Capillary Action

1. Adhesive forces increase the surface tension of the liquid. The surface tension reduces the surface area of the liquid, thereby pulling the liquid up the tube. The liquid climbs until adhesive and cohesive forces are balanced by gravity on the liquid Why soda or water rise up in a straw? How Plants and trees get water and dissolved nutrients from the soil? Capillary Action The rise of liquids up in narrow tubes is called capillary action

2. a. • Both viscosity and surface tension decrease with increasing temperature. • Viscosity increases with increasing temperature while surface tension decreases. • Viscosity decreases with increasing temperature while surface tension increases. • Both viscosity and surface tension increase with increasing temperature.

3. a. • Both viscosity and surface tension decrease with increasing temperature. • Viscosity increases with increasing temperature while surface tension decreases. • Viscosity decreases with increasing temperature while surface tension increases. • Both viscosity and surface tension increase with increasing temperature.

4. b. • Viscosity increases as intermolecular forces increase while surface tension decreases. • Viscosity decreases as intermolecular forces increase while surface tension increases. • Both viscosity and surface tension increase as intermolecular forces increase. • Both viscosity and surface tension decrease as intermolecular forces increase.

5. The polarity of the substance is necessary to determine which force is most important. • Viscosity and surface tension are not related to adhesive and cohesive forces. • adhesive forces • cohesive forces

6. The polarity of the substance is necessary to determine which force is most important. • Viscosity and surface tension are not related to adhesive and cohesive forces. • adhesive forces • cohesive forces

7. 4) Phase Changes

8. Energy Changes Associated with Changes of State • The heat added to the system at the melting and boiling points goes into pulling the molecules farther apart from each other. • The temperature of the substance does not change during the phase change. Heating Curve for 1.00 mol of water at constant pressure of 1 atm (Ti = -25 oC, Tf = 125 oC)

9. “C”: is the heat capacity; “s” : is the specific heat; DT = Tf – Ti unit depends on the unit of “C” or “s”

10. Energy Changes Associated with Changes of State • Heat of Fusion or Enthalpy of fusion ( ): Energy required to change a solid at its melting point to a liquid. • Heat of Vaporization or Enthalpy of vaporization ( ): Energy required to change a liquid at its boiling point to a gas. • Heat of Sublimation or Enthalpy of Sublimation ( ): Energy required to change a solid at its melting point to a gas.

11. evaporation, endothermic • melting (or fusion), endothermic • sublimation, endothermic • melting (or fusion), exothermic

12. evaporation, endothermic • melting (or fusion), endothermic • sublimation, endothermic • melting (or fusion), exothermic

13. Give it some thoughts • We use cubes of ice to cool water, how this works? • We feel cool when we step out of a swimming pool or a warm shower. Why? • Refrigeration, How?

14. Sample Exercise Calculating DH fro Temperature and Phase Changes Calculate the enthalpy change upon converting 1.00 mole of ice at -25 oC to water vapor (steam) at 125 oC under a constant pressure of 1 atm. The specific heats of ice, water, and steam are 2.09, 4.18, 1.84 J/g.K, respectively. For water DHfus = 6.01 and DHvap = 40.67 kJ/mol Note: DHvap >> DHfus Why?

15. Strategy: Consider both temperature and phase changes AB: -25 to 0 oC (solid) BC: 0 (solid) to 0 oC (liquid) CD: 0 (liquid) to 100 oC (liquid) DE: 100 (liquid) to 100 oC (vapor) EF:100 (vapor) to 125 oC (liquid) s= 2.09 (ice), 4.18 (liq), 1.84 (vap) J/g.K, DHfus = 6.01 and DHvap = 40.67 kJ/mol

16. Practice Exercise What is the enthalpy change during the process in which 100.0 g of water at 50.0 oC is cooled to ice at -30.0 oC? Use the specific heats and enthalpies of phase change given in the previous example. Answer: -20.9 kJ – 33.4 kJ -6.27 kJ = -60.6 kJ

17. What Happens? Pgas = 0 vacuum ? Imagine t = 0 After some time (t)

18. If no molecules exist in the gas phase, there is zero pressure At any temperature, some molecules can escape from the surface of a liquid by evaporation vacuum As more molecules escape the liquid, the pressure exerted by the vapor in the space above the liquid will begin to increase After some time After some to time, liquid and vapor reach a state of dynamic equilibrium: liquid molecules evaporate and vapor molecules condense at the same rate. The gas pressure remains constant as long as the temperature remains constant. This constant pressure is called the“vapor pressure” Equilibrium vapor pressure over a liquid. At equilibrium molecules enter and leave the liquid at the same rate

19. Depends on the strength of the attractive forces The average K.E of surface molecules > energy needed to escape the surface, molecules evaporate The average K.E of surface molecules < energy needed to escape the surface, molecules do not evaporate The weaker the attractive forces the larger is the vapor pressure As the temperature increases the vapor pressure increases Explanation of Vapor Pressure on the Molecular Level The distribution of average kinetic energies of surface molecules as function of temperature

20. Value of boiling point depends on the external pressure. External pressure higher than 1 atm causes the water to boil at a temperature higher than 100 oC In pressure cooker, water boils at temp> 100 oC and therefore food gets cooked fast Volatility In an open container, liquids (such as gasoline) that evaporate readily are said to be volatile Vapor Pressure, and Temperature • The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure. • The normal boiling point is the temperature at which its vapor pressure is 760 torr (1 atm).

21. [2] , At P1 and T1 , At P2 and T2 [3] Subtracting Eq. [3] from Eq. [2], we get If we know (P2, T2); (P1, T1) we can calculate DHvap The Clausius-Clapeyron Equation [1] , At any P and T Slope = Dlnp / D(1/T) lnp2 Dlnp lnp1 D(1/T) 1/T2 1/T1

22. Phase Diagrams Phase diagrams display the state of a substance at various pressures and temperatures and the places where equilibria exist between phases. Let us have a closer look

23. Critical point (B): the temperature and the pressure are called the critical temperature and the critical pressure. It is the highest temp. and pressure at which a liquid can exist. Above the critical point the liquid and vapor are indistinguishable from each other. The greater the attractive force the higher is the critical point Solid-Liq equilibrium Liq-gas equilibrium Solid-liq-gas equilibrium Solid lines: are called the co-existence curves or the interface lines. Each point along the AB, AD, and AC lines, is the B.P, M.P, and sublimation point at a given pressure, respectively. The normal B.P, M.P, and sublimation points are those along AB, AD, and AC at pressure of 1 atm, respectively.

24. Phase Diagram of Water and Carbon Dioxide No normal B.P or normal F.P Normal F.P The critical point P=5.11 atm T=-56.4 oC Carbon dioxide cannot exist in the liquid state at pressures below 5.11 atm; when heated at 1 atm it does not melt but sublimes. The critical point P=217.7 atm T=374.4 oC (T.P) (T.P) (T.P) Normal B.P Normal sublimation point Note the high critical temperature and critical pressure: these are due to the strong hydrogen bonding between water molecules. The low critical temperature and critical pressure for CO2 make supercritical CO2 a good solvent for extracting nonpolar substances (such as caffeine). The slope of the solid–liquid line is negative. This means that M.P decreases with increasing pressure (liq is more compact than ice