   # The molar enthalpy of vaporization of water at 373 K and 1.00 atm is 40.7 kJ/mol. What fraction of this energy is used to change the internal energy of the water, and what fraction is used to do work against the atmosphere? (Hint: Assume that water vapor is an ideal gas.) ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 9, Problem 144CP
Textbook Problem
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## The molar enthalpy of vaporization of water at 373 K and 1.00 atm is 40.7 kJ/mol. What fraction of this energy is used to change the internal energy of the water, and what fraction is used to do work against the atmosphere? (Hint: Assume that water vapor is an ideal gas.)

Interpretation Introduction

Interpretation

Amount of energy used to change internal energy of water and what amount has used to do work against atmosphere to be discussed.

Concept Introduction

Internal energy change: Change in internal energy of the system is sum of heat added to the system and work done by the system.

ΔE = q+w

Where,

q - heat of vaporisation

w-workdonebythesystem

Ideal gas law: This law expressed by following equation is,

PV = nRT

Where,

P - PressureV - volumen - Number of molesR - GasconstantT -Temperature

### Explanation of Solution

Explanation

Record the given data

Enthalpy of vaporization of water = 40.7 kJ/mol (373 K and 1atm)

Enthalpy of vaporization of water is recorded as shown.

To calculate volume of water required to 1 mol vapor.

Considering water is an ideal solution.

V373=nRTP=1.00 mol×0.8206 L atmK mol ×373K1.00 atm

= 30.6 L

By plugging the value of moles of water, universal gas law constant and temperature into ideal gas equation, the volume of water required to one mole vapor has calculated.

To calculate work required to vaporization.

Given: density of water = 1.00g/cm3

Therefore one mole H2O occupies 18cm3 or 0.0180 L

w = - PΔV

w =1.00 atm (30.6 L- 0.0180 L) = 30.6 L atm

w= -30

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