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? 101.325 kPa In P₁ P2 AHvap 1 R T₂ T₁ Hvap P₁ = 1.00 atm x 101.325 kPa/atm T₁ = 80.1 °C + 273.15 = 353.25 K 1 P₂ = 83.4 kPa = 3.08 x 104 J/mol In 101.325 325 83.4 = 3.08 x 104 8.314 1 T₂ 2 R = 8.314 J/mol K 1 353.25 - => 0.19468 = 3.7046 x 103 (1/T₂ - 2.831 x 10-³) 2 5.2551 x 10-5 = 1/T₂ - 2.831 x 10-³ 2.8836 x 10-3 = 1/T₂ T₂ = 346.8 K = 73.6 °C T2
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? 101.325 kPa In P₁ P2 AHvap 1 R T₂ T₁ Hvap P₁ = 1.00 atm x 101.325 kPa/atm T₁ = 80.1 °C + 273.15 = 353.25 K 1 P₂ = 83.4 kPa = 3.08 x 104 J/mol In 101.325 325 83.4 = 3.08 x 104 8.314 1 T₂ 2 R = 8.314 J/mol K 1 353.25 - => 0.19468 = 3.7046 x 103 (1/T₂ - 2.831 x 10-³) 2 5.2551 x 10-5 = 1/T₂ - 2.831 x 10-³ 2.8836 x 10-3 = 1/T₂ T₂ = 346.8 K = 73.6 °C T2
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Question
Transcribed Image Text:Estimating Temperature (or Vapor Pressure)
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?
101.325 kPa
In
P₁
P2
AHvap
1
R T₂ T₁
P₁ = 1.00 atm x 101.325 kPa/atm
T₁ = 80.1 °C + 273.15 = 353.25 K
1
In 101.325
83.4
P₂ = 83.4 kPa
P2
AHvap = 3.08 x 104 J/mol
= 3.08 x 104
8.314
1
T2
R = 8.314 J/mol K
1
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
T₂ = 346.8 K = 73.6 °C
T2
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