=3-2=1 T and P covary [select T, then P determined select P, then T determined] dp H H dt TV RT g vap 2 27

28 phase diagrams solid gas equilibrium line (sublimation, deposition) CO 2 2 phases, f=3-2=1 T and P covary T s and P s for solidgas equilibrium dp H H dt TV RT eq 28 sub 2

critical point and triple point Triple point: for a pure substance, there is only one point (value of T and P) where all three phases (solid, liquid, and gas) can simultaneously exist in equilibrium

29 critical point and triple point Triple point: for a pure substance, there is only one point (value of T and P) where all three phases (solid, liquid, and gas) can simultaneously exist in equilibrium Critical point: point (value of T and P) above which liquid and gas become one phase (fluid or supercritical fluid) movie: benzene critical point A B originally from: jchemed.chem.wisc.edu/jcesoft/cca/samples/cca2benzene.html 29

30 why does ice float? H 2 0 is polar and can form hydrogen bonds (macho intermolecular forces) High surface tension and capillarity Hydrogen bonds form very open structure in solid H 2 O (ice) giving ice a lower density than H 2 O liquid. ICE FLOATS!! 30

31 ice bomb!!!! 31

32 remember for CO2 : P increases gas liquid solid CO 2 (s) CO 2 () slope dp dt s dp dt H melt s 0; V V V 0 s dp dt s H TV 0 melt melt ; low P gas raise P liquid (more dense) CO 2 raise P more solid (most dense) 32

33 phase diagram for water remember for CO 2 : P increases gas liquid solid but for H 2 O as P increases: gas solid liquid WHY? dp dt for ice s 0 water slope dp dt s dp dt s H melt ; H melt 0 TV melt dp Vs V Vmelt 0 0 dt s 33

34 ice skater myth Does the weight of an ice skater create a pressure that melts ice to form a liquid groove for skate? 1atm V Vs P atm Tmelt ln 1 ) T H fusion V V 0 pressure ' melts' ice s E&R(3 rd ) Problem P bar for -4º thin blade ; 78kg172lb; -1.5º NO, not even if they are quite weighty! (not enough pressure and further details of water-ice phase diagram) e.g. Rosenberg, Robert (December 2005). "Why is ice slippery?". Physics Today:

Rosenberg, Robert (December 2005). "Why is ice slippery?". Physics Today: 50 54. Figure 1.

melting temperature by only a few degrees.

the ice as skaters move around must account for ice's slipperiness at the wide variety of subzero temperatures

35 Rosenberg, Robert (December 2005). "Why is ice slippery?". Physics Today: Figure 1. An ice skater exerts pressures on the order of a few hundred atmospheres on the ice surface, enough to reduce the melting temperature by only a few degrees. Premelting the development of a liquid-like surface layer at temperatures below freezing and frictional heating of the ice as skaters move around must account for ice's slipperiness at the wide variety of subzero temperatures found in nature. (Ice Skating, by Hy Sandham, 1885, courtesy of the Library of Congress The nature of the liquid-like layer is not clear from experimental measurements, so theorists have tried to clarify the 35 situation.

effects of inert gas (increased total pressure) on vapor pressure E&R sec. 8.7 pure H 2 O at 298K P H20 = 0.

36 effects of inert gas (increased total pressure) on vapor pressure E&R sec. 8.7 pure H 2 O at 298K P H20 = bar 2 H2O increases at P increase P must increase to restore H O( g) 2 ( ) H O 2 P ( ) HO ( ) ( g) Ptotal P ( ) RT ln V dp H2O P P T V total ( ) ( g) P=1 bar [or+ N 2 (g) + O 2 (g)] normal H 2 O in P total =1 bar [H 2 O (g) + N 2 (g) + O 2 (g)] at 298K P H20 = bar 36

37 E&R section 8.7 (effect of inert gas on vapor pressure HW #46*) H 2 O at 300 K P H20= atm add air (inert N 2 + O 2 ) to raise P total =1 atm new P H20= atm (P/P ) H20 =

38 End of Lecture 38

39 triple point triple point: simultaneous equilibrium of gas, liquid solid 3 phases, f=3-p f=3-3=0 T and P fixed 39

40 vary T and P through critical point fluid 40

Chemistry 163B, Winter 2014 Lectures Introduction to Phase Diagrams

Chemistry 163B, Winter 2014 Lectures Introduction to Phase Diagrams Chemistry 63B, Winter 04 qualitative factors in phase changes Chemistry 63B One-Component hase Diagram Basics freezing/fusion vaporization condensation ENDOHERMIC EXOHERMIC sublimation deposition vapor