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Learn about vapor pressure, gas-liquid equilibrium, and the relationship between temperature, pressure, and composition in liquids. Explore Raoult's Law and Henry's Law to predict partial pressures and equilibrium coefficients.
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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 T1 , 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, p1* (at the given temp T1). Note the vapor pressure is the maximum pressure the vapor can attain. ChE 201 Spring 2003/2004 Dr. F. Iskanderani
At T1 P of vapor = p* at T1 Time 1 2 3 100 equilibrium ChE 201 Spring 2003/2004 Dr. F. Iskanderani
Vapor pressure and liquids Now change the temp to a higher temperature T2. The system will reach equilibrium , and the vapor will have a new vapor pressure , p2* > p1* At T1 At T2 vapor vapor Liquid Liquid At equilibrium, vapor will reach p2* At equilibrium, vapor will reach p1* ChE 201 Spring 2003/2004 Dr. F. Iskanderani
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
Liquid solid Vapor
1 ln p* =m ( ) + b T .A. ln p* =( ) + B T+C • 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) (See appendix G on page 669) V – nb + ChE 201 Spring 2003/2004 Dr. F. Iskanderani
Vl d(p*) = Vg T – nb + n dP T Change of vapor pressure with pressure Under normal conditions the effect of P on the vapor pressure, p* is small ChE 201 Spring 2003/2004 Dr. F. Iskanderani
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 = V1 x1 + V2 x2 + .. +.. • This eq is good for components with similar structure such as hydrocarbons ChE 201 Spring 2003/2004 Dr. F. Iskanderani
Saturation and Equilibrium • For a mixture of pure vapor and a non-condensable gas • example : water vapor + air At T1 saturation Time 0 2 3 100 200 ChE 201 Spring 2003/2004 Dr. F. Iskanderani
Water vaporizes until equilibrium at T1 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 = pair + pwater vapor • At saturation Ptotal = pair + 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
Dry air + vapor Water liquid 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 P = 1atm, T= 65oC P = 1atm, T= 65oC Dry air Inject some liquid water When the air is fully saturated with the vapor, the partial pressure of the vapor = p* = ChE 201 Spring 2003/2004 Dr. F. Iskanderani
saturated Water Water Water ChE 201 Spring 2003/2004 Dr. F. Iskanderani
p2 n2 ptot ntot If ideal gas holds, then: (Dalton’s Law) pair V = nair R T pw V = nw R T Remember : ptot=pw+ pair and ntot= nw+nair OR if we take the vapor as 1 and the gas as 2: p2 V = n2 R T ……. (1) p1 V = n1 R T ……..(2) ptot= p1 + p2 and ntot= n1 + n2 p2= ptot – p1 and n2= ntot - n1 Divide eq (2) by (1) p1 = n1 & p1 =n1
At saturation : pw = pw* And the equations also hold ChE 201 Spring 2003/2004 Dr. F. Iskanderani
Example: What is the min volume (m3) of dry air needed to evaporate 6.0 kg of ethyl alcohol, if the total pressure remains constant at 100 kPa.
Remember : ptot=p1+ p2 and ntot= n1+n2 Therefore, 2.07 kgmol of dry air at 20oC and 100 kPa, has a volume of: V = 2.07x 8.314 x 293 100 Then n2 = 2.07 kgmol
O2 theoretically required = 9.5 gmoles To calculate O2 entering: )Note: air is saturated with the vapor) nO2
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
ChE 201 Spring 2003/2004 Dr. F. Iskanderani
Henry’s Law : pA= HA xA ( Good for xi 0) ptot = pA + pB , Then: yA= pA/ptot = HA xA/ptotand yB= pB/ptot = HB xB/ptot Raoult’s Law: (Good for xA 1) pA = pA*. xA and pB = pB* . xB where pA+pB=ptot Again, yi=pi/ptot THEN, Ki = yi/xi = pi*/ptot where Ki is the equilibrium constant ChE 201 Spring 2003/2004 Dr. F. Iskanderani
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) • Calculate the bubble point temperature of a liquid mixture given the total pressure and liquid composition • Calculate the dew point temperature of a liquid mixture given the total pressure and vapor composition ChE 201 Spring 2003/2004 Dr. F. Iskanderani
Typical problems that involve the use of equilibrium constant Ki • 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 • 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
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