Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Thermodynamics of Interfaces And you thought this was just for the chemists...

Similar presentations


Presentation on theme: "1 Thermodynamics of Interfaces And you thought this was just for the chemists..."— Presentation transcript:

1 1 Thermodynamics of Interfaces And you thought this was just for the chemists...

2 2 Terms Intensive Variables  P: pressure   Surface tension  T: Temperature (constant)   Chemical potential Intensive Variables  P: pressure   Surface tension  T: Temperature (constant)   Chemical potential Extensive Variables  S: entropy  U: internal energy  N: number of atoms  V: volume   Surface area Key Concept: two kinds of variables Intensive: do not depend upon the amount (e.g., density) Extensive: depend on the amount (e.g., mass)

3 3 Phases in the system Three phases  liquid; gaseous; taut interface Subscripts  ‘’ indicates constant intensive parameter  ‘g’; ‘l’; ‘a’; indicate gas, liquid, and interface Gaseous phase ‘g’ Interface phase ‘a’ Liquid phase ‘l’

4 4 Chemical Potential  refers to the per molecule energy due to chemical bonds. Since there is no barrier between phases, the chemical potential is uniform   g =  a =  l =  [2.21]  refers to the per molecule energy due to chemical bonds. Since there is no barrier between phases, the chemical potential is uniform   g =  a =  l =  [2.21]

5 5 Fundamental Differential Forms We have a fundamental differential form (balance of energy) for each phase  TdS g = dU g + P g dV g -  dN g (gas)[2.22]  TdS l = dU l + P l dV l -  dN l (liquid)[2.23]  TdS a = dU a -  d  (interface)[2.24] The total energy and entropy of system is sum of components  S = S a + S g + S l [2.25]  U = U a + U g + U l [2.26] We have a fundamental differential form (balance of energy) for each phase  TdS g = dU g + P g dV g -  dN g (gas)[2.22]  TdS l = dU l + P l dV l -  dN l (liquid)[2.23]  TdS a = dU a -  d  (interface)[2.24] The total energy and entropy of system is sum of components  S = S a + S g + S l [2.25]  U = U a + U g + U l [2.26]

6 6 How many angels on a pin head? The inter-phase surface is two- dimensional, The number of atoms in surface is zero in comparison to the atoms in the three-dimensional volumes of gas and liquid:  N = N l + N g [2.27] The inter-phase surface is two- dimensional, The number of atoms in surface is zero in comparison to the atoms in the three-dimensional volumes of gas and liquid:  N = N l + N g [2.27]

7 7 FDF for flat interface system If we take the system to have a flat interface between phases, the pressure will be the same in all phases (ignoring gravity), which we denote P The FDF for the system is then the sum of the three FDF’s  TdS = dU + P dV -  dN -  d  (system)[2.27] If we take the system to have a flat interface between phases, the pressure will be the same in all phases (ignoring gravity), which we denote P The FDF for the system is then the sum of the three FDF’s  TdS = dU + P dV -  dN -  d  (system)[2.27]

8 8 Gibbs-Duhem relationship For an exact differential, the differentiation may be shifted from the extensive to intensive variables maintaining equality. TdS = dU + P dV -  dN -  d  (system) S a dT =  d  [2.29] or Equation of state for the surface phase (analogous to Pv = nRT). Relates temperature dependence of surface tension to the magnitude of the entropy of the surface. For an exact differential, the differentiation may be shifted from the extensive to intensive variables maintaining equality. TdS = dU + P dV -  dN -  d  (system) S a dT =  d  [2.29] or Equation of state for the surface phase (analogous to Pv = nRT). Relates temperature dependence of surface tension to the magnitude of the entropy of the surface.

9 9 Laplace’s Equation from Droplet in Space Now consider the effect of a curved air-water interface.  P g and P l are not equal   g =  l =  Fundamental differential form for system TdS = dU + P g dV g + P l dV l -  (dN g +dN l ) -  d  [2.31] Now consider the effect of a curved air-water interface.  P g and P l are not equal   g =  l =  Fundamental differential form for system TdS = dU + P g dV g + P l dV l -  (dN g +dN l ) -  d  [2.31]

10 10 Curved interface Thermo, cont. Considering an infinitesimally small spontaneous transfer, dV, between the gas and liquid phases  chemical potential terms equal and opposite  the total change in energy in the system is unchanged (we are doing no work on the system)  the entropy constant TdS = dU + P g dV g + P l dV l -  (dN g +dN l ) -  d  [2.31] Holding the total volume of the system constant, [2.31] becomes  (P l - P g )dV -  d  = 0[2.32] Considering an infinitesimally small spontaneous transfer, dV, between the gas and liquid phases  chemical potential terms equal and opposite  the total change in energy in the system is unchanged (we are doing no work on the system)  the entropy constant TdS = dU + P g dV g + P l dV l -  (dN g +dN l ) -  d  [2.31] Holding the total volume of the system constant, [2.31] becomes  (P l - P g )dV -  d  = 0[2.32]

11 11 Droplet in space (cont.) where P d = P l - P g We can calculate the differential noting that for a sphere V = (4  r 3 /3) and  = 4  r 2 [2.34] which is Laplace's equation for the pressure across a curved interface where the two characteristic radii are equal (see [2.18]). where P d = P l - P g We can calculate the differential noting that for a sphere V = (4  r 3 /3) and  = 4  r 2 [2.34] which is Laplace's equation for the pressure across a curved interface where the two characteristic radii are equal (see [2.18]).

12 12 Simple way to obtain La Place’s eq.... Pressure balance across droplet middle  Surface tension of the water about the center of the droplet must equal the pressure exerted across the area of the droplet by the liquid  The area of the droplet at its midpoint is  r 2 at pressure P d, while the length of surface applying this pressure is 2  r at tension  P d  r 2 = 2  r  [2.35] so P d =2  /r, as expected Pressure balance across droplet middle  Surface tension of the water about the center of the droplet must equal the pressure exerted across the area of the droplet by the liquid  The area of the droplet at its midpoint is  r 2 at pressure P d, while the length of surface applying this pressure is 2  r at tension  P d  r 2 = 2  r  [2.35] so P d =2  /r, as expected

13 13 Vapor Pressure at Curved Interfaces Curved interface also affects the vapor pressure  Spherical water droplet in a fixed volume  The chemical potential in gas and liquid equal  l =  g [2.37] and remain equal through any reversible process d  l = d  g [2.38] Curved interface also affects the vapor pressure  Spherical water droplet in a fixed volume  The chemical potential in gas and liquid equal  l =  g [2.37] and remain equal through any reversible process d  l = d  g [2.38]

14 14 Fundamental differential forms As before, we have one for each bulk phase  TdS g = dU g + P g dV g -  g dN g (gas)[2.39]  TdS l = dU l + P l dV l -  l dN l (liquid)[2.40] Gibbs-Duhem Relations:  S g dT = V g dP g - N g d  g (gas)[2.41]  S l dT = V l dP l - N l d  l (liquid)[2.42] As before, we have one for each bulk phase  TdS g = dU g + P g dV g -  g dN g (gas)[2.39]  TdS l = dU l + P l dV l -  l dN l (liquid)[2.40] Gibbs-Duhem Relations:  S g dT = V g dP g - N g d  g (gas)[2.41]  S l dT = V l dP l - N l d  l (liquid)[2.42]

15 15 Some algebra…  S g dT = V g dP g - N g d  g (gas)[2.41]  S l dT = V l dP l - N l d  l (liquid)[2.42] Dividing by N g and N l and assume T constant  v g dP g = d  g (gas)[2.43]  v l dP l = d  l (liquid)[2.44] v indicates the volume per mole. Use d  g = d  l [2.38] to find  v g dP g = v l dP l [2.45] which may be written (with some algebra)  S g dT = V g dP g - N g d  g (gas)[2.41]  S l dT = V l dP l - N l d  l (liquid)[2.42] Dividing by N g and N l and assume T constant  v g dP g = d  g (gas)[2.43]  v l dP l = d  l (liquid)[2.44] v indicates the volume per mole. Use d  g = d  l [2.38] to find  v g dP g = v l dP l [2.45] which may be written (with some algebra)

16 16 Using Laplace’s equation... or since v l is four orders of magnitude less than v g, so suppose (v g - v l )/v l  v g /v l Ideal gas, P g v g = RT, [2.49] becomes or since v l is four orders of magnitude less than v g, so suppose (v g - v l )/v l  v g /v l Ideal gas, P g v g = RT, [2.49] becomes

17 17 Continuing... Integrated from a flat interface (r =  ) to that with radius r to obtain where P  is the vapor pressure of water at temperature T. Using the specific gas constant for water (i.e., = R/v l ), and left- hand side is just P d, the liquid pressure: Integrated from a flat interface (r =  ) to that with radius r to obtain where P  is the vapor pressure of water at temperature T. Using the specific gas constant for water (i.e., = R/v l ), and left- hand side is just P d, the liquid pressure:

18 18 Psychrometric equation  Allows the determination of very negative pressures through measurement of the vapor pressure of water in porous media. For instance, at a matric potential of - 1,500 J kg -1 (15 bars, the permanent wilting point of many plants), P g /P  is 0.99.  Allows the determination of very negative pressures through measurement of the vapor pressure of water in porous media. For instance, at a matric potential of - 1,500 J kg -1 (15 bars, the permanent wilting point of many plants), P g /P  is 0.99.

19 19 Measurement of P g /P   A thermocouple is cooled while its temperature is read with a second thermocouple.  At the dew point vapor, the temperature decline sharply reduces due to the energy of condensation of water.  Knowing the dew point T, it is straightforward to obtain the relative humidity  see Rawlins and Campbell in the Methods of Soil Analysis, Part 1. ASA Monograph #9, 1986  A thermocouple is cooled while its temperature is read with a second thermocouple.  At the dew point vapor, the temperature decline sharply reduces due to the energy of condensation of water.  Knowing the dew point T, it is straightforward to obtain the relative humidity  see Rawlins and Campbell in the Methods of Soil Analysis, Part 1. ASA Monograph #9, 1986

20 20 Temperature Dependence of   Often overlooked that all the measurements we take regarding water/media interactions are strongly temperature dependent.  Surface tension decreases at approximately one percent per 4 o C!  Often overlooked that all the measurements we take regarding water/media interactions are strongly temperature dependent.  Surface tension decreases at approximately one percent per 4 o C!


Download ppt "1 Thermodynamics of Interfaces And you thought this was just for the chemists..."

Similar presentations


Ads by Google