please help?
2020-05-06 10:41 am
A certain substance has a heat of vaporization of 66.99 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.00 times higher than it was at 363 K? T=......K?
回答 (1)
2020-05-06 11:13 am
✔ 最佳答案
liquid ⇌ vapor K = v.p.K₁/K₂ = (v.p.1/v.p.2) = 1/5
T₁ = 363 K
T₂ = T
ΔH = 66.99 kJ/mol = 66990 J/mol
R = 8.314 J / (mol / K)
ln(K₁/K₂) = (ΔH/R) [(1/T₂) - (1/T₁)]
ln(1/5) = (66990/8.314) [(1/T) - (1/363)]
(1/T) - (1/363) = (8.314/66990) ln(1/5)
1/T = (8.314/66990) ln(1/5) + (1/363)
T = 1 / [(8.314/66990) ln(1/5) + (1/363)]
T = 391 K
收錄日期: 2021-04-24 07:52:41
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