Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
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Chapter 5, Problem 5A.10AE

(i)

Interpretation Introduction

Interpretation: Under the given conditions, the solubility of CO2 gas in water has to be calculated.

Concept introduction: Henry’s law states that the partial vapor pressure of a solute B is directly proportional to the mole fraction of solute B.  Hence, for dilute solution the compositions can be determined by Henry’s law, which is expressed as,

  pB=KB×χB

(i)

Expert Solution
Check Mark

Answer to Problem 5A.10AE

The solubility of CO2 in water at pCO2=0.10atm is 3.37×10-3kg-1mol_.

Explanation of Solution

Carbon dioxide is dissolving in water, which is in small amount.  Therefore, CO2 acts as a solute in water.  Since the amount of CO2 in water is small, it means solution is very dilute.  The Henry’s law for CO2 is expressed as,

  pCO2=KCO2×χCO2                                                                                         (1)

Where,

  • pCO2 is the partial vapor pressure of CO2.
  • χCO2 is the mole fraction of CO2.
  • KCO2 is the Henry’s law constant.

Since, χCO2 is the mole fraction of CO2, it represents the number of moles or amount of CO2 dissolved in water.  Therefore solubility of CO2 in water can be expressed in terms of mole fraction of CO2.

Therefore, the equation (1) can be written as,

  pCO2=KCO2×bCO2                                                                                         (2)

Where,

  • bCO2 is the solubility of CO2 gas in water.

Rearrange the equation (2) in terms of bCO2.

  bCO2=pCO2KCO2                                                                                                   (3)

It is given that,

The value of Henry’s law constant, KCO2 is 3.01×103kPakgmol1.

The Partial pressure, pCO2 is 0.1atm.

The conversion of atm to kPa is done as,

  1atm=101.325kPa

Therefore the conversion of 0.1atm to kPa is done as,

  0.1atm=0.1×101.325kPa=10.1325kPa

Substitute KCO2=3.01×103kPakgmol1 and pCO2=10.1325kPa in equation (3)

   bCO2=10.1325kPa3.01×103kPakgmol1=3.37×10-3kg-1mol_

Therefore, the solubility of CO2 in water at pCO2=0.1atm is 3.37×10-3kg-1mol_.

(ii)

Interpretation Introduction

Interpretation: Under the given conditions, the solubility of CO2 gas in water has to be calculated.

Concept introduction: Henry’s law states that the partial vapor pressure of a solute B is directly proportional to the mole fraction of solute B.  Hence, for dilute solution the compositions can be determined by Henry’s law, which is expressed as,

  pB=KB×χB

(ii)

Expert Solution
Check Mark

Answer to Problem 5A.10AE

The solubility of CO2 in water at pCO2=1atm is 33.7×10-3kg-1mol_.

Explanation of Solution

Carbon dioxide is dissolving in water, which is in small amount.  Therefore, CO2 acts as a solute in water.  Since the amount of CO2 in water is small, it means solution is very dilute.  The Henry’s law for CO2 is expressed as,

  pCO2=KCO2×χCO2                                                                                         (1)

Where,

  • pCO2 is the partial vapor pressure of CO2.
  • χCO2 is the mole fraction of CO2.
  • KCO2 is the Henry’s law constant.

Since, χCO2 is the mole fraction of CO2, it represents the number of moles or amount of CO2 dissolved in water.  Therefore solubility of CO2 in water can be expressed in terms of mole fraction of CO2.

Therefore, the equation (1) can be written as,

  pCO2=KCO2×bCO2                                                                                         (2)

Where,

  • bCO2 is the solubility of CO2 gas in water.

Rearrange the equation (2) in terms of bCO2.

  bCO2=pCO2KCO2                                                                                                   (3)

It is given that,

The value of Henry’s law constant, KCO2 is 3.01×103kPakgmol1.

The Partial pressure, pCO2 is 1atm.

The conversion of atm to kPa is done as,

  1atm=101.325kPa

Substitute the value of KCO2=3.01×103kPakgmol1 and pCO2=101.325kPa in equation (3)

   bCO2=101.325kPa3.01×103kPakgmol1=33.7×10-3kg-1mol_

Therefore, the solubility of CO2 in water at pCO2=1atm is 33.7×10-3kg-1mol_.

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Chapter 5 Solutions

Atkins' Physical Chemistry

Ch. 5 - Prob. 5A.4DQCh. 5 - Prob. 5A.5DQCh. 5 - Prob. 5A.1AECh. 5 - Prob. 5A.1BECh. 5 - Prob. 5A.2AECh. 5 - Prob. 5A.2BECh. 5 - Prob. 5A.3AECh. 5 - Prob. 5A.3BECh. 5 - Prob. 5A.4AECh. 5 - Prob. 5A.4BECh. 5 - Prob. 5A.5AECh. 5 - Prob. 5A.5BECh. 5 - Prob. 5A.6AECh. 5 - Prob. 5A.6BECh. 5 - Prob. 5A.7AECh. 5 - Prob. 5A.7BECh. 5 - Prob. 5A.8AECh. 5 - Prob. 5A.8BECh. 5 - Prob. 5A.9AECh. 5 - Prob. 5A.9BECh. 5 - Prob. 5A.10AECh. 5 - Prob. 5A.10BECh. 5 - Prob. 5A.11AECh. 5 - Prob. 5A.11BECh. 5 - Prob. 5A.1PCh. 5 - Prob. 5A.3PCh. 5 - Prob. 5A.4PCh. 5 - Prob. 5A.5PCh. 5 - Prob. 5A.6PCh. 5 - Prob. 5A.7PCh. 5 - Prob. 5B.1DQCh. 5 - Prob. 5B.2DQCh. 5 - Prob. 5B.3DQCh. 5 - Prob. 5B.4DQCh. 5 - Prob. 5B.5DQCh. 5 - Prob. 5B.6DQCh. 5 - Prob. 5B.7DQCh. 5 - Prob. 5B.1AECh. 5 - Prob. 5B.1BECh. 5 - Prob. 5B.2AECh. 5 - Prob. 5B.2BECh. 5 - Prob. 5B.3AECh. 5 - Prob. 5B.3BECh. 5 - Prob. 5B.4AECh. 5 - Prob. 5B.4BECh. 5 - Prob. 5B.5AECh. 5 - Prob. 5B.5BECh. 5 - Prob. 5B.6AECh. 5 - Prob. 5B.6BECh. 5 - Prob. 5B.7AECh. 5 - Prob. 5B.7BECh. 5 - Prob. 5B.8AECh. 5 - Prob. 5B.8BECh. 5 - Prob. 5B.9AECh. 5 - Prob. 5B.9BECh. 5 - Prob. 5B.10AECh. 5 - Prob. 5B.10BECh. 5 - Prob. 5B.11AECh. 5 - Prob. 5B.11BECh. 5 - Prob. 5B.12AECh. 5 - Prob. 5B.12BECh. 5 - Prob. 5B.1PCh. 5 - Prob. 5B.2PCh. 5 - Prob. 5B.3PCh. 5 - Prob. 5B.4PCh. 5 - Prob. 5B.5PCh. 5 - Prob. 5B.6PCh. 5 - Prob. 5B.9PCh. 5 - Prob. 5B.11PCh. 5 - Prob. 5B.13PCh. 5 - Prob. 5C.1DQCh. 5 - Prob. 5C.2DQCh. 5 - Prob. 5C.3DQCh. 5 - Prob. 5C.1AECh. 5 - Prob. 5C.1BECh. 5 - Prob. 5C.2AECh. 5 - Prob. 5C.2BECh. 5 - Prob. 5C.3AECh. 5 - Prob. 5C.3BECh. 5 - Prob. 5C.4AECh. 5 - Prob. 5C.4BECh. 5 - Prob. 5C.1PCh. 5 - Prob. 5C.2PCh. 5 - Prob. 5C.3PCh. 5 - Prob. 5C.4PCh. 5 - Prob. 5C.5PCh. 5 - Prob. 5C.6PCh. 5 - Prob. 5C.7PCh. 5 - Prob. 5C.8PCh. 5 - Prob. 5C.9PCh. 5 - Prob. 5C.10PCh. 5 - Prob. 5D.1DQCh. 5 - Prob. 5D.2DQCh. 5 - Prob. 5D.1AECh. 5 - Prob. 5D.1BECh. 5 - Prob. 5D.2AECh. 5 - Prob. 5D.2BECh. 5 - Prob. 5D.3AECh. 5 - Prob. 5D.3BECh. 5 - Prob. 5D.4AECh. 5 - Prob. 5D.4BECh. 5 - Prob. 5D.5AECh. 5 - Prob. 5D.5BECh. 5 - Prob. 5D.6AECh. 5 - Prob. 5D.1PCh. 5 - Prob. 5D.2PCh. 5 - Prob. 5D.3PCh. 5 - Prob. 5D.4PCh. 5 - Prob. 5D.5PCh. 5 - Prob. 5D.6PCh. 5 - Prob. 5D.7PCh. 5 - Prob. 5E.1DQCh. 5 - Prob. 5E.2DQCh. 5 - Prob. 5E.3DQCh. 5 - Prob. 5E.4DQCh. 5 - Prob. 5E.1AECh. 5 - Prob. 5E.1BECh. 5 - Prob. 5E.2AECh. 5 - Prob. 5E.2BECh. 5 - Prob. 5E.3AECh. 5 - Prob. 5E.3BECh. 5 - Prob. 5E.4AECh. 5 - Prob. 5E.4BECh. 5 - Prob. 5E.5AECh. 5 - Prob. 5E.5BECh. 5 - Prob. 5E.1PCh. 5 - Prob. 5E.2PCh. 5 - Prob. 5E.3PCh. 5 - Prob. 5F.1DQCh. 5 - Prob. 5F.2DQCh. 5 - Prob. 5F.3DQCh. 5 - Prob. 5F.4DQCh. 5 - Prob. 5F.5DQCh. 5 - Prob. 5F.1AECh. 5 - Prob. 5F.1BECh. 5 - Prob. 5F.2AECh. 5 - Prob. 5F.2BECh. 5 - Prob. 5F.3AECh. 5 - Prob. 5F.3BECh. 5 - Prob. 5F.4AECh. 5 - Prob. 5F.4BECh. 5 - Prob. 5F.5AECh. 5 - Prob. 5F.5BECh. 5 - Prob. 5F.6AECh. 5 - Prob. 5F.6BECh. 5 - Prob. 5F.7AECh. 5 - Prob. 5F.7BECh. 5 - Prob. 5F.8AECh. 5 - Prob. 5F.8BECh. 5 - Prob. 5F.1PCh. 5 - Prob. 5F.2PCh. 5 - Prob. 5F.3PCh. 5 - Prob. 5F.4PCh. 5 - Prob. 5.1IACh. 5 - Prob. 5.2IACh. 5 - Prob. 5.3IACh. 5 - Prob. 5.4IACh. 5 - Prob. 5.5IACh. 5 - Prob. 5.6IACh. 5 - Prob. 5.8IACh. 5 - Prob. 5.9IACh. 5 - Prob. 5.10IA
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY