Steam distillation takes place at point of intersection of the curves at which
pA = p – pW \ p = pA + pW
Total pressure = vapour pressure of component + vapour pressure of water
p–pW pA
Vapour pressure Temperature
WORKED EXAMPLES
4.1 The vapour pressure of ethyl ether at 0 °C is 185 mm Hg. Latent heat of vaporization is 92.5 cal/g.
Calculate vapour pressure at 20 °C and 35 °C. Molecular weight of ethyl ether = 74.
l = (92.5 ¥ 74) = 6845 cal/g mole R = 1.99 cal/g mole K T0 = 273 K, T1 = 293 K, T2 = 308 K At 20 °C, p ÈØ6845 1È Ø1ÈØÉÙ 2.303ÉÙ 273É Ùlog 185ÊÚ Ê Ú Ê Ú Therefore, p = 437 mm Hg.
At 35 °C, p ÈØÈØ6845 1È Ø1 log 185ÉÙ ÉÙ É Ù ÊÚ 2.303Ê Ú Ê Ú Hence, p = 773 mm Hg.
4.2 It is proposed to purify benzene from small amount of non-volatile
solutes by subjecting it to distillation with saturated steam under
atmospheric pressure of 745 mm Hg. Calculate the temperature at which the distillation will proceed and the weight of steam accompanying 1 g of benzene vapour.
Temperature, °C Vapour pressure of Vapour pressure Total pressure, benzene, mm Hg of water,
mm Hg mm Hg 60 390 65 460 68 510 69 520 150 540 190 650 215 725 225 745 distillation temperature, 69 °C 70 550 235 785 ppW pB Vapour pressure Temperature
Basis: 1 g mole of mixed vapour.
g mole of C H66 Vapour pressure of C H66 520 2.31g mole of water Vapour pressure of water 225
gC H66 2.31 78 10.01gwater 18
g water 0.1g benzene
4.3 It is decided to purify myristic acid (C13H27COOH) by steam distillation under 740 mm Hg pressure. Calculate the temperature and the weight of steam to be used per kg of acid.
At 99 °C: Vapour pressure of H2O = 740 mm Hg Vapour pressure of acid = 0.032 mm Hg
\ The distillation temperature can be taken as 99 °C Basis: 1 kmole of mixed vapour. Moles of the acid
0.032 : ÈØ
Moles of water = 1 kmole = 18 kg Weight of steam
18
ÈØ 1840 kgkg of acidÉÙ
ÊÚ
4.4 The acid given in example 4.3 is to be distilled at 200 °C by use of superheated steam. It may be
assumed that the relative saturation of the steam with acid vapours will be 80% (a) Find the weight of steam required per kg of acid distilled at 740 mm Hg. (b) Calculate the weight of steam per kg of acid if 26 inches of Hg vacuum is maintained.
(a) Vapour pressure of acid at 200 °C = 14.5 mm Hg Partial pressure of acid (80% saturation) = (14.5 ´ 0.8) = 11.6 mm Hg Basis: 1 kmole of mixed vapour
11.6
Weight of acid = ÈØ
ÉÙ´ 1 = 0.0157 kmole = 3.58 kgÊÚ
Weight of water = (1 – 0.0157) = 0.9843 kmole = 17.70 kg Steam 17.7 = ÈØ ÉÙ = 4.95 kgkg of acid ÊÚ (b) 26 inches of Hg. vacuum = 740 – (26 ´ 25.4) = 80 mm Hg 11.6 Weight of acid = ÈØ ÉÙ´ 1 = 0.145 kmole = 33.1 kgÊÚ
Weight of water (steam) = 0.855 kmole = 15.4 kg Steam 15.4
=
ÈØ = 0.465 kg
kg of acid ÉÙ
ÊÚ
Discussion on examples 4.3 and 4.4: When ordinary steam is used at atmospheric pressure, 1840 kg
of steam is needed. When superheated steam at 200 °C is used, 4.95 kg is needed. Due to low pressure and hence low boiling point under vacuum, quantity of steam needed is 0.465 kg only.
4.5 Calculate the total pressure and the composition of the vapours
in contact with a solution at 100 °C containing 35% Benzene, 40% Toluene and 25% Xylene by weight. At 100 °C, the vapour pressures of benzene (molecular weight : 78) is 1340 mm Hg, toluene (Molecular weight : 92) is 560 mm Hg and Xylene (molecular weight : 106) is 210 mm Hg.
Basis: 100 kg of solution.
Component Mol. Vapour Weight Weight, Mole wt pressure, % kmole fraction mm Hg in liquid phase Partial Mole pressure, fraction
mm Hg in vapour phase Benzene (C6H6) Toluene (C7H8) Xylene (C8H10) 78 1340 35 35/78 = 0.449 0.401 0.401 ´ 1340 = 536 0.673 92 560 40 40/92 = 0.435 0.388 0.388 ´ 560 = 217 0.272 106 210 25 25/106 = 0.236 0.211 0.211 ´ 210 = 44 0.055 Total 100 1.12 1.000 797 1.000
Total pressure = 797 mm Hg. 4.6 An aqueous solution of NaNO3 having 10 g moles of salt/1 kg of water boils at 108.7 °C at 760 mm Hg. Assume that the relative vapour pressure of the solution is independent of temperature. Find the vapour pressure of the solution at 30 °C and the boiling point elevation.
Since the solution boils at 108.7 °C, the vapour pressure of solution = 760 mm Hg. Vapour pressure of water at 108.7 °C = 1030 mm Hg (from Steam Tables)
k Vapour pressure of solution 760 0.74.Vapour pressure of solvent 1030
Vapour pressure of water at 30 °C = 31.8 mm Hg (from Tables) Vapour pressure of solution at 30 °C = (31.8 ´ 0.74) = 23.5 mm Hg,
from Cox chart
Boiling point of water at 23.5 mm Hg = 24.8 °C ÈØ
ÉÙ
ÊÚ Boiling point elevation
= (Boiling point of solution – Boiling point of solvent) = 30 – 24.8 = 5.2 °C
4.7 The following table gives vapour pressure data:
Temperature °C 69 70 75 80 85 90 95 99.2
Hexane (A), mm Hg 760 780 915 1060 1225 1405 1577 1765 Heptane (B), mm Hg 295 302 348 426 498 588 675 760
Assuming Raoult’s law is valid, use the above data to calculate for each of the above temperature the mole percent ‘x’ of hexane in the liquid and the mole percent ‘y’ of hexane in vapour at 760 mm Hg.
pA = xA PA; pB = xBPB; (xA + xB) = 1
where PA, PB are vapour pressures of hexane (A) and heptane (B) respectively.
pA + pB = P; xB = (1 – xA) (xAPA) + (1 – xA)PB = P
xA(PA – PB) = (P – PB)
xA 760 + (1 – xA)295 = 760
At 69 °C, (Boiling point of hexane)
xA(760 – 295) = (760 – 295); \ xA = 1 y 760 = 1 A = 760 At 70 °C xA(780 – 302) = (760 – 302); \ xA = 0.96 pA = (0.96 ´ 780) = 748.8; \ yA = 748.8/760 = 0.985 At 75 °C xA(915 – 348) = (760 – 348); \ xA = 0.73 pA = (0.73 ´ 915) = 668; \ yA = 668/760 = 0.880 At 80 °C xA(1060 – 426) = (760 – 426); \ xA = 0.53 pA = (0.53 ´ 1060) = 562; \ yA = 562/760 = 0.740 At 85 °C xA(1225 – 498) = (760 – 498); \ xA = 0.36 pA = (0.36 ´ 1225) = 441; \ yA = 441/760 = 0.58 At 90 °C xA(1405 – 588) = (760 – 588); \ xA = 0.21 pA = (0.21 ´ 1405) = 295; \ yA = 295/760 = 0.39 At 95 °C xA(1577 – 675) = (760 – 675); \ xA = 0.0945 pA = (0.0945 ´ 1577) = 149; \ yA = 149/760 = 0.196 At 99.2 °C Boiling point of heptane
xA(1765 – 760) = (760 – 760); \ xA = 0 = yA 4.8 A solution of methanol in water containing 0.158 mole fraction alcohol boils at 84.1 °C (760 mm Hg). The resulting vapour contains 0.553 mole fraction of alcohol. How does the actual composition of the vapour compare with composition calculated from Raoult’s Law? Temperature, °C 80 100
Vapour pressure, mm Hg 1340 2624
To get vapour pressure at 84.1 °C, the equation ln p = A – B/(T – 43)
can be used, ( p, vapour pressure in mm Hg and T, temperature in Kelvin A and B are constants) Solving A = 18.2 and B = 3420;
Vapour pressure at 84.1 ºC = 1520 mm Hg.
According to Raoult’s law; pA = xAPA = (0.158 ´ 1520) = 240;
yA = 240 = 0.316 760 % Error = Actual value Calculated value ËÛ ÌÜ ´ 100ÍÝ 0.553 - 0.316 = ÈØ ÉÙ ´ 100 = 42.8%ÊÚ
4.9 Methane burns to form CO2 and water. If 1 lb mole is burnt with 10% excess pure O2 and the resulting gas mixture is cooled and dried, calculate (a) volume of dry exit gas at 70 °F and 750 mm Hg. (b) Partial pressure of O2 in exit (c) Weight of water removed.
Basis : 1 lb mole of methane.
CH4 + 2O2 ® CO2 + 2H2O
O2 supplied = 2 ´ 1.1 = 2.2 lb moles
Gases leaving after drying: CO2 : 1.0 lb mole; O2 : 0.2 lb mole.
530 ÈØÈ Ø760(a) Volume of dry exit gas = 1.2 ´ 359 ´
492ÉÙÉ Ù750ÊÚÊ Ú
= 470.26 ft3
(b) Partial pressure of O2 in exit = 0.2 750= 125 mm Hg1.2 (c) Water removed = 2 lb moles = 36 lb
4.10 Bottled liquid gas is sold. Determine (a) the pressure of the system;
(b) vapour composition. The composition in mole % in liquid phase is given as follows:
n-Butane : 50%, Propane : 45%, and Ethane : 5%
Vapour pressure (in bar) at 30 °C are n-butane: 3.4, propane: 10.8 and ethane: 46.6
Gas n-Butane Propane Ethane Total
Mole % 50 45 5 100
Vapour pressure at 30 °C (bar) 3.4 10.8 46.6 Partial pressure (bar) 1.70 4.86 2.33 8.89 (3.4 ´ 0.5) (10.8 ´ 0.45) (46.6 ´ 0.05) Vapour composition % 19.12 54.67 26.21 100
4.11 A solvent recovery system delivers a gas saturated with benzene vapour which analyzes on a
benzene free basis as follows:
CO : 15%, O2 : 4% and N2 : 81%. This gas is at 21.1 °C and 750 mm Hg. It is compressed to 5 atm and cooled to 21.1 °C after compression. How many kilograms of benzene are condensed by this process per 1000 m3 of original mixture? The vapour pressure of benzene at 21.1 °C is 75 mm Hg.
Basis: 1000 m3 of original mixture
1000 750ÈØÈØÈ Ø273Moles of original mixture =
22.414ÉÙÉÙÉ Ù294.1ÊÚÊÚÊ Ú
= 40.87 kmoles 75
Benzene present originally = 40.87 ´
ÈØ
ÉÙ = 4.087 kmolesÊÚ
mm Hg = 75/760 = 0.0987 atm; (other gas is Tie element) 0.0987
ÈØBenzene after compression = 36.783 ´ ÉÙ
Ê50.0987Ú = 0.741 kmole
Benzene condensed = (4.087 – 0.741) ´ 78 = 261 kg
4.12 N2 from a cylinder is bubbled through acetone at 840 mm Hg and 323 K at the rate of 0.012 m3/h. The N2 saturated with acetone vapour, leaves at 760 mm Hg and 308 K at 0.023 m3/h. Find the vapour pressure of acetone at 308 K.
0.012= 5 × 10–4 kmole/h 22.414
760 323Molar flow rate of N2 = ¦µ¦ µ
840 §¶§ ¶ ¨·¨ · (N2 + CH3COCH3) = 0.023= 9.09 × 10–4 kmole/h1.013 308 22.414 1.013 ¦µ¦µ §¶§¶ ¨·¨·
Let y be the mole fraction of N2 in leaving stream. Then, 9.09 × 10–4y = 5 × 10–4
Solving, we get y = 0.55
Thus, mole fraction of acetone = 0.45
(PT) × (y) = (Partial pressure of acetone) = (VP)acetone (y)acetone (760)(0.45) = (VP) (1.0) = 342 mm Hg
4.13 Determine the pressure of the system and equilibrium VP at 30 oC. Assuming all ethane is removed, estimate the pressure and vapour phase composition components of the system at 30 oC.
Compound Mole fraction, x, in liquid phase n-butane 0.50
n-propane 0.45 Ethane 0.05
Compound
n-butane n-propane Ethane Total
Mole fraction, VP at 30 °C, Partial pressure, Y, mole fraction x, in liquid phase mm Hg mm Hg × 100
0.50 3.4 1.7 19.12 0.45 10.8 4.86 54.67
0.05 46.6 2.33 26.21 8.89 100.00 y (Mole fraction) = Partial pressure Total pressure 1.7 = ÈØ ÉÙ × 100 = 19.12 for n-butaneÊÚ 54.67 for n-propane
26.21 for ethane. When ethane is removed, total moles will be 0.95 kmole.
Compound Mole fraction, VP at 30 °C, Partial pressure, Y, mole fraction x, in liquid phase mm Hg mm Hg × 100
n-butane 0.50/0.95 = 0.5263 3.4 1.789 25.91 n-propane 0.4737 10.8 4.86 74.09
Total 6.904 100.00
4.14 Estimate the liquid phase composition of a mixture of benzene and toluene at 80 oC when their gas phase compositions are 80% benzene and 20% toluene. Vapour pressures of benzene and toluene are 1340 mm Hg and 560 mm Hg respectively at 80 oC.
Vapour pressure of benzene is 1340 mm Hg at 80 oC Vapour pressure toluene is 560 mm Hg at 80 oC At equilibrium, partial pressure of benzene = (Mole fraction, xB, of benzene) × (1340)
= (Total pressure) × (Mole fraction in vapour phase) = PT (0.8)
i.e. (xB) × (1340) = PT × (0.8) Similarly for toluene,
xT × (560) = PT × (0.2)
PT × (0.8) + PT × (0.2) = PT We also know that, xB + xT = 1 4 × (1 – xB) × (560) = 4 × PT× (0.2)
xB × (1340) = PT × (0.8) 4 (1 – xB) 560 = xB(1340) 2240 – 1340xB – 2240xB = 0 Solving, xB = 0.6
4.15 A certain quantity of an organic solvent (molecular weight 125 and density 1.505 g/cc) is kept in
displacing all the air. The flask is closed, evacuated and the contents reach equilibrium at 30 °C. Vapour pressure of solvent at 30 °C is 240 mm Hg. It is observed that only 10 ml of liquid solvent is present finally.
(i) What is the pressure in the flask at equilibrium?
(ii) What is the total mass in grams of organic liquid in the flask?
(iii) What fraction of the organic liquid present in the flask is in vapour phase at equilibrium? Equilibrium vapour pressure = 240 mm Hg at 30 °C and the pressure in the system is 240 mm Hg. Volume of vapour space = 4000 – 10 = 3990 ml
T = 303 K; P = 240 mm Hg Moles at NTP = 3990 240 273 1= 0.05065 g mole 303 t tt 760 22414 Molecular weight = 125 Weight = 0.05065 × 125 = 6.331 g
Mass of liquid left behind = 10 ml × 1.505 g/cc = 15.05 g Total weight = 15.05 + 6.331 = 21.381 g Mole fraction of vapour = (6.331/21.381) = 0.2961
4.16 Benzene and toluene form an ideal solution. If vapour pressure of benzene is 70 mm Hg and that
of toluene is 20 mm Hg and their mole fractions in liquid phase are 0.7 and 0.3 respectively, calculate their vapour phase composition.
Let A be benzene and B be toluene
Let PP and PT be partial pressure and total pressure respectively PPA = (PT) (yA) = xA PA PPB = (PT)(yB) = xB PB PT(yA + yB) = xA (PA) + (1 – xA) PB = PP PT = PPA + PPB PT = (0.7)(70) + (0.3)(20) = 49 + 6 = 55 mm Hg Hence, yA= (0.7)(70)/55 = 0.89 yB = (0.3)(20)/55 = 0.11
4.17 Vapour pressure of pure benzene and toluene are 960 mm Hg and 380 mm Hg respectively at
86.0 °C. Calculate the liquid phase and vapour phase composition at that temperature. Total pressure = 760 mm Hg
Let x and y be the liquid and vapour phase compositions respectively. Suffix B and T denote benzene and toluene respectively Then, by Raoult’s law,
PTot = (xB) (PB) + (xT) (PT) 760 = (xB) (960) + (1 – xB) 380 Solving,
xB = 0.655
xT = 0.345
Px
yB = BB 960 t0.655= 0.827PTot760
In the same way, yPPx 380t 0.345= 0.173
T = TT760Tot
4.18 Vapour pressure of benzene is 3 atm and that of toluene is 1.333 atm. A liquid fuel containing
0.4 mole benzene and 0.6 mole toluene is vaporized. Estimate the mole fraction of benzene in vapour phase.
Partial pressure of benzene = (3)(0.4) = 1.2 Partial pressure of toluene = (1.333)(0.6) = 0.8 Total pressure = 2.0 atm
ybenzene = 1.2/2.0 =0.6
EXERCISES
4.1 The vapour pressure of benzene can be calculated from the Antoine equation.
ln ( p*) = 15.9008 – [2788.51/(–52.36 + T)]
where p* is in mm Hg and T is in K. Determine the latent heat of vaporization of benzene at its normal boiling point of 353.26 K, and compare with the experimental value.
4.2 Prepare a Cox chart for ethyl acetate. The vapour pressure of ethyl acetate is 200 mm Hg abs. at
42 °C and 5.0 atm at 126.0 °C. By using the chart estimate the boiling point of ethyl acetate at 760 mm Hg and compare with the experimental value (77.1 °C).
4.3 The following data gives the vapour pressure (VP) for benzene (A)– toluene (B) system. Compute
the vapour-liquid equilibrium data at a total pressure of 760 mm Hg.
VPA, 760 811 882 957 1037 1123 1214 1310 1412 1530 1625 1756 — mm Hg VPB, — 314 345 378 414 452 494 538 585 635 689 747 760 mm Hg
4.4 The following data gives the vapour pressure data for a binary system. Compute the VLE vapour–
liquid equilibrium data at a total pressure of 760 mm Hg. Vapour pressure of A, mm Hg 760 860 1002 1160 1262 Vapour pressure of B, mm Hg 433 498 588 698 760
4.5 It is desired to purify nitrobenzene by steam distillation under a pressure of 760 mm Hg.
Distillation takes place at 99 °C. Vapour pressure of nitrobenzene is 20 mm Hg. Estimate the weight of nitro benzene associated per kg of steam in distillate. Assuming the vaporization efficiency to be 75%, estimate the weight of nitrobenzene per kg of steam in distillate.
Hg respectively. Calculate the total pressure and composition of the vapour in contact with a liquid containing 30 weight % of methyl alcohol and 70 weight % of ethyl alcohol at 100 °C.
4.7 A liquefied fuel has the following analysis: C2H6 : 2%, n-C3H8 : 40%, i-C4H10 : 7%, n-C4H10 : 47% and C5H12 : 4%. It is stored in cylinders for sale. (a) Calculate the total pressure in cylinder at 21 °C and the composition of the vapour evolved. (b) Find the total pressure at 21 °C if all C2H6 were removed.
Vapour pressure data (mm Hg):
C2H6 : 28500, n-C3H8 : 6525, i-C4H10 : 2250, n-C4H10 : 1560 and C5H12 : 430.
4.8 Calculate the total pressure and the composition of the vapours and liquid in molar quantities at
100 °C for a solution containing 45% benzene, 40% toluene and 15% xylene by weight. At 100 °C, the vapour pressures are benzene (C6H6) (molecular weight: 78) : 1340 mm Hg; Toluene (C7H8) (molecular weight: 92) : 560 mm Hg, xylene (C7H8) (molecular weight: 106) : 210 mm Hg.
4.9 The partial pressure of actetaldehyde in a solution containing 0.245 kg of CH3CHO in 20 kg of water is 190 mm Hg at 93.5 °C. Calculate the partial pressure of CH3CHO over 0.15 molal solution in water.
4.10 Ethyl acetate at 30 °C exerts a vapour pressure of 110 mm Hg. Calculate the composition of the
saturated mixture of ethyl acetate and air at a temperature of 30 °C and an absolute pressure of 900 mm Hg pressure. Express the composition by (a) volume %, and (b) weight %.