Vapor pressure and liquids Vapor : A gas that exists below its critical point Gas : gas that exists...

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

vapor

liquid

vapor vapor vapor vapor

At T1

Time 1 2 3 100 equilibrium

P of vapor = p* at T1


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*

Liquid

vapor

At T1

At equilibrium, vapor will reach p1*

Liquid

vapor

At T2

At equilibrium, vapor will reach p2*


Curve gives all points (T, p*) at which Liquid and Vapor exist in equilibrium. Therefore vapor can exist at any temperature. (Example)


Vapor

Liquidsolid

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


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

VlVg

=


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


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

Dry air +

water vapor

Dry air +

water vapor

Dry air +

water vapor

At T1

Time 0 2 3 100 200

saturation

• 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

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

+ vaporWater liquid

Dry air Inject some liquid water

P = 1atm, T= 65oC

P = 1atm, T= 65oC

When the air is fully saturated with the vapor, the partial

pressure of the vapor = p* =


Water Water

saturated

Water


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 = n1p2 n2 ptot ntot

At saturation :

pw = pw*

And the equations also hold


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 293100 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

Henry’s Law : pA= HA xA ( Good for xi 0) ptot = pA + pB , Then:

yA= pA/ptot = HA xA/ptot and 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 constantChE 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)

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


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


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:


Vapor pressure and liquids Vapor : A gas that exists below its critical point Gas : gas that exists...

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