Benzene (l) has a vapor pressure of 0.1269 bar at 298.15 K and an enthalpy of va

Question

Benzene (l) has a vapor pressure of 0.1269 bar at 298.15 K and an enthalpy of vaporization of 30.72kJmol^-1. The Cpm of the vapor and liquid phases at that temperature are 82.4 and 136.0 J K^-1 mol^-1, respectively. Calculate the vapor pressure of C6H6(l) at 340.0 K assuming

a. that the enthalpy of vaporization does not change with temperature. Answer: 0.583 bar

b. that the enthalpy of vaporization at temperature T can be calculated from the equation deltaHvaporization (T)= deltaHvaporization (T0) + deltaCp(T-T0) assuming that deltaCp does not change with temperature. Answer: 0.553 bar

The answers are given, I need a detailed solution.

Explanation / Answer

a)ln(p1/p2)=dH/R (1/T2 - 1/T1)

ln(0.1269/p2)=30.72*10^3/ 8.314 (1/340 - 1/298.15) =-1.53

0.1269/p2=0.22

p2=0.1269/0.22=0.583 bar

b)deltaHvaporization (T)= deltaHvaporization (T0) + deltaCp(T-T0)

= 30720+( 82.4-130)(1/340-1/298.15)

=28.48 KJ/mole

ln(p1/p2)=dH/R (1/T2 - 1/T1)

ln(0.1269/p2)=34.17*10^3/ 8.314 (1/340 - 1/298.15) =-1.414

0.1269/p2=0.2432

p2=0.1269/0.2432=0.523 bar

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

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