Ethanol has a heat of vaporization of 38.56 kJ/molk and a normal boiling point of 78.4 ∘C. What is the vapor pressure of ethanol at 19 ∘C∘C?

Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter9: Liquids, Solids, And Materials
Section: Chapter Questions
Problem 29QRT: The vapor pressure of ethanol, C2H5OH, at 50.0 C is 233 mmHg, and its normal boiling point at 1 atm...
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Ethanol has a heat of vaporization of 38.56 kJ/molk and a normal boiling point of 78.4 ∘C.

What is the vapor pressure of ethanol at 19 ∘C∘C?
Expert Solution
Step 1

Given that,

Ethanol has a heat of vaporization of Hvap = 38.56 kJ/mol = (38.56×103) J/mol = 38560 J/mol.

The boiling point of ethanol is T1 = 78.4oC = (78.4+273) K = 351.4 K.

We have to calculate the the vapor pressure of ethanol at 19oC i.e., T2 = 19oC = (273+19) K = 292 K.

Introduction: We know that boiling point is the temperature at which vapour pressure becomes equal to the atmospheric pressure.

FormulalnP2P1 = HvapRT2-T1T1T2, where R is the universal constant = 8.314 J/(mol.K).

 

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