Use the Antoine equation (for pure components) log_10 P^0 = A - B/T+ C where T =

Question

Use the Antoine equation (for pure components) log_10 P^0 = A - B/T+ C where T = temperature [degree C] P degree - saturated vapour pressure [mmHg] Vapour Pressure of pure components Use Excel to prepare a graph. The vapour pressures [kPa] of pure benzene and pure toluene should be on the ordinate of the graph. Plot the values of the pressure on a logarithmic scale. Do not plot the (logarithm of the pressure) on a linear scale The inverse temperature [K^-1] (that is equivalent to the range from 80 degree C to 120 degree C) should be on the abscissa. The graph should have two lines, one for the vapour pressure of pure benzene and one for the vapour pressure of pure toluene. The constants for the Antoine equation are given in Felder and Rousseau Table B4. P-x diagram for a binary mixture of benzene and toluene at 95 degree C. Determine the mole fractions of the two components in the vapour phase that are in equilibrium with a liquid whose mole fraction of toluene is 0.3.

Explanation / Answer

For Benzene, log Psat= 7.2009-1415.8(248.028+t) , t in deg.c

For toluene log Psat= 6.95087- 1342.31/(219.987+t), t in deg.c

the data in the temperature range of 80 to120 deg.c is generated and the plot is shown below

at 95 deg.c, vapo pressures : Benzene =1184.5 mm Hg and toluene = 489

P= x1P1sat + (1-x1) P2sat, where P1sat= Vapor pressur of Benzene and P2sat= Vapor pressure of toluene.

given toluene mole fraction = 0.3 and Benzene mole fractinos =1-0.3=0.7

at x1= 0.7 and x2 =0.3, P= 975.88 ( from the above data(

but y1P= x1p1sat and y1P= 0.7*1184.5=829.15 mm Hg

y1= 829.15/975.88=0.85 and y1= 1-0.85= 0.15

x1 x2= (1-x1) P= (x1psat+x2p2sat) (mm Hg) 0 1 489.0942332 0.1 0.9 558.6361925 0.2 0.8 628.1781518 0.3 0.7 697.7201111 0.4 0.6 767.2620704 0.5 0.5 836.8040296 0.6 0.4 906.3459889 0.7 0.3 975.8879482 0.8 0.2 1045.429908 0.9 0.1 1114.971867 1 0 1184.513826

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