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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

Respuesta :

Answer:

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

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