Chapter 8: Phase Equilibria (Solution)

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Problem 8.1: Show that the following equations provide the criteria of equilibrium under certain constraints. (a) dUS,V = 0 (b) dSH,P = 0 (c) dHS,P = 0

Ans:

Explanation:

Problem 8.2: For each of the following non-reactive equilibrium systems, determine the number of available degrees of freedom. (a) Two miscible materials in vapor-liquid equilibrium with vapor composition specified at a given temperature and pressure. (b) A mixture of methane and air in contact with a solid adsorbent at atmospheric pressure and a specified temperature. (c) Liquid water in equilibrium with a mixture of water vapor and nitrogen. (d) Two partially miscible liquid phases and a vapor phase in equilibrium with them at a constant pressure. (e) A liquid mixture of benzene and toluene in equilibrium with its vapor at 1 bar. (f) A vapor phase consisting of ammonia and air and a liquid phase consisting of ammonia and water at a given temperature. (g) A liquid mixture of components A and C in equilibrium with a liquid solvent B in which only C is soluble at a given temperature and pressure.

Ans: (a) 2, overdefined (b) 1 (c) 2 (d) 0 (e) 1 (f) 2 (g) 1

Explanation:

Problem 8.3: Determine the mole fraction of methane, xi, dissolved in a light oil at 200 K and 20 bar. Henry’s law is valid for the liquid phase, and the gas phase may be assumed to be an ideal solution. At these conditions, Henry’s law constant for methane in oil = 200 bar, fugacity coefficient of pure methane gas = 0.90 and mole fraction of methane in the gas phase, y1 = 0.95.

Ans: 0.0855

Explanation:

Problem 8.4: The vapor pressures of benzene and toluene are given below.

T (K)353.1358363368373378383383.6
P_A^S  (kPa)101.3116.9135.4155.7179.1204.2233.0240.0
P_B^S  (kPa)39.646.054.063.374.286.099.0101.3

Calculate the equilibrium data for the system at 101.3 kPa and formulate an equation for the equilibrium diagram in terms of average relative volatility.

Ans: y = 2.45x/(1 + 1.45x)

T (K)353.1358363368373378383383.6
x10.780.5810.4110.2580.1300.0170
y10.900.7770.6320.4560.2610.0390

Explanation:

Problem 8.5: At 303 K, the vapor pressures of benzene (A) and toluene (B) are 15.75 kPa and 4.89 kPa respectively. Determine the partial pressures and weight composition of the vapour in equilibrium with a liquid mixture consisting of equal weights of the two components.

Ans: 8.5239 kPa, 76.3 % benzene

Explanation:

Problem 8.6: An equimolar mixture of benzene and toluene is contained in a piston/cylinder arrangement at a temperature T. What is the maximum pressure below which the mixture exists as a vapor phase alone? At the given T, the vapour pressures of benzene and toluene are 203.9 kPa and 85.3 kPa, respectively. Assume that Raoult’s law is valid.

Ans: 120.3 kPa

Explanation:

Problem 8.7: Two substances A and B are known to form ideal liquid solutions. A vapor mixture containing 50 % (mol) A and 50 % (mol) B is at 311 K and 101.3 kPa. This mixture is compressed isothermally until condensation occurs. At what pressure does condensation occur and what is the composition of the liquid that forms? The vapor pressures of A and B are 142 kPa and 122 kPa respectively.

Ans: 131.24 kPa, 46.2 % A

Explanation:

Problem 8.8: Air is cooled to 80 K at 101.3 kPa. Calculate the composition of the liquid and vapor phases at this condition assuming that the mixture behaves ideally. The vapor pressure of nitrogen and oxygen at 80 K are 135.74 kPa and 30.04 kPa respectively.

Ans: N2: 67.4 % (liquid), 90.34 % (vapor)

Explanation:

Problem 8.9: The binary system, acetone (1)–acetonitrile (2) conforms closely to Raoult’s law. Using the vapour pressure data given below plot the following (a) P-x1 and P-y1 curves at 323 K (b) T-x1 and T-y1 curves at 53.32 kPa.

T (K)311.45315319323327331335.33
P_1^S  (kPa)53.3261.0970.9181.9794.36108.2124.95
P_2^S  (kPa)21.2524.6128.9033.7939.3545.6253.32
Ans: (a)

P (kPa)33.7943.42653.06262.69872.33481.79
x100.20.40.60.81
y100.37750.61790.78440.90661

(b)

T (K)311.45315319323327331335.33
x110.7870.5810.4050.2540.1230
y110.9020.7730.6230.4490.2500

Explanation:

Problem 8.10: Assuming Raoult’s law to be valid for the system benzene (1)–ethyl benzene (2) and the vapor pressures are given by the Antoine equations

\ln\left(P_1^S\right)=13.8858-\frac{2788.51}{T-52.41} \ln\left(P_2^S\right)=14.0045-\frac{3279.47}{T-60.00}

where P is in kPa and T is in K. Construct the following: (a) The P-x-y diagram at 373 K (b) The T-x-y diagram at 101.3 kPa.

Ans: (a)

P (kPa)34.0263.0392.04121.04150.05179.06
x100.20.40.60.81
y100.5680.7780.8880.9951

(b)

T (K)353.3363.3373.3383.3393.3403.3409.4
x110.6860.4580.2870.1560.0530
y110.9250.8160.6660.4640.1980

Explanation:

Problem 8.11: A liquid mixture containing 65 % (mol) benzene and 35 % (mol) toluene is subjected to flash vaporization at 363 K and 101.3 kPa. The vapor pressure of benzene at this temperature is 136.09 kPa and the vapor pressure of toluene is 54.21 kPa. Flash vaporization is essentially an equilibrium stage operation. Calculate (a) The exit vapor composition (b) The exit liquid composition (c) The mole percent of the feed that is vaporized.

Ans: (a) 77.3 % benzene (b) 57.5 % benzene (c) 37.9 % benzene

Explanation:

Problem 8.12: For the system n-pentane (1) – n-heptane (2), the vapor pressures are given by the Antoine equation

\ln\left(P\right)=A-\frac B{T-C}

Where P is in kPa and T is in K. The constants are as follows.

SystemABC
n-pentane13.81832477.0740.00
n-heptane13.85872911.3256.56

Assuming that the solution formed is ideal, calculate: (a) The composition of the liquid and vapor in equilibrium at 95 kPa and 336.2 K. (b) The composition of the vapor in equilibrium with a liquid containing 34 % (mol) pentane and the equilibrium temperature at P = 95 kPa. (c) The total pressure and the vapor composition in equilibrium with a liquid of composition x1 = 0.44 at T = 333.2 K.

Ans: (a) x1 = 0.3138, y1 = 0.7730 (b) 334.4 K, 79.6 % pentane (c) 110.25 kPa, 85.7 % pentane

Explanation:

Problem 8.13: 8.13 Using Gibbs–Duhem equations prove that if one constituent of a mixture exhibits positive deviation from ideal behavior the other constituent also shows positive deviation.

Ans:

Explanation:

Problem 8.14: Prove that a solution exhibiting negative deviation from ideal behavior is formed with an evolution of heat.

Ans:

Explanation:

Problem 8.15: Using van Laar method calculate the vapor-liquid equilibrium compositions for acetone (1)– chloroform (2) system at a pressure of 101.3 kPa. At this pressure, the system forms an azeotrope of composition 66.6 % (mol) chloroform which boils at 337.7 K. The vapor pressures of the pure components are given below.

T (K)318323333343353
P_1^S  (kPa)68.0481.65114.71158.61214.73
P_2^S  (kPa)58.5170.1198.58135.82187.01

The normal boiling points of acetone and chloroform are respectively 329.5 K and 334.1 K. (Hint: The ratio of vapor pressures remains almost constant. Use the method employed in Example 8.13 for calculating y for arbitrarily chosen x values.)

Ans:

T (K)334.4336.5337.7337335331.5329.5
x100.20.3340.40.60.81
y100.1650.3340.42770.68580.87501

Explanation:

Problem 8.16: Show that the van Laar equation and Margules equation are consistent with the Gibbs–Duhem equations.

Ans:

Explanation:

Problem 8.17: The toluene–acetic acid mixture forms an azeotrope containing 62.7 % (mol) toluene and having a minimum boiling point of 378.6 K at 101.3 kPa. The vapor pressure data are:

T (K)343353363373383393
Toluene (kPa)26.9838.6153.9374.27
Acetic acid (kPa)18.1326.9639.0555.6077.41105.83

The normal boiling point of toluene and acetic acid are respectively 383.9 K and 391.7 K. (a) Calculate the van Laar constants A and B (b) Plot ln γ1 and ln γ2 as ordinate against mole fraction of toluene.

Ans: (a) A = 1.0624, B = 1.0217 (b)

x100.20.40.60.81
lnγ11.06240.62990.37060.16210.03990
Lnγ200.04350.171300.37930.66401.0217

Explanation:

Problem 8.18: Under atmospheric pressure, the acetone–chloroform azeotrope boils at 337.8 K and contains 33.5 % (mol) acetone. The vapor pressures of acetone and chloroform at 337.8 K are respectively 132.62 kPa and 113.96 kPa. (a) Calculate the composition of the vapor in equilibrium with a liquid analyzing 11.1 % (mol) acetone. How does it compare with the experimental value of 6.5 % acetone in the vapor? (b) What is the total pressure at this condition?

Ans: (a) 7.19 % acetone (b) 108.35 kPa

Explanation:

Problem 8.19: Ethyl alcohol and hexane form an azeotrope at 33.2 % (mol) ethanol. It boils at 331.9 K at 101.3 kPa. At 331.9 K, the vapor pressures are 44.25 kPa for ethanol and 72.24 kPa for hexane. Determine: (a) The van Laar constants (b) The vapor composition for a solution containing 50 % (mol) hexane boiling at 331.9 K(c) The total pressure for the conditions in part (b).

Ans: (a) A = 1.6625, B = 2.7475 (b) 67.81 % hexane (c) 101.56 kPa

Explanation:

Problem 8.20: At atmospheric pressure, ethyl acetate and ethyl alcohol form an azeotrope containing 53.9 % (mol) acetate boiling at 345 K. Determine: (a) The van Laar constants (b) The azeotropic composition and the total pressure if the mixture forms an azeotrope boiling at 329.5 K (c) The composition of the vapor in equilibrium with a liquid of composition 60 % (mol) alcohol and boiling at 329.5 K

Vapor pressure (kPa)at 345 Kat 329.5 K
Ethyl alcohol78.2439.72
Ethyl acetate84.7747.98
Ans: (a) A = 0.8940, B = 0.8426 (b) 60.3 % acetate, 54.72 kPa (c) 48.64 % acetate

Explanation:

Problem 8.21: An organic liquid solution containing A (molecular weight 46) and B (molecular weight 78) forms an azeotrope containing 52 % by weight A at 333 K and 101.3 kPa. Vapor pressures of A and B are 69.31 kPa and 68 kPa respectively. Determine the van Laar constants.

Ans: A = 0.9376, B = 3.0119

Explanation:

Problem 8.22: For the acetone (1)–diethylamine (2) system the activity coefficient values as a function of concentration are given below:

x10.10.20.30.70.80.9
γ11.11281.08401.06171.00951.004110.0009
γ21.00151.00611.01331.06391.08121.1000

Using the above data estimate the van Laar constants for the system. (Hint: A = ln γ1 as x1 → 0 and B = ln γ2 as x2 → 0)

Ans: A = 0.1365, B = 0.1122

Explanation:

Problem 8.23: Find the van Laar constants for the binary system benzene (1)–ethanol (2) using the following data

x10.10.30.450.70.9
P_1^S  (kPa)73.3168.6463.9867.9881.31
P_2^S  (kPa)75.9869.6467.9869.3179.98
Ans: A = 1.7492, B = 1.4446

Explanation:

Problem 8.24: The T-x-y data for the system metaxylene (1)–propionic acid (2) at 101.3 kPa is given below:

T (K)412.6411.1408.4407.1408.8411.7
x10.0350.0900.2580.6520.8770.989
x20.0640.1330.3140.6210.8210.975

Does the system form an azeotrope? Give reasons.

Ans: Yes

Explanation:

Problem 8.25: For the conditions in Example 8.19, calculate (a) the equilibrium temperature and vapor composition for x1 = 0.32 and P = 101.3 kPa and (b) the equilibrium temperature and liquid composition for y1 = 0.57 and P = 101.3 kPa.

Ans: (a) 335.53 K, 80.9 % acetone (b) 351.96 K, 4.07 % acetone

Explanation:

Problem 8.26: For the isobutanol–water system, it is found from VLE measurements that the composition of the vapor and liquid in equilibrium at 101.3 kPa and 364.7 K are 28.6 % (mol) and 1.4 % (mol) i-butanol respectively. The vapor pressure of i-butanol at this temperature is 53.32 kPa. The activity coefficient of water may be taken as 1.0012. Compute the van Laar constants.

Ans: A = 3.8297, B = 2.3540

Explanation:

Problem 8.27: Ethanol–water mixture forms an azeotrope boiling at 351.4 K under a pressure of 101.3 kPa and its composition is 89.4 % (mol) ethanol. The vapor pressures of ethanol and water at 351.4 K are 100 kPa and 44 kPa respectively. Using van Laar method and assuming that the ratio of vapor pressures remains constant calculate the composition of the vapour in equilibrium with a liquid containing 80 % ethanol.

Ans: 83.16 % alcohol

Explanation:

Problem 8.28: Two liquids P and Q form an azeotrope containing 58 % (mol) P at 101.3 kPa pressure. At the azeotropic temperature, the vapour pressures of P and Q are 200 kPa, 125.3 kPa respectively. Construct the equilibrium (y–x) diagram. List all the assumptions made.

Ans:

x100.20.40.60.81
y101.3550.32990.61230.92121

Explanation:

Problem 8.29: For the binary system n-pentanol (1)–n-hexane (2), determine the activity coefficients at 313 K in an equimolar mixture. The Wilson parameters are given as follows:

a12 = 7194.18 J/mola21 = 697.52 J/mol
V1 = 109.2×10-6 m3/molV2 = 132.5×10-6 m3/mol
Ans: γ1 = 1.3551, γ2 = 1.682

Explanation:

Problem 8.30: The Wilson parameters for acetone (1)–methanol (2) are obtained from the following values:

a12 = -712.51 J/mola21 = 2487.71 J/mol
V1 = 74.05×10-6 m3/molV2 = 40.73×10-6 m3/mol

The vapour pressures are given by the Antoine equations:

\ln\left(P_1^S\right)=14.39155-\frac{2795.817}{T-43.2} \ln\left(P_2^S\right)=16.59381-\frac{3644.297}{T-33.435}

Calculate the total pressure and vapor composition in equilibrium with a liquid containing 31 % (mol) acetone at 333 K.

Ans: 108.4 kPa, 43.46 % acetone

Explanation:

Problem 8.31: For the 2-propanol (1)–water (2) system, the following Wilson parameters are reported.

a12 = 1833.74 J/mola21 = 5183.26 J/mol
V1 = 76.92×10-6 m3/molV2 = 18.07×10-6 m3/mol

The vapor pressures can be calculated by the Antoine equations, which are given below:

\ln\left(P_1^S\right)=16.678-\frac{3640.20}{T-53.54} \ln\left(P_2^S\right)=16.2887-\frac{3816.44}{T-46.13}

where T is in K and the vapor pressures are in kPa. Calculate: (a) Equilibrium pressure and vapor composition at T = 353.15 K and x1 = 0.25 (b) Equilibrium pressure and liquid composition for T = 353.15 K and y1 = 0.60 (c) Equilibrium temperature and vapor composition for P = 101.3 kPa and x1 = 0.85 (d) Equilibrium temperature and liquid composition for P = 101.3 kPa and y1 = 0.40.

Ans: (a) 91.50 kPa, 53.77 % propanol (b) 96.75 kPa, 43.91 % propanol (c) 353.84 K, 81.5 % propanol (d) 360.615 K, 6.38 % propanol

Explanation:

Problem 8.32: A solution of hydrocarbons contains n-propane 5.0 %, n-butane 30.0 %, n-pentane 40.0 % and n-hexane 25.0 %. Compute the bubble point and the dew point at 350 kPa. The K-values can be taken from the DePriester nomographs.

Ans: 329.7 K, 356.9 K

Explanation:

Problem 8.33: A solution has the following composition in mol percent: ethane 0.25 %, propane 25.00 %, isobutane 18.5 %, n-butane 56.0 % and isopentane 0.25 %. For a pressure of 10 bars, calculate (a) The bubble point (b) The dew point (c) The composition of the liquid and vapor when 40 % of the mixture is vaporized.

Ans: (a) 330 K (b) 340.6 K (c)

EthanePropaneIsobutanen-butaneIsopentane
Mole % (liquid)0.4836.318.1844.980.13
Mole % (vapor)0.0917.4518.7463.350.38

Explanation:

Problem 8.34: A stream of gas in a natural gasoline plant has the following composition by volume: ethane 10 %, propane 14 %, isobutane 19 %, n-butane 54 % and isopentane 3 %. (a) Calculate the pressure necessary to condense this gas completely at 311 K. (b) For a condenser operating at the pressure in part (a), calculate the temperature at which condensation starts and the temperature at which 50 % (mol) of the vapor gets condensed. Also, calculate the composition of the first liquid to condense and the composition of the liquid and vapor phases at 50 % condensation.

Ans: (a) 930.3 kPa (b) 337.4 K

EthanePropaneIsobutanen-butaneIsopentane
Composition (condensate)1.8 %6.2 %17.3 %67.1 %7.50 %

330.2 K

EthanePropaneIsobutanen-butaneIsopentane
Composition (liquid)3.22 %9.31 %19.55 %63.38 %4.54 %
Composition (vapor)16.81 %18.7 %18.46 %44.57 %1.46 %

Explanation:

Problem 8.35: Determine the composition of the vapor in equilibrium with the liquid and the pressure of the system at 313 K for a liquid mixture of 5 % (mol) methane, 10 % (mol) ethane, 30 % (mol) propane, 25 % (mol) isobutane, and 30 % (mol) n-butane. Determine the pressure and composition of the liquid in equilibrium with a vapor mixture of the above composition.

Ans: 2205 kPa

MethaneEthanePropaneIsobutanen-butane
Composition41.9120.2521.968.577.31

700 kPa

MethaneEthanePropaneIsobutanen-butane
Composition0.21.9116.1430.8850.88

Explanation:

Problem 8.36: Calculate the pressure at which condensation starts and the pressure at which condensation is complete when a vapor mixture of the following composition is subjected to condensation at a temperature of 300 K: 20 % (mol) ethylene, 20 % (mol) ethane, 40 % (mol) propane and 20 % (mol) n-butane.

Ans: 861.4 kPa, 2446 kPa

Explanation:

Problem 8.37: A vapor mixture containing 15 % ethane, 20 % propane, 60 % isobutane, and the rest n-butane is subjected to partial condensation so that 75 % of the vapor is condensed. If the condenser temperature is 300 K determine the pressure.

Ans: 758 kPa

Explanation:

Problem 8.38: An equimolar mixture of propane (1) and n-butane (2) is partially condensed so that 50 % (mol) of the mixture is in the liquid state at 311 K. Using the DePriester nomograph determine the following: (a) The pressure (b) The vapor and liquid compositions (c) The pressure at which condensation begins at the constant temperature of 311K.

Ans: (a) 717 kPa (b) propane: 63.93 %, n-butane: 36.07 % (c) propane: 36.07 %, n-butane: 63.93 %

Explanation:

Problem 8.39: Verify whether the following data are consistent.

x100.20.40.60.81.0
γ10.5760.6550.7480.8560.9501.000
γ21.000.9850.9300.8140.6260.379
Ans: consistent

Explanation:

Problem 8.40: Calculate the constants A and B in the van Laar equation from the following data. Check whether the data are consistent. (assume P = 101.3 kPa)

x10.280.400.600.675
y10.4200.5160.6560.710
T (K)331.5330.4329.3328.3
P_1^S  (kPa)109.16104.77101.1797.70
P_2^S  (kPa)77.1773.4469.9867.04
Ans: A = 0.685, B = 0.785, consistent

Explanation:

Problem 8.41: The following data were reported for vapor-liquid equilibrium for the ethanol-water system at 298 K. Test whether the data are thermodynamically consistent.

x10.1220.1630.2260.3200.3370.4370.4400.5790.830
y10.4740.5310.5620.5820.5890.6200.6190.6850.849
P (kPa)5.576.026.386.766.807.027.047.307.78

The vapor pressures of ethanol and water are, respectively, 7.86 kPa and 3.17 kPa.

Ans: inconsistent

Explanation:

Problem 8.42: The following vapor-liquid equilibrium data were obtained for the water (1)–nitric acid (2) system at 293 K.

wt % HNO31009080706050
\overline{p_1}  (kPa)00.0610.1800.3770.6571.004
\overline{p_2}  (kPa)6.3983.4691.3980.4110.1190.065

Test the above data for thermodynamic consistency.

Ans: consistent

Explanation:

Problem 8.43: From the data activity coefficient versus mole fraction for the system acetone (1)– dichloroethylene (2) given below check their thermodynamic consistency.

x10.0230.0530.3570.5160.8830.979
γ10.6080.7110.8540.9170.9871.0
γ10.9930.9740.9340.8910.7810.694
Ans: Consistent

Explanation:

Problem 8.44: The partial pressure of ether at 303 K for the ether (1)–acetone (2) system is given as follows:

x11.000.870.500.30
\overline{p_1}  (kPa)86.10576.1151.9835.59

The vapor pressure of pure acetone at 303 K is 37.72 kPa. (a) Calculate the activity coefficient of ether and plot the logarithm of the activity coefficient as a function of concentration. (b) Predict the partial pressure of acetone in the corresponding solution for which the activity coefficients of ether have been determined.

Ans:

x110.870.500.30
γ111.0161.20741.3778
x110.870.500.300
  \overline{p_2} 07.4520.0531.6337.72

Explanation:

Problem 8.45: Vapour–liquid equilibrium data for the system methanol (1)–benzene (2) at 313 K are given below:

x100.1410.2270.3040.4020.4680.5520.6430.7020.7500.8340.8780.8960.9151.00
y100.5070.5240.5310.5400.5430.5480.5660.5780.5780.6410.6700.7230.7531.00
P (kPa)24.4646.5247.4548.3248.5448.7348.7848.8147.6147.6546.0144.5243.3542.9935.12

Use the area test to determine the thermodynamic consistency of the data.

Ans: Inconsistent

Explanation:

Problem 8.46: The activity coefficient of thallium in amalgams at 293 K are given below:

x00.00330.01680.04860.09860.16800.27010.4240
γ1.01.0421.2311.7762.8114.3216.1967.707

Determine the activity coefficient of mercury as a function of composition.

Ans:

x100.00330.01680.04860.09860.1680.27010.424
γ110.99990.99820.98540.97460.93130.85350.7688

Explanation:

Problem 8.47: The following data refers to the system water (1)–n-propyl alcohol (2) at 298 K.

x200.010.020.050.100.20
γ212.512.311.39.926.053.12

What is the activity coefficient of water in a 10 % (mol) n-propyl alcohol solution?

Ans: 1.0493

Explanation:

Problem 8.48: At 323 K, the vapor pressures of pure ether and pure ethyl alcohol are 170.13 and 29.47 kPa respectively. The total pressures versus liquid composition data are given below:

x10.0650.140.2110.2930.3830.4830.5870.7130.854
P (kPa)53.3674.3992.40107.24120.43132.62142.98151.42161.07

Using the Gibbs–Duhem equation compute from these data the partial pressures of ether and alcohol over liquid solutions of various compositions at 323 K.

Ans:

x10.0650.140.2110.2930.3830.4830.5870.7130.854
  \overline{p_1} 24.2047.9066.6784.56100.31114.08125.48137.19150.67

Explanation:

Problem 8.49: The data given below refer to the boiling points of the ethanol (1)–benzene (2) system at 100 kPa and the vapor pressures of pure ethanol and benzene at these temperatures.

x100.040.110.280.430.610.800.890.941.00
P_1^S  (kPa)107.1689.4474.6467.5866.2467.5872.6479.7186.64100
P_2^S  (kPa)10086.3774.9169.0467.8469.0473.1878.9184.2494.77

Calculate the van Laar constants from these data assuming γ to be independent of temperature. Also, find γ1 and γ2 from the van Laar equations.

Ans: A = 1.7405, B = 1.4012

x100.040.110.280.430.610.800.890.941.0
γ15.74.82313.69902.20641.59021.22261.05611.01441.0041.00
γ201.00341.02511.16031.38801.84172.64063.18693.5524.06

Explanation:

Problem 8.50: The total pressure versus solution concentration data for the system dioxane (1)–water (2) at 353 K is given below:

x20.10.20.30.40.60.70.80.9
P (kPa)63.4570.1874.1176.1176.7175.9173.3166.84

The vapor pressures of pure water and dioxane at this temperature are 47.33 and 51.05 kPa. Calculate: (a) The van Laar constants (b) The constants in the Margules equation (c) The vapor composition in equilibrium with a liquid containing 60 % water by weight and the total pressure over this solution using the van Laar method.

Ans: (a) 1.7951, 1.4679 (b) 1.7951, 1.4679 (c) 65.21 % water, 65.82 kPa

Explanation:

Problem 8.51: Benzene (1)–cyclohexane (2) form an azeotrope at 0.525 mole fraction benzene at a temperature of 350.8 K and 101.3 kPa. At this temperature, the vapor pressure of benzene is 99.3 kPa and that of cyclohexane is 98 kPa. Using the van Laar model estimate the activity coefficients at x1 = 0.2 and 0.9. Using this activity coefficient information calculate the equilibrium pressure and the vapor compositions at 350.8 K for the two liquid compositions.

Ans: x1 = 0.20, γ1 = 1.072, γ2 = 1.0059, P = 100.15 kPa, y1 = 0.2126; x1 = 0.90, γ1 = 1.0007, γ2 = 1.0815, P = 100.03 kPa, y1 = 0.894

Explanation:

Problem 8.52: The azeotrope of the n-propanol–water system has a composition 56.83 % (mol) water with a boiling point of 360.9 K at a pressure of 101.3 kPa. At this temperature, the vapour pressures of water and propanol are respectively 64.25 kPa and 69.71 kPa. Evaluate the activity coefficients for a solution containing 20 % water through the van Laar equations.

Ans: γ1 = 1.0288, γ2 = 2.6100

Explanation:

Problem 8.53: The pressure exerted over the binary system ethanol–methylcyclohexane containing 40.5 % (mol) ethanol at 308 K is 20.31 kPa. The vapor phase contained 54.7 % (mol) ethanol. The vapor pressures at 308 K are 13.74 kPa for ethanol and 9.81 kPa for methylcyclohexane. What are the total pressure and composition of the vapor in equilibrium with a liquid containing 60 % (mol) ethanol at 308 K?

Ans: 20.22 kPa, 53.57 % ethanol

Explanation:

Problem 8.54: A binary liquid mixture of components A and B containing 80 % (mol) A is in equilibrium with a vapor containing 84.3 % (mol) A at 101.3 kPa and 339 K. Estimate the pressure and composition of the vapor in equilibrium with a liquid containing 50 % A at 339 K. The vapor pressures of A and B at this temperature are 106.6 kPa and 79.97 kPa respectively.

Ans: 93.3 kPa, 57.1 % A

Explanation:

Problem 8.55: At 333 K, compounds A and B each have vapor pressures of 106.63 kPa. The mixture of A and B forms an azeotrope at 333 K and 133.29 kPa and has a composition of 50 % A. (a) Calculate the equilibrium pressure and vapor composition over a liquid solution containing 25 % A. (b) If A and B have equal latent heats of vaporization, how do you expect the azeotropic composition to respond to an increase in temperature?

Ans: (a) 128.6 kPa, 34.25 % A (b) no change

Explanation:

Problem 8.56: At 353 K, compounds A and B each have vapor pressures of 93.30 kPa. At this temperature mixtures of A and B form azeotrope containing 50 % (mol) A and exerts a pressure of 127.96 kPa. Calculate the equilibrium pressure and vapor composition at 353 K over a liquid solution containing 25 % (mol) A.

Ans: 123.21 kPa, 38.54 % A

Explanation:

Problem 8.57: For the binary mixture of A and B the activity coefficients are given by

\ln\left(\gamma_1\right)=Ax_2^2;\;\;\;\ln\left(\gamma_2\right)=Ax_1^2

The vapor pressures of A and B at 353 K are 119.96 kPa and 79.97 kPa respectively. Does an azeotrope exist at 353 K? If so, what is the azeotropic pressure and composition for A = 0.6?

Ans: Azeotrope exit at x_1=0.5\left(1+\frac1A\ln\frac{P_1^S}{P_2^S}\right) , 121.8672 kPa, y1 = 0.8379

Explanation:

Problem 8.58: It is proposed to purify benzene from small amounts of non-volatile impurities by subjecting it to distillation with saturated steam at 99.3 kPa. Calculate the temperature at which distillation will proceed and the weight of steam accompanying 1 kg benzene. The vapor pressure data is given in Example 8.28.

Ans: 342 K, 0.0975 kg

Explanation:

Problem 8.59: At 383 K, a saturated solution of aniline in water contains 7.95 % aniline by weight, and a saturated solution of water in aniline contains 88.05 % aniline by weight. The vapor pressures of pure aniline and of water at 383 K are 9.22 kPa and 143.10 kPa respectively. Construct the Px-y diagram for the mixture at 383 K.

Ans: For aniline-rich phase

P (kPa)9.2242.47575.73108.99142.24146.23
xwater (liquid phase)00.10.20.30.40.412
ywater (vapor phase)00.8050.9030.9410.9610.963

For water-rich phase

P (kPa)146.23145.06144.67144.08143.49143.10
xwater (liquid phase)0.9840.9900.9920.9950.9981.00
ywater (vapor phase)0.9630.9770.9810.9880.9951.00

Explanation:

Problem 8.60: Construct T-x-y diagram for the ether (1)–water (2) system at 101.3 kPa from the following data.

T (K)P (kPa)Water phase x1Ether phase x1 P_2^S  (kPa) P_1^S  (kPa)
307101.30.01230.94565.3799.58
313126.630.01160.94167.39122.78
323176.670.01030.934812.26170.08
333241.200.00930.927119.85230.05
343323.650.00750.921231.00305.72
353428.400.00690.915847.31398.62
363558.570.00580.910770.00510.55
373713.15101.3647.31

Assume that Raoult’s law is valid for ether in the ether phase and for water in the water phase.

Ans: For water-rich phase:

T (K)307313323333343353363
xether (liquid phase)0.01230.009140.00560.00340.00180.000980.00037
yether (vapor phase)0.92950.92770.87960.80470.69450.53340.3092

Explanation:

Problem 8.61: Dimethylaniline is distilled with steam at 90 kPa to free it from non-volatile impurities. Assuming it to be completely immiscible with water determine (a) The distillation temperature (b) The composition of the vapour produced. The vapor pressure data are following:

T (K)343.2374.8399
PS (aniline), kPa1.3335.33213.329
PS (water), kPa31.19106.91285.86

The plot of ln PS versus 1/T may be assumed linear.

Ans: (a) 369.2 K (b) 4.78 % aniline

Explanation:

Problem 8.62: A stream contains 30 % (mol) toluene, 40 % (mol) ethyl benzene, and 30 % (mol) water. Assuming that mixtures of ethylbenzene and toluene obey Raoult’s law and they are completely immiscible in water, calculate the following for a total pressure of 101.3 kPa: (a) The bubble-point temperature and the composition of the vapor (b) The dew-point temperature and the composition of the liquid. The vapor pressure data are given below:

T (K)352.6361.0366.5383.2388.8
PS, kPa (water)46.2964.3379.42
PS, kPa (Toluene)38.4950.6560.7899.27116.50
PS, kPa (ethyl benzene)16.4122.7927.3547.6156.73
Ans: (a) 361.6 K, 21.9 % toluene, 13.1 % ethylbenzene, 65 % water (b) 387.8 K, 26.8 % toluene, 73.2 % ethyl benzene

Explanation:

Problem 8.63: n-Heptane (1) and water (2) are essentially immiscible as liquids. A vapor mixture containing 65 % (mol) water at 373 K and 101.3 kPa is cooled slowly at constant pressure until condensation is complete. Construct a plot for the process showing temperature versus equilibrium mole fraction of heptane in the residual vapor. For n-heptane,

\ln\left(P_1^S\right)=13.8777-\frac{2918.738}{T-56.404}

where P is in kPa and T in K.

Ans:

T (K)361.53359.45357.23354.83353.82352.40
yheptane (vapor phase)0.350.400.450.500.520.548

Explanation:

Problem 8.64: Toluene (1) and water (2) are essentially immiscible in the liquid state. Determine the dewpoint temperature and the composition of the first drops of liquid formed when the vapor mixtures of these species containing (a) 23 % (mol) toluene (b) 77 % (mol) toluene at 101.3 kPa. What are the bubble-point temperature and the composition of the last drop of vapour in each case? The vapour pressure of toluene is

\ln\left(P_1^S\right)=14.00976-\frac{3103.010}{T-53.413}

The vapor pressure of water is

\ln\left(P_2^S\right)=16.287-\frac{3816.44}{T-46.11}
Ans: (a) 366 K, pure water (b) 374.9 K, pure toluene; last drop of vapor contains 44.36 % toluene

Explanation:

Problem 8.65: Components 1 and 2 are insoluble in the liquid phase. Estimate the dew-point temperature and the compositions of the first drop of liquid formed when vapor mixtures of components 1 and 2 containing (a) 75 % (mol) component 1; (b) 25 % (mol) component 1 are cooled at a constant pressure of 101.33 kPa. The vapor pressures of the pure components in kPa are given against temperature in Kelvin in the following table.

T358363368373378383388393398.6
P_1^S 23.5333.5639.5846.5954.6263.9774.3286.34101.33
P_1^S 57.8270.1384.53101.33120.79143.28169.02198.51239.54
Ans: (a) 388.8 K, pure component 1 (b) 365.5 K, pure component 2

Explanation:

Problem 8.66: An experimental determination of the vapor-liquid equilibrium state of ether (1) and acetone (2) gave the following results: x1 = 0.3, y1 = 0.42, T = 313 K and P = 105 Pa
The saturation vapor pressures of the pure components at 313 K are: ether = 1.21×105 Pa and acetone = 0.56×105 Pa. The vapor phase can be assumed ideal. (a) Calculate the liquid-phase activity coefficients. (b) What is the value of excess Gibbs free energy GE/RT for the liquid phase?

Ans: γ1 = 1.1570, γ2 = 1.4796 (b) 0.318

Explanation: