Chemistry
Chemistry
3rd Edition
ISBN: 9780073402734
Author: Julia Burdge
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 13, Problem 36QP
Interpretation Introduction

Interpretation: The amount of nitrogen gas released in litres is to be calculated.

Concept introduction:

The solubility of gas depends upon the pressure of a gas.

The relation between solubility and pressure of a gas is expressed as follows:

c=kP

Here, P is the pressure of the gas, k is the henry’s constant, and c is concentration of gas.

Expert Solution & Answer
Check Mark

Answer to Problem 36QP

Solution: 0.28 L

Explanation of Solution

Given: The solubility of N2

in blood is 5.6×104 mol/L.

The total volume of blood in body is 5.0 L.

In liquid, the solubility of a gas is directly proportional to the pressure of the gas, which is expressed as follows:

c=kP …… (1)

Here, k is the Henry’s law constant, P is the pressure, and c is the molar concentration of gas.

Substitute 5.6×104 mol/L

for c and 0.80 atm

for P in equation (1) as follows:

5.6×104 mol/L=k(0.80 atm)k=(5.6×104 mol/L)(0.80 atm)=7.0×104 mol/Latm

So, the Henry’s law constant is 7.0×104 mol/Latm.

At 4.0 atm, the concentration of nitrogen in blood is calculated as follows:

Substitute 7.0×104 mol/Latm

for k and 4.0 atm

for P in equation (1) as follows:

c=(7.0×104 mol/Latm)(4.0 atm)=2.8×103 mol/L

So, the concentration of nitrogen in blood is 2.8×103 mol/L.

At 0.80 atm,

The number of moles of nitrogen in blood is given as follows:

Moles N2=molarity×liter of solution …… (2)

Substitute 5.6×104 mol/L

for molarity

and 5.0 L

for liter of  solution

in equation (2) as follows:

Moles N2=(5.6×104 mol/L)×(5.0 L)=2.8×103 mol

At 4.0 atm,

The number of moles of nitrogen in blood is given as follows:

Moles N2=molarity×liter of solution …… (3)

Substitute 2.8×103 mol/L

for molarity

and 5.0 L

for liter of solution

in equation (3) as follows:

Moles N2=(2.8×103 mol/L)×(5.0 L)=1.4×102 mol

The amount of nitrogen released in number of moles is calculated as follows:

Moles N2=(1.4×102 mol)(2.8×103 mol)=1.1×102 mol

The ideal gas equation is represented as follows:

PV=nRT …… (4)

Here, P

is the pressure, V

is the volume, n

is the number of moles of nitrogen, R

is the gas constant, and T

is the temperature.

T(K)=T(oC)+273=37+273=310 K

Substitute 1.1×102 mol

for n, 310 K

for T, 0.0821 L.atm/mol.K

for R, and 1.0 atm

for P

in equation (4) as follows:

(1.0 atm)V=(1.1×102 mol)(0.0821 L.atm/mol.K)(310 K)V=(1.1×102 mol)(0.0821 L.atm/mol.K)(310 K)(1.0 atm)=0.28 L

Conclusion

The amount of nitrogen gas released in litres is 0.28 L.

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Chapter 13 Solutions

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