Question

The normal boiling point of benzene is 80.10 °C.
At 25.00 °C, the vapor pressure of benzene is 0.1253 atm.
What is the vapor pressure of benzene when the temperature is 75.00 °C?

Answer 1

Step 1:

T1 = 80.1 oC

=(80.1 + 273)K

= 353.1 K

T2 = 25 oC

=(25 + 273)K

= 298 K

P1 = 1 atm

P2 = 0.1253 atm

use:

ln(P2/P1) = (ΔH/R)*(1/T1 - 1/T2)

ln(0.1253/1) = ( ΔH/8.314)*(1/353.1 - 1/298.0)

-2.077 = (ΔH/8.314)*(-5.236*10^-4)

ΔH = 32978 J/mol

Step 2:

T1 = 80.1 oC

=(80.1 + 273)K

= 353.1 K

T2 = 75 oC

=(75 + 273)K

= 348 K

P1 = 1 atm

ΔH = 32978 J/mol

use:

ln(P2/P1) = (ΔH/R)*(1/T1 - 1/T2)

ln(P2/1) = (32978.0/8.314)*(1/353.1 - 1/348.0)

ln(P2/1) = 3967*(-4.15*10^-5)

P2 = 0.8482 atm

Answer: 0.8482 atm