2. Vapor pressure and liquids Vapor : A gas that exists below its critical point Gas : gas that exists above its critical point ِNote : A gas can not condense in the process If we have some liquid ( say water) in a closed container at some T 1, then after some time, some vapor will exist above the liquid. This vapor will reach equilibrium (with the liquid). The vapor will have a pressure = vapor pressure, p 1 * (at the given temp T 1 ). Note the vapor pressure is the maximum pressure the vapor can attain. ChE 201 Spring 2003/2004 Dr. F. Iskanderani

3. vapor liquid vapor At T 1 Time 1 2 3 100 equilibrium P of vapor = p* at T 1 ChE 201 Spring 2003/2004 Dr. F. Iskanderani

4. Vapor pressure and liquids Now change the temp to a higher temperature T 2. The system will reach equilibrium, and the vapor will have a new vapor pressure, p 2 * > p 1 * Liquid vapor At T 1 At equilibrium, vapor will reach p 1 * Liquid vapor At T 2 At equilibrium, vapor will reach p 2 * ChE 201 Spring 2003/2004 Dr. F. Iskanderani

5. Curve gives all points (T, p*) at which Liquid and Vapor exist in equilibrium. Therefore vapor can exist at any temperature. (Example) ChE 201 Spring 2003/2004 Dr. F. Iskanderani

6. Vapor Liquid solid

7. Change of Vapor pressure with Temperature p* vs T is a curve ( It is not a straight line) A plot of ln p* vs 1/T for moderate temperatures linear Another form of this eq is the Antoine Equation the vapor pressure can be found from tables, charts or empirical equations (the Antoine equation) V –nb + ln p* =m ( 1 T ) + b (See appendix G on page 669) ln p* =(. A. T+C ) + B ChE 201 Spring 2003/2004 Dr. F. Iskanderani

8. Change of vapor pressure with pressure Under normal conditions the effect of P on the vapor pressure, p* is small d(p*) –nb +ndPT T VlVl VgVg = ChE 201 Spring 2003/2004 Dr. F. Iskanderani

9. Liquid Properties Liquid mixtures are more complex than gases P V T behaviour prediction is difficult If we can assume liquids are ideal liquids, then: V avg = V 1 x 1 + V 2 x 2 +.. +.. This eq is good for components with similar structure such as hydrocarbons ChE 201 Spring 2003/2004 Dr. F. Iskanderani

10. Saturation and Equilibrium For a mixture of pure vapor and a non- condensable gas example : water vapor + air Dry air liquid Dry air + water vapor At T 1 Time 0 2 3 100 200 ChE 201 Spring 2003/2004 Dr. F. Iskanderani saturation

11. Water vaporizes until equilibrium at T 1 is reached. At any condition before saturation, the vapor is partially saturated and its partial pressure is < p* At saturation, air is fully saturated with the vapor and the partial pressure of the vapor is = p* Total pressure of gas mixture = p air + p water vapor At saturation P total = p air + p* water vapor When the mixture of gas and vapor is at saturation, we say thast the mixture is at the dew point Q: If we lower the temperature, what will happen? A: The vapor will condense

12. Dew point for a mixture of pure vapor and a non- condensable gas is the temp at which the vapor just starts to condense if cooled at constant pressure Dry air + vapor Water liquid Dry air Inject some liquid water P = 1atm, T= 65 o C When the air is fully saturated with the vapor, the partial pressure of the vapor = p* = ChE 201 Spring 2003/2004 Dr. F. Iskanderani

13. Water saturated Water ChE 201 Spring 2003/2004 Dr. F. Iskanderani

14. If ideal gas holds, then: (Dalton’s Law) p air V = n air R T p w V = n w R T Remember : p tot =p w + p air and n tot = n w +n air OR if we take the vapor as 1 and the gas as 2: p 2 V = n 2 R T ……. (1) p 1 V = n 1 R T ……..(2) p tot = p 1 + p 2 and n tot = n 1 + n 2 p 2 = p tot – p 1 and n 2 = n tot - n 1 Divide eq (2) by (1) p 1 = n 1 & p 1 = n 1 p 2 n 2 p tot n tot

15. At saturation : p w = p w * And the equations also hold ChE 201 Spring 2003/2004 Dr. F. Iskanderani

16. Example: What is the min volume (m 3 ) of dry air needed to evaporate 6.0 kg of ethyl alcohol, if the total pressure remains constant at 100 kPa.

17. Remember : p tot =p 1 + p 2 and n tot = n 1 +n 2 Therefore, 2.07 kgmol of dry air at 20 o C and 100 kPa, has a volume of: V = 2.07x 8.314 x 293 100 Then n 2 = 2.07 kgmol

19. O 2 theoretically required = 9.5 gmoles To calculate O 2 entering: )Note: air is saturated with the vapor) n O2

20. Vapor-Liquid Equilibria for Multicomponent Systems Use Raoult’s Law and Henry’s Law to predict the partial pressure of a solute and a solvent. List typical problems that involve the use of equilibrium coefficient Ki We have 2 components A and B present in 2 phases ( V & L). At equilibrium, A in the liquid phase is in equilibrium with A in the Vapor phase. Equilibrium is a function of T,P and composition of the mixture. ChE 201 Spring 2003/2004 Dr. F. Iskanderani

22. Henry’s Law : p A = H A x A ( Good for x i  0) p tot = p A + p B, Then: y A = p A /p tot = H A x A /p tot and y B = p B /p tot = H B x B /p tot Raoult’s Law: (Good for x A  1) p A = p A *. x A and p B = p B *. x B where p A +p B =p tot Again, y i =p i /p tot THEN, K i = y i /x i = p i */p tot where K i is the equilibrium constant ChE 201 Spring 2003/2004 Dr. F. Iskanderani

23. Typical problems that involve the use of the equilibrium constant Ki ( Note : These cases will be studied in detail in the Separation Processes I course next year) 1. Calculate the bubble point temperature of a liquid mixture given the total pressure and liquid composition 2. Calculate the dew point temperature of a liquid mixture given the total pressure and vapor composition ChE 201 Spring 2003/2004 Dr. F. Iskanderani

24. Typical problems that involve the use of equilibrium constant Ki 3. Calculate the related equilibrium V-L compositions over a range of mole fractions from 0 to 1 as a function of T given the total pressure 4. Calculate the composition of the V and L streams and their respective quantities when a liquid of a given composition is partially vaporized at a given T and P ChE 201 Spring 2003/2004 Dr. F. Iskanderani

25. The Phase Rule ( for systems in equilibrium only) F = C - P + 2, where: P = number of phases that can exist in the system C = number of components in the system F = number of degrees of freedom (i.e., number of independent properties to be specified to determine all the intensive properties of each phase Examples: ChE 201 Spring 2003/2004 Dr. F. Iskanderani