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In For benzene (C6H6), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa? P₁ AHvap P₁ = 1.00 atm x 101.325 kPa/atm = 101.325 kPa 1 1 P2 R T2 T₁ T₁ = 80.1 °C + 273.15 = 353.25 K P2 = 83.4 kPa AH vap = 3.08 x 104 J/mol R = 8.314 J/mol K In 101.325 = 3.08 x 104 1 1 => 83.4 8.314 T₂ 353.25 0.19468 = 3.7046 x 103 (1/T2 - 2.831 x 10-3) 5.2551 x 10-5 = 1/T2 - 2.831 x 10-3 2.8836 x 10-3 = 1/T2 T₁ = 346.8 K = 73.6 °C

In For benzene (C6H6), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa? P₁ AHvap P₁ = 1.00 atm x 101.325 kPa/atm = 101.325 kPa 1 1 P2 R T2 T₁ T₁ = 80.1 °C + 273.15 = 353.25 K P2 = 83.4 kPa AH vap = 3.08 x 104 J/mol R = 8.314 J/mol K In 101.325 = 3.08 x 104 1 1 => 83.4 8.314 T₂ 353.25 0.19468 = 3.7046 x 103 (1/T2 - 2.831 x 10-3) 5.2551 x 10-5 = 1/T2 - 2.831 x 10-3 2.8836 x 10-3 = 1/T2 T₁ = 346.8 K = 73.6 °C

General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter6: Thermochemisty
Section: Chapter Questions
Problem 6.83QP: The cooling effect of alcohol on the skin is due to its evaporation. Calculate the heat of...
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Can you explain this problem? Where did the P1 came from? And can you redo the calculation?

In For benzene (C6H6), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa? P₁ AHvap P₁ = 1.00 atm x 101.325 kPa/atm = 101.325 kPa 1 1 P2 R T2 T₁ T₁ = 80.1 °C + 273.15 = 353.25 K P2 = 83.4 kPa AH vap = 3.08 x 104 J/mol R = 8.314 J/mol K In 101.325 = 3.08 x 104 1 1 => 83.4 8.314 T₂ 353.25 0.19468 = 3.7046 x 103 (1/T2 - 2.831 x 10-3) 5.2551 x 10-5 = 1/T2 - 2.831 x 10-3 2.8836 x 10-3 = 1/T2 T₁ = 346.8 K = 73.6 °C
Transcribed Image Text:In For benzene (C6H6), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa? P₁ AHvap P₁ = 1.00 atm x 101.325 kPa/atm = 101.325 kPa 1 1 P2 R T2 T₁ T₁ = 80.1 °C + 273.15 = 353.25 K P2 = 83.4 kPa AH vap = 3.08 x 104 J/mol R = 8.314 J/mol K In 101.325 = 3.08 x 104 1 1 => 83.4 8.314 T₂ 353.25 0.19468 = 3.7046 x 103 (1/T2 - 2.831 x 10-3) 5.2551 x 10-5 = 1/T2 - 2.831 x 10-3 2.8836 x 10-3 = 1/T2 T₁ = 346.8 K = 73.6 °C
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