Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 21, Problem 16P

(a)

To determine

The amount of iron in 1.00mole of iron.

(a)

Expert Solution
Check Mark

Answer to Problem 16P

1.00mole of iron consists 0.05585kg of iron.

Explanation of Solution

Given Info: The cross-sectional area of iron wire is 5.00×106m2 and the electric current is 30.0A .

Amount of iron in one mole is given as,

M=55.85g/mol

It is the standard value for iron wire.

Convert the amount of iron into kg/mol .

M=55.85g/mol×(103kg/mol1g/mol)=0.05585

Thus, 1.00mole of iron consists 0.05585kg of iron.

Conclusion:

Therefore, 1.00mole of iron consists 0.05585kg of iron.

(b)

To determine

The molar density of iron.

(b)

Expert Solution
Check Mark

Answer to Problem 16P

The molar density of iron is 1.41×105mol/m3 .

Explanation of Solution

Given Info: The cross-sectional area of iron wire is 5.00×106m2 and the electric current is 30.0A .

The formula for the molar density is,

MD=ρiM

Here,

ρi is the density of iron.

M is the moles.

Substitute 7.86kg/m3 for ρi and 0.05585kg/mol for M in above equation to find the MD .

MD=7.86kg/m30.05585kg/mol=1.41×105mol/m3

Thus, the molar density of iron is 1.41×105mol/m3 .

Conclusion:

Therefore, the molar density of iron is 1.41×105mol/m3 .

(c)

To determine

The number density of iron atoms.

(c)

Expert Solution
Check Mark

Answer to Problem 16P

The number density of iron atoms is 8.49×1028atoms/m3

Explanation of Solution

Given Info: The cross-sectional area of iron wire is 5.00×106m2 and the electric current is 30.0A .

The formula for the number density is,

n=AMD

Here,

A is the Avogadro number.

MD is the molar density.

Substitute 1.41×105mol/m3 for MD and 6.02×1023atoms/mol for A in above equation to find n .

n=(6.02×1023atoms/mol)(1.41×105mol/m3)=8.49×1028atoms/m3

Thus, the number density of iron atoms is 8.49×1028atoms/m3 .

Conclusion:

Therefore, the number density of iron atoms is 8.49×1028atoms/m3 .

(d)

To determine

The number density of two conduction iron atoms.

(d)

Expert Solution
Check Mark

Answer to Problem 16P

The number density of two conduction iron atoms is 1.7×1029atoms/m3

Explanation of Solution

Given Info: The cross-sectional area of iron wire is 5.00×106m2 and the electric current is 30.0A .

The formula for the number density of two conduction atoms is,

n=2AMD

Here,

A is the Avogadro number.

MD is the molar density.

Substitute 1.41×105mol/m3 for MD and 6.02×1023atoms/mol for A in above equation to find n .

n=2(6.02×1023atoms/mol)(1.41×105mol/m3)=1.7×1029atoms/m3

Thus, the number density of two conduction iron atoms is 1.7×1029atoms/m3

Conclusion:

Therefore, the number density of two conduction iron atoms is 1.7×1029atoms/m3

(e)

To determine

The drift speed of the conduction electrons.

(e)

Expert Solution
Check Mark

Answer to Problem 16P

The drift speed of the conduction electrons is. 2.2×104m/s .

Explanation of Solution

Given Info: The cross-sectional area of iron wire is 5.00×106m2 and the electric current is 30.0A .

Formula to calculate the drift speed is,

Vd=InqA

Substitute 30.0A for I , 1.6×1019C for q , 1.7×1029atoms/m3 for n and 5.00×106m2 for A in above equation to find the Vd

Vd=30.0A(1.7×1029atoms/m3)×(1.6×1019C)×(5.00×106m2)=2.2×104m/s

Conclusion:

Therefore, the total drift speed of the conduction electrons is. 2.2×104m/s .

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Chapter 21 Solutions

Principles of Physics: A Calculus-Based Text

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