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
Question
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Chapter 4, Problem 4.5PR

(a)

Interpretation Introduction

Interpretation:

The volume of oxygen is given up by 100cm3 of blood flowing from the lungs in the capillary has to be determined.

(a)

Expert Solution
Check Mark

Answer to Problem 4.5PR

The volume of oxygen is given up by 100cm3 of blood flowing from the lungs in the capillary has been calculated to be 14.62275cm3.

Explanation of Solution

Heamoglobin concentration in normal blood is 150gdm3 which implies that in 1L of blood there is 150g of Hb.  So in 100cm3 of blood, the amount of Hb will be

  150g1000×100=15g.

1g of Heamoglobin binds about 1.34cm3 of oxygen per gram.  So 15g of Hb will bind

  15×1.34cm3=20.1cm3.

Which also implies that 100cm3 of blood binds 20.1cm3 of oxygen.

Heamoglobin in the lungs is about 97% saturated with oxygen, while in the capillary is only 75% saturated.

According to the question, 100cm3 of blood is flowing from lungs to capillary.

  LungCapillary

In lung, 100cm3 of blood will bind 19.497cm3 of oxygen,

  20.1cm3×97100=19.497cm3.

In capillary, 100cm3 of blood from lung will bind 14.62275cm3 of oxygen,

  19.497cm3×75100=14.62275cm3.

The amount of oxygen taken up by the body tissues will be (19.49714.62275)cm3=4.87425cm3.

Therefore, the volume of oxygen is given up by 100cm3 of blood flowing from the lungs in the capillary has been calculated to be 4.87425cm3.

(b)

Interpretation Introduction

Interpretation:

The mass of nitrogen dissolved in 100g of water saturated with air at 4.0atmand20C has to be calculated.

Concept Introduction:

Henry’s law:

The vapour pressure of a volatile solute A is proportional to its mole fraction in a solution. 

Mathematically, it can be represented as

pA=KH'XA

Where,

KH'= Henry’s law constant

XA= Mole fraction of a volatile solute A

pA= Vapour pressure of a volatile solute A

Another version of Henry’s law can be mathematically represented as

C=KHp

Where,

KH'= Henry’s law constant

C= Concentration of the dissolved gas

p= partial pressure of the gas

(b)

Expert Solution
Check Mark

Answer to Problem 4.5PR

The mass of nitrogen dissolved in 100g of water saturated with air at 4.0atmand1.0atm has been calculated to be 56.2167μg and 14.0544μg respectively.

Explanation of Solution

According to the given question, in 1g of water the amount of nitrogen dissolves is 0.18μg. So in 100g of water the amount of dissolved nitrogen will be 0.18μg×100=18μg.

Air is 78.08 mole per cent nitrogen which implies that in 1atm pressure of air, the partial pressure of nitrogen is .7808.  So in 4atm pressure of air, the partial pressure of nitrogen will be .7808×4=3.1232atm.

Substituting all the values in Henry’s law and solving for C,

  C=KHp=18μgatm×3.1232atm=56.2167μg.

Therefore, the mass of nitrogen dissolved in 100g of water saturated with air at 4.0atmand20C has been calculated to be 56.2167μg.

Similarly, at 1atm

  C=KHp=18μgatm×0.7808atm=14.0544μg.

Therefore, the mass of nitrogen dissolved in 100g of water saturated with air at 1.0atmand20C has been calculated to be 14.0544μg.

(c)

Interpretation Introduction

Interpretation:

The increase in nitrogen concentration in fatty tissue in going from 1atm to 4atm has to be determined.

Concept Introduction:

Henry’s law:

The vapour pressure of a volatile solute A is proportional to its mole fraction in a solution.

Mathematically, it can be represented as

pA=KH'XA

Where,

KH'= Henry’s law constant

XA= Mole fraction of a volatile solute A

pA= Vapour pressure of a volatile solute A

Another version of Henry’s law can be mathematically represented as

C=KHp

Where,

KH'= Henry’s law constant

C= Concentration of the dissolved gas

p= partial pressure of the gas

(c)

Expert Solution
Check Mark

Answer to Problem 4.5PR

The increase in nitrogen concentration in fatty tissue in going from 1atm to 4atm has been calculated to be 172.6492μg.

Explanation of Solution

According to the question, nitrogen is 4 times as soluble in fatty tissues as in water.

So at 4atm,

  (Solubility)Fatty=4(Solubility)water=4×56.2167μg=224.8668μg.

So at 1atm,

  (Solubility)Fatty=4(Solubility)water=4×14.0544μg=56.2167μg.

The increase in nitrogen concentration in fatty tissue in going from 1atm to 4atm will be

(224.866856.2167)μg=172.6492μg.

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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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