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The vapor pressure of ethanol is 100 mmHg at 34.9°C. What is its vapor pressure at 63.5 °C? (AHvap for ethanol = 39.3 kJ/mol) P₁ AHvap (1 In P2 R T2 T₁ P₁ = 100 mm Hg/760 mm Hg atm-1 = 0.1316 atm T₁ = 34.9+273.15 = 308.1 K T₂ = 63.5+ 273.15 = 336.7 K = AH vap = 3.93 x 104 J/mol In P₁ = AH AHvap R 1 1 – 1 T₁₂ T 2 R 8.314 J/mol K P2 In = x => (0.3316-3831410873001)- P2 In (0.1316) - In(P2) = 4.727 x 103 (-2.75697 x 10-4) -In(P2) = 0.72477 => P2 = e-0.72477 => In(P2) = -0.72477 => P2 = 0.4844 atm = 368 mm Hg Search

The vapor pressure of ethanol is 100 mmHg at 34.9°C. What is its vapor pressure at 63.5 °C? (AHvap for ethanol = 39.3 kJ/mol) P₁ AHvap (1 In P2 R T2 T₁ P₁ = 100 mm Hg/760 mm Hg atm-1 = 0.1316 atm T₁ = 34.9+273.15 = 308.1 K T₂ = 63.5+ 273.15 = 336.7 K = AH vap = 3.93 x 104 J/mol In P₁ = AH AHvap R 1 1 – 1 T₁₂ T 2 R 8.314 J/mol K P2 In = x => (0.3316-3831410873001)- P2 In (0.1316) - In(P2) = 4.727 x 103 (-2.75697 x 10-4) -In(P2) = 0.72477 => P2 = e-0.72477 => In(P2) = -0.72477 => P2 = 0.4844 atm = 368 mm Hg Search

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The vapor pressure of ethanol is 100 mmHg at 34.9°C. What is its vapor pressure at 63.5 °C? (AHvap for ethanol = 39.3 kJ/mol) P₁ AHvap (1 In P2 R T2 T₁ P₁ = 100 mm Hg/760 mm Hg atm-1 = 0.1316 atm T₁ = 34.9+273.15 = 308.1 K T₂ = 63.5+ 273.15 = 336.7 K = AH vap = 3.93 x 104 J/mol In P₁ = AH AHvap R 1 1 – 1 T₁₂ T 2 R 8.314 J/mol K P2 In = x => (0.3316-3831410873001)- P2 In (0.1316) - In(P2) = 4.727 x 103 (-2.75697 x 10-4) -In(P2) = 0.72477 => P2 = e-0.72477 => In(P2) = -0.72477 => P2 = 0.4844 atm = 368 mm Hg Search
Transcribed Image Text:The vapor pressure of ethanol is 100 mmHg at 34.9°C. What is its vapor pressure at 63.5 °C? (AHvap for ethanol = 39.3 kJ/mol) P₁ AHvap (1 In P2 R T2 T₁ P₁ = 100 mm Hg/760 mm Hg atm-1 = 0.1316 atm T₁ = 34.9+273.15 = 308.1 K T₂ = 63.5+ 273.15 = 336.7 K = AH vap = 3.93 x 104 J/mol In P₁ = AH AHvap R 1 1 – 1 T₁₂ T 2 R 8.314 J/mol K P2 In = x => (0.3316-3831410873001)- P2 In (0.1316) - In(P2) = 4.727 x 103 (-2.75697 x 10-4) -In(P2) = 0.72477 => P2 = e-0.72477 => In(P2) = -0.72477 => P2 = 0.4844 atm = 368 mm Hg Search
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