1. Thermodynamics of Water - 1

2. Take notes!

3. 1. Quick review from 121A To solve any thermo problem for dry air… Consider whether the Gas Law alone will help!

5. review … If that’s not enough… Consider whether the First Law of Thermodynamics will help (and maybe the Gas Law)

7. review … And if that’s not enough… Consider whether the Second Law of Thermodynamics will help

9. review … We look at various special cases: 1)Isothermal processes (dT = 0) 2)Isobaric processes (dp = 0) 3)Isosteric processes (d  = 0) 4)Adiabatic processes (d  = 0)

10. TD Diagrams For an ideal gas, we have three unknowns: pTpT But the Eqn. of State allows us to reduce to two: p,  or p,Tor ,T

11. TD Diagrams We often represent processes on diagrams with axes (p,  ) or (p,T)or( ,T) The (p,  ) diagram is often used CAR p. 234 Fig. VIII-8

12. TD Diagrams Shows a cyclic process Area enclosed = work done during process Area =  pd  =  w

13. TD Diagrams CAR p. 248-249 Figs. VIII-12-14 Show isothermal, isosteric, isobaric processes

14. TD Diagrams Tsonis p. 39 (handout) For a p  -diagram: Isotherms are “equilateral hyperbolas” Adiabats are too, but are steeper (Fig. a)

15. TD Diagrams Note… The 2 nd Law is often derived via analysis of the Carnot Cycle – a cycle involving two adiabatic processes and two isothermal processes (p. 260, Fig. VIII-19).

16. TD Diagrams For a pT-diagram: Isochores (constant  ) are straight lines (Fig. b) Adiabats are “equilateral hyperbolas”

17. TD Diagrams In p  T-space: Fig. d Adiabats are “equilateral hyperbolas”

18. WATER!!!

19. Water Water is hugely important and interesting! We will look at: Vapor alone (briefly…it’s a gas) Liquid alone (briefly) Ice alone (briefly)

20. Water We will look at: The coexistence of water in two states. Mainly… Vapor & liquid Liquid and ice

21. Water Concepts… Vapor pressure (e) Saturation vapor pressure (e s ) – Actually … e sw = SVP over water Latent heats

22. Water And … The Clausius-Clapeyron Equation – Gives the relationship for e s (T)

23. Water Consider pure water … CAR p. 274 Figs. IX-1-3 show the p  T- surface for water. Find the three phases: solid, liquid, vapor/gas

24. Water Note the projections/slices in the pT-plane (Fig. 2) and the p  -plane (3). Note the triple line/triple point of water: the temperature and pressure at which all three phases coexist (T t, p t ).

25. Water In the pT-plane there is the triple point. p t = 6.11 mb T t = 0  C = 273 K Note the critical point where  (vapor) =  (liquid)!

26. Water p c = 220,598 mb (yikes!) T t = 647 K T > T t means we have a gas T < T t means we have a vapor

27. Water vapor alone… Water vapor is an ideal gas and obeys its Equation of State: “p  =RT” - which we write as: where e = vapor pressure (and everything else should be obvious!) e  v =R v T

28. Water vapor alone… At saturation, e  e s – the saturation vapor pressure (SVP) - in which case: e s  v =R v T

30. Liquid or ice … For liquid water:  w = 10 -3 m 3 /kg  density For ice:  w = 1.091 x 10 -3 m 3 /kg  density

31. Specific heats… VaporC pv 1846 VaporC vv 1384.5 C pv - C pv = R v LiquidCwCw 4187 Solidcici 2106

32. 2. Phase changes Look at CAR p. 275 Fig. IX-3… Imagine starting with vapor @ T o in a piston Compress the vapor (p ,   )  point “b” at which vapor is saturated (e  e s ) Further reduction in volume  liquid appears – the two phases coexist

33. Phase changes Continued reduction in volume occurs @ constant pressure and constant temperature (line “bc”) but not constant volume (since the vapor-liquid mix is NOT an ideal gas) At “c”, all water is in liquid form Now we need much larger pressure increases to get volume changes

34. Phase changes In going from “b” to “c”, the vapor is compressed and work is done on the vapor. Heat is liberated during the process … latent heat (latent heat of condensation here) In this case in the atmosphere, the surrounding “air” would be warmed by this – not the water substance.

35. Phase changes Whenever water goes to a state of reduced molecular energy, latent heat is released Thus, latent heat is heat released (or absorbed) during the process: L =  Q p

36. Phase changes Remember enthalpy? Specific enthalpy is: h = u + p  With: dh = c p dTcalled sensible heat

37. Phase changes h = u + p  dh = du + pd  +  dp dh = du + pd  since isobaric! dh =  qfrom the 1 st Law! Thus: L =  Q p  l =  q p = dh Latent heat  enthalpy change

38. Latent heats… Vaporization (vapor-liquid) l v or l wv 2.5 x 10 6 J/kg Fusion (liquid-ice) l i or l iw 3.34 x 10 5 J/kg Sublimation (vapor-ice) l i or l iv 2.5 x 10 6 J/kg

39. Latent heats Latent heats are “reversible” – Example: l wv = - l vw Also they are “additive” – Example: l iv = l iw + l wi

40. Clausius-Clapeyron Unsaturated water vapor: e  v =R v T Saturated water vapor:e s = e s =(T) We’d like to know how e s varies with T

41. Clausius-Clapeyron Following CAR, we use Gibbs (free) energy to derive a relationship. p.268…g = u – Ts + p  s = entropy

42. Clausius-Clapeyron So:dg = du –Tds – sdT + pd  +  dp Total work is:  w = Tds – du(VII-82) By substitution:  w tot = – dg – sdT + pd  +  dp (VII-100) Here, pd  = expansion work And, – dg – sdT +  dp = other work done during in the process

43. Clausius-Clapeyron Importantly, for an isothermal, isobaric process: dg = 0 Gibbs free energy is unchanged

44. Clausius-Clapeyron In a phase change (say vapor  liquid), temperature & pressure are unchanged. See Fig. IX-1 Hence:dg = 0 Or:g 1 = g 2 (e.g., g vapor = g liquid )

45. Clausius-Clapeyron From Fig. IX-4 At one (p,T): g 1 = g 2 At another (p,t) = (p+dp,T+dT) (g+dg) 1 = (g+dg) 2 Thus: dg 1 = dg 2

46. Clausius-Clapeyron Again in a phase change (say vapor  liquid), the only work done is expansion work. So from slide 42,  w tot = pd  And VII-100 gives: – dg – sdT + pd  +  dp = pd   – dg – sdT +  dp = 0  dg =  dp – sdTin a phase change

47. Clausius-Clapeyron dg =  dp – sdT Going back to having two values (p,T) and (p+dp,T=dT) – remember we are varying T to determine the variation of e s with T – we have: dg 1 = dg 2  1 dp – s 1 dT =  2 dp – s 2 dT

48. Clausius-Clapeyron   1 dp – s 1 dT =  2 dp – s 2 dT  (  1 -  2 )dp = (s 1 – s 2 )dT  dp = (s 1 – s 2 ) dT (  1 -  2 )

49. Clausius-Clapeyron From slide 37: T(s 1 – s 2 ) = l =  q p = dh  dp = l dT T(  1 -  2 ) Clausius- clapeyron equation

50. Clausius-Clapeyron Special case: vaporization Then:p  e s And thus:

51. Clausius-Clapeyron If we know how l wv varies with T, we can integrate! For vapor  liquid: We know:  w (slide 30) We know:  v = R v T/e sw (slide 28)

52. Clausius-Clapeyron And thus:

53. Clausius-Clapeyron If we assume l wv is constant, we can solve: e sw = (e s ) t @ triple point

54. Clausius-Clapeyron CAR also derives expressions for the other processes.