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Thermodynamics of Water - 1

Thermodynamics of Water - 1. Take notes!. 1. Quick review from 121A. To solve any thermo problem for dry air… Consider whether the Gas Law alone will help!. review …. If that’s not enough… Consider whether the First Law of Thermodynamics will help (and maybe the Gas Law). review ….

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Thermodynamics of Water - 1

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  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!

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

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

  6. review … We look at various special cases: • Isothermal processes (dT= 0) • Isobaric processes (dp= 0) • Isosteric processes (d = 0) • Adiabatic processes (d = 0)

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

  8. 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

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

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

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

  12. TD Diagrams Note… • The 2nd 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).

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

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

  15. WATER!!!

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

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

  18. Water Concepts… • Vapor pressure (e) • Saturation vapor pressure (es) • Actually … esw = SVP over water • Latent heats

  19. Water And … • The Clausius-Clapeyron Equation • Gives the relationship for es(T)

  20. 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

  21. 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 (Tt, pt).

  22. Water • In the pT-plane there is the triple point. • pt = 6.11 mb • Tt = 0C = 273 K • Note the critical point where (vapor) = (liquid)!

  23. Water • pc = 220,598 mb (yikes!) • Tt = 647 K • T > Tt means we have a gas • T < Tt means we have a vapor

  24. 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=RvT

  25. Water vapor alone… • At saturation, e  es – the saturation vapor pressure (SVP) - in which case: esv=RvT

  26. Liquid or ice … • For liquid water: • w = 10-3 m3/kg  density • For ice: • w = 1.091 x 10-3 m3/kg  density

  27. Specific heats…

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

  29. 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

  30. 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.

  31. 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 = Qp

  32. Phase changes • Remember enthalpy? • Specific enthalpy is: h = u + p • With: dh = cpdT called sensible heat

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

  34. Latent heats…

  35. Latent heats • Latent heats are “reversible” • Example: lwv= - lvw • Also they are “additive” • Example: liv= liw + lwi

  36. Clausius-Clapeyron • Unsaturated water vapor: ev=RvT • Saturated water vapor: es= es=(T) • We’d like to know how esvaries with T

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

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

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

  40. Clausius-Clapeyron • In a phase change (say vapor  liquid), temperature & pressure are unchanged. • See Fig. IX-1 • Hence: dg = 0 • Or: g1 = g2 (e.g., gvapor= gliquid)

  41. Clausius-Clapeyron • From Fig. IX-4 • At one (p,T): g1 = g2 • At another (p,t) = (p+dp,T+dT) (g+dg)1= (g+dg)2 • Thus: dg1= dg2

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

  43. 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 es with T – we have: • dg1 = dg2 •  1dp – s1dT = 2dp – s2dT

  44. Clausius-Clapeyron •  1dp – s1dT = 2dp – s2dT • (1 -2)dp = (s1 – s2)dT •  dp= (s1 – s2) dT (1 - 2)

  45. Clausius-Clapeyron • From slide 37: T(s1– s2) = l = qp = dh •  dp = l dT T(1 - 2) Clausius-clapeyron equation

  46. Clausius-Clapeyron • Special case: vaporization • Then: p  es • And thus:

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