Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
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Chapter 4, Problem 4D.6E
Interpretation Introduction

Interpretation:

It should be calculated that the equilibrium concentration of carbon dioxide in fat given that the Henry’s law constant is 8.6×104Torr and the partial pressure of carbon dioxide is 55kPa.

Concept introduction:

Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

Partial pressure of a gas in terms of its mole fraction and total pressure is,

PA=χA×PTOTAL

Henry’s law states that the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.

p=KHxwhere,KH:Henry'sconstantx:molefractionp:partialpressure

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture. Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  moleculefractionofA,(χA=numbersof moles ofmolecule A(nA)total number of moles(nA+nB+nC

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Chapter 4 Solutions

Elements Of Physical Chemistry

Ch. 4 - Prob. 4D.1STCh. 4 - Prob. 4D.2STCh. 4 - Prob. 4D.3STCh. 4 - Prob. 4D.4STCh. 4 - Prob. 4E.1STCh. 4 - Prob. 4E.2STCh. 4 - Prob. 4F.1STCh. 4 - Prob. 4A.1ECh. 4 - Prob. 4A.2ECh. 4 - Prob. 4A.3ECh. 4 - Prob. 4A.4ECh. 4 - Prob. 4A.5ECh. 4 - Prob. 4A.6ECh. 4 - Prob. 4B.1ECh. 4 - Prob. 4B.2ECh. 4 - Prob. 4B.3ECh. 4 - Prob. 4B.4ECh. 4 - Prob. 4B.5ECh. 4 - Prob. 4B.6ECh. 4 - Prob. 4B.7ECh. 4 - Prob. 4B.8ECh. 4 - Prob. 4C.1ECh. 4 - Prob. 4C.2ECh. 4 - Prob. 4C.3ECh. 4 - Prob. 4C.4ECh. 4 - Prob. 4C.5ECh. 4 - Prob. 4C.6ECh. 4 - Prob. 4C.7ECh. 4 - Prob. 4D.1ECh. 4 - Prob. 4D.2ECh. 4 - Prob. 4D.3ECh. 4 - Prob. 4D.4ECh. 4 - Prob. 4D.5ECh. 4 - Prob. 4D.6ECh. 4 - Prob. 4D.7ECh. 4 - Prob. 4D.8ECh. 4 - Prob. 4D.9ECh. 4 - Prob. 4D.10ECh. 4 - Prob. 4D.11ECh. 4 - Prob. 4D.12ECh. 4 - Prob. 4D.13ECh. 4 - Prob. 4E.1ECh. 4 - Prob. 4E.2ECh. 4 - Prob. 4E.3ECh. 4 - Prob. 4E.4ECh. 4 - Prob. 4F.1ECh. 4 - Prob. 4F.2ECh. 4 - Prob. 4F.3ECh. 4 - Prob. 4.1DQCh. 4 - Prob. 4.2DQCh. 4 - Prob. 4.3DQCh. 4 - Prob. 4.4DQCh. 4 - Prob. 4.5DQCh. 4 - Prob. 4.6DQCh. 4 - Prob. 4.8DQCh. 4 - Prob. 4.9DQCh. 4 - Prob. 4.10DQCh. 4 - Prob. 4.11DQCh. 4 - Prob. 4.12DQCh. 4 - Prob. 4.13DQCh. 4 - Prob. 4.14DQCh. 4 - Prob. 4.15DQCh. 4 - Prob. 4.1PCh. 4 - Prob. 4.2PCh. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Prob. 4.5PCh. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - Prob. 4.11PCh. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Prob. 4.16PCh. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Prob. 4.22PCh. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Prob. 4.29PCh. 4 - Prob. 4.30PCh. 4 - Prob. 4.1PRCh. 4 - Prob. 4.2PRCh. 4 - Prob. 4.3PRCh. 4 - Prob. 4.4PRCh. 4 - Prob. 4.5PRCh. 4 - Prob. 4.6PR
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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