Water has a vapor pressure of 23.8 mm Hg at 25°C and a heat of vaporization of 40.657 kJ/mol. Using the Clausius-Clapeyron equation given below, determine the vapor pressure of water at 96°C.

ln

P2

P1

=

-ΔHvap

R

1

T2

-

1

T1

P = 559.553 mmHg

Explanation:

Clasius-Clapeyron:

Ln (P2/P1) = - ΔHv/R [ 1/T2 - 1/T1 ]

∴ P1 = 23.8 mmHg = 3.173 KPa

∴ T1 = 25°C ≅ 298 K

∴ ΔHv = 40.657 KJ/mol

∴ R = 8.314 E-3 KJ/K. mol

∴ T2 = 96°C ≅ 369 K

⇒ Ln P2/P1 = - (40.657 KJ/mol/8.314 E-3 KJ/K, mol) [ (1/369 K) - (1/298 K) ]

⇒ Ln P2/P1 = - (4890.185 K) [ - 6.457 E-4 K-1 ]

⇒ Ln P2/P1 = 3.1575

⇒ P2/P1 = 23.511

⇒ P2 = (23.511) (3.173 KPa)

⇒ P2 = 74.601 KPa = 559.553 mmHg