Ticketing Center Pet x W 3743 x C using Raoult\'s Law Fo x MICHEM-ENO 120.11 xy

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Ticketing Center Pet x W 3743 x C using Raoult's Law Fo x MICHEM-ENO 120.11 xy d wileypuus Moodle UMass Ami C Secure https ledugenwileyplus.com/edugen/student/mainfr.unl ley PLUS Felder, Rousseau, Bullard, Elementary Principles of Chemical Processes, 4e CHEMICAL ENGR FUNDMENTALs (CHEMENG120) Read, Study & Practice ont Open Assignment ENT RESOURCES Problem 6.69 Benzene/Toluene Pxy mixture of 0 gram moles of benzene and 2.00 gram-moles of toluene are placed in a closed cyinder equipped with a piston The cylinder is immersed in a bath of boiling water that maintains the temperature at 100.0°C. The force exerted on the piston can be varied to adjust the cylinder pressure to any desired values. The pressure is initially 130o mm Hg and is gradualy lowered to 600 mm Hg. Use the Po diagram below to convince yourself that the cylinder initially contains only liquid benzene and toluene and to answer the following questions. 115 110 1400 H 105 1300 1200 G 100 1100 95 E 1000 85 75 70 65 0 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0 Mole fraction benzene Mole fraction benzene Vension 4.22.2. ent I Privacy. Policy I e2000-201zuohn Sons unc. All Rights Reserved A Division of zehn wiler s sons Inc.

Explanation / Answer

The mole fraction of the initial solution will be

xb = liquid mole fraction of benzene = 8/(2+8) = 0.8

xt = liquid mole fraction of toluene = 2/(2+8) = 0.2

As given in the question, the container is kept at 100C. So if we read the Temperature vs mole fraction graph at 100C & move in the horizontal direction, the xb will 0.25 & yb (vapor mole fraction of benzene) will be 0.5

Problem 1.

yb = 0.5 & xb = 0.25

So if we read the Pressure vs mole fraction graph, first vapor will be formed at 750 mm Hg & mole fraction will be xb = 0.25. At xb = 0.25 if we move vertically & intersectt the liquid graph & move horizontally then we get 750 mm Hg. This is the pressure when first vapor drop will be formed.

Problem 2

The last drop will be when all the liquid will be vaporised. So from the question we know that at xb =0.25 and move vertically upwards till we intersect vapor line & then move horizontally, we will get the 650 mm Hg.

Problem 3

To find the vapor & liquid composition at a particular pressure, i.e. 1150 mm Hg, we will move horizontally at that pressure till we intersect the liquid & vapor lines.

From the intersection point of the liquid line, move vertically downwards & we get the liquid composition xb = 0.75

From the intersection point of the vapor line, move vertically downwards & we get the vapor composition yb = 0.85

Ratio of mole vapor / mole liquid = 0.85 / 0.75 = 1.133

Problem 4

We will use PV = nRT equation for calculating the volume of the cylinder.

n = we will assume the moles to be 1 mole in the cylinder & R = 62.363 L mm Hg K-1 mol-1

So at pressure 1300 mm Hg, the temperature will be will be 80C = 353 K (as at 1300 mm Hg xb = 1, so the temperature will be 85C)

So V1300mmHg = 1*62.363*353/1300 = 16.93 liter

At pressure 1150 mm Hg, temperature will be 85C = 358K

So V1150mmHg = 1*62.363*358/1150 = 19.41 liter

At pressure 600 mmHg, temperature will be 108C = 381K

So V600mmHg = 1*62.363*381/600 = 39.6 liter

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