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Continuation of CHAPTER 11 CHEM 1212. Prentice Hall © 2003Chapter 11 Vapor Pressure Vapor pressure is the pressure exerted when the liquid and vapor are.

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Presentation on theme: "Continuation of CHAPTER 11 CHEM 1212. Prentice Hall © 2003Chapter 11 Vapor Pressure Vapor pressure is the pressure exerted when the liquid and vapor are."— Presentation transcript:

1 Continuation of CHAPTER 11 CHEM 1212

2 Prentice Hall © 2003Chapter 11 Vapor Pressure Vapor pressure is the pressure exerted when the liquid and vapor are in dynamic equilibrium. Dynamic Equilibrium - the point when as many molecules escape the surface as strike the surface.

3 Prentice Hall © 2003Chapter 11 Explaining Vapor Pressure on the Molecular Level Vapor Pressure

4 If equilibrium is never established then the liquid evaporates. Volatile substances evaporate rapidly.  vapor pressure of substance is high The higher the temperature, the higher the average kinetic energy, the faster the liquid evaporates. Vapor Pressure and Volatility

5 Chapter 11

6 Liquids boil when the external pressure equals the vapor pressure. Temperature of boiling point increases as pressure increases. Vapor Pressure and Boiling Pt.

7 Two ways to get a liquid to boil: increase temperature or decrease pressure. –Pressure cookers operate at high pressure. At high pressure the boiling point of water is higher than at 1 atm. Therefore, there is a higher temperature at which the food is cooked, reducing the cooking time required. Normal boiling point is the boiling point at 760 mmHg (1 atm). Vapor Pressure and Boiling Pt.

8 Enthalpy of Vaporization aka heat of vaporization (  H vap ) Is the amount of heat needed to convert a liquid to a vapor at its normal boiling point

9 What you have learned so far……. Some compounds are volatile (evaporates easily) because… (a) the molecules are held together by weak intermolecular forces  therefore it doesn’t take much E to break the bonds  and so they evaporate easily. Compounds that are volatile have lower boiling points.

10 Chapter 11

11 Clausius-Clapeyron Equation ln P = -  H vap + C RT Where: T = absolute temperature R = gas constant (8.314 J/K-mol)  H = heat of vaporization P = vapor pressure C = constant

12 IDEAL GAS CONSTANT R = 0.08206 liter. atm/K. Mole OR R = 8.314 J/K. mole

13 Vapor Pressure and Boiling Pt. Sample Problem If the normal boiling point of water is 100. o C, what will be its boiling point at 735 torr? The heat of vaporization of water is 40.67 kJ/mole.

14 Clausius-Clapeyron Equation To determine the effect of changing temperature or vapor pressure, use the following equation: ln [P 1 /P 2 ] = -  H vap [1/T 1 - 1/T 2 ] R

15 Clausius-Clapeyron Equation Or to remove the negative sign: ln [P 1 /P 2 ] =  H vap [1/T 2 - 1/T 1 ] R

16 Clausius-Clapeyron Equation So your very first equation in Chem 1212 is: ln [P 1 /P 2 ] =  H vap [1/T 2 - 1/T 1 ] R

17 Problem CCl 4 has a vapor pressure of 213 torr at 40 o C and 836 torr at 80 o C. What is the heat of vaporization of CCl 4 ?

18 Extra Problem The melting point of potassium is 63.2 o C. Molten potassium has a vapor pressure of 10.0 torr at 443 o C and a vapor pressure of 400.0 torr at 708 o C. A.) Calculate the heat of vaporization of liquid potassium. B.) Calculate the normal boiling point of potassium. C.) Calculate the vapor pressure of liquid potassium at 100 o C.


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