Solutions to some of the problems of this chapter require vapor pressure as a fu

Question

Solutions to some of the problems of this chapter require vapor pressure as a function of temperature for species which constitute systems in VLE. Table B.2. Appendix B, lists parameter values for the Antoine equation. In P^sat/kPa = A - B/t/degree C + C Assuming the validity of Raoult's law. do the following calculations for the benzene (1)/toluene(2)system: (a) Given x_1 = 0.33 and T = 100 degree C, find y_1 and P. (b) Given y_1 = 0.33 and T = 100 degree C, find x_1 and P. (c) Given x_1 = 0.33 and P = 120 kpa, find y_1 and T. (d) Given y_1 = 0.33 and P = 120 kPa, find x_1 and T. (e) Given T = 105 degree C and P = 120 kPa, find x_1 and y_1. (f) For part (e), if the overall mole fraction of benzene is z_1 = 0.33. what molar fraction of the two-phase system is vapor? (g) Why is Raoult's law likely to be an excellent VLE model for this system at the stated (or computed) conditions?

Explanation / Answer

I do not have access to the appendix section of your text; however, I will use values of A, B and C obtained from internet sources.

(a) and (b) For benzene(1), A = 6.90565, B = 1211.033 and C = 220.790; for toluene(2), A = 6.95464, B = 1344.800 and C = 219.482.

Therefore, at 100°C, we have for benzene(1) and toluene(2),

P1sat = exp[6.90565 – 1211.0363/(100 + 220.790] = exp[6.90565 – 3.77516] = exp(3.13049) = 22.8852 kPa

P2sat = exp[6.95464 – 1344.800/(100 + 219.482)] = exp[6.95464 – 4.20931] = 15.5697 kPa.

(a) x1 = 0.33, T = 100°C

As per Raoult’s law, P1 = x1.P1sat = (0.33)*(22.8852 kPa) = 7.552 kPa 7.55 kPa.

Again, x1 + x2 = 1

===> 0.33 + x2 = 1

===> x2 = 0.67

Therefore, P2 = x2.P2sat = (0.67)*(15.5697 kPa) = 10.432 kPa 10.43 kPa.

P = P1 + P2 = (7.55 + 10.43) kPa = 17.98 kPa (ans).

y1 = P1/P = 7.55 kPa/17.98 kPa = 0.419 0.42 (ans).

(b) Given y1 = 0.33; therefore, y2 = 1 – y1 = (1-0.33) = 0.67

Also, y1P/P1sat + y2P/P2sat = 1

===> 0.33*P/(22.8852 kPa) + 0.67*P/(15.5697 kPa) = 1

===> (0.0144*P + 0.0430*P) kPa-1 = 1

===> 0.0574*P kPa-1 = 1

===> P = 1/(0.0574) kPa = 17.42 kPa (ans)

y1 = P1/P

===> 0.33 = P1/17.42 kPa

===> P1 = 0.33*17.42 kPa = 5.7486 kPa

P1 = x1.P1sat

===> (5.7486 kPa) = x1.(22.8852 kPa)

===> x1 = 5.7486 kPa/22.8852 kPa = 0.25 (ans)

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.