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A mixture of 4.00 gram-moles of benzene and 6.00 gram-moles of toluene are placed in a...

A mixture of 4.00 gram-moles of benzene and 6.00 gram-moles of toluene are placed in a closed cylinder 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 1300 mm Hg and is gradually lowered to 600 mm Hg. Use the Pxy diagram below to convince yourself that the cylinder initially contains only liquid benzene and toluene and to answer the following questions. Carefully read the diagram. The solution can be sensitive to small errors.

At what pressure does the first vapor bubble form?    in mm Hg. What is its composition?    mol benzene/mol

At what pressure does the last droplet of liquid evaporate? in mm Hg. What is its composition?    mol benzene/mol

What are the liquid and vapor compositions in equilibrium with each other when the pressure is 800 mm Hg?

Vapor:    mol benzene/mol

Liquid:    mol benzene/mol .

What is the ratio (mole vapor/mole liquid) at this point?

Estimate the volume of the cylinder contents when the pressure is 1300 mm Hg, 800 mm of Hg, and 600 mm Hg.

1300 mm Hg: in L

800mm of Hg: in L

600 mm Hg: in L

Solutions

Expert Solution

First of all, we need to start with of the diagram Pxy of the mixture.

*Since T is constant, we calculate the slopes of both components; it could be calculated using the Antoine equation.

Where A, B and C are constant

Benzene

A= 6.90565

B=1211.033

C= 220.79

Toluene

A=6.95464

B=1344.8

C= 219.482.

Psat Bezene = 1350.49mmHg and Psat toluene= 556.32mmHg

*We assume the values of the composition of the liquid phase of the component more volatile (x1) is between 0 and 1.

*We calculate the P of the system using Raoult's Law by placing it in X1 function

* With each value of x1 we calculate the corresponding pressure

* We calculate all x1 and calculate the y1 by Raoult

*We plot x1 vs. P and y1 vs. P, obtaining the bubble and dew curves
respectively

X1=0.6 vap bubble Pressure = 874 mm Hg

Dew Pressure = 1033 mmHg

P=800 mmHg

X=0.3 Vap See in the dew curve

X= 0.51 liq See in the Bubble curve

queen_honey_blossom answered 2 years ago
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