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

Answers

one state would change   like louisiana to texas. thats a change

explanation:

it is called acid bonding

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

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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