Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 21, Problem 19SP

When 100 coins are tossed, there is one way in which all can come up heads. There are 100 ways in which only one tail comes up. There are about 1 × 10 29 ways that 50 heads can come up. One hundred coins are placed in a box with only one head up. They are shaken and then there are 50 heads up. What was the change in entropy of the coins caused by the shaking?

Expert Solution & Answer
Check Mark
To determine

The change in entropy of coins when 100 coins placed in the box with only one head up is shaken and then there are 50 coins with heads up.

Answer to Problem 19SP

Solution:

8.6×1022 J/K

Explanation of Solution

Given data:

There is one way in which all heads come up if 100 coins are tossed.

There are100 ways in which only one tail comes up if 100 coins are tossed.

There are about 1×1029 ways in which 50 heads come up if 100 coins are tossed.

A total of 100 coins are placed in a box with one head up.

The coins are shaken until there are 50 heads up.

Formula used:

Entropy is calculated by the formula:

S=kBlnΩ

Here, Ω is the number of ways in which the system can achieve the state, S represents the entropy, and kB is the Boltzmann constant.

The number of ways in which a number of things can be arranged is calculated by the formula:

Pnr=n!(nr)!

Here, n is the total number of objects, r is the number of places in which n objects have to be arranged, and Pnr is the number of ways in which n objects can be arranged at r distinct places.

Explanation:

Recall the expression for calculating the number of ways in which the 100coins can be arranged with one head up:

Pnr=n!(nr)!

Understand that thenumber of ways is to be calculated for the case in which one coin with its head facing up can be arranged at100 different places.

Substitute 100 for n and 1 for r

P1001=100!(1001)!=100!99!=100

Understand that the number of ways in which 100 coins can be arranged with one coin withits head facing up is equal to the number of ways in which one coin with its head facing up can be arranged at100 different places. Therefore,

Ω1=P1001

Substitute 100 for P1001

Ω1=100

Here, Ω1 is the number of ways in which 100 coins can be arranged with one coin facing headup.

Recall the expression for entropy when 100 coins are arranged with one head up:

S1=kBlnΩ1

Here, S1 is the entropy when 100 coins are arranged with one head up.

Consider the standard value of Boltzmann constant is 1.38×1023 J/K.

Substitute 1.38×1023 J/K for kB and 100 for Ω1

S1=(1.38×1023 J/K)ln100=6.36×1023 J/K

Recall the expression for entropy when 100 coins are arranged with 50 heads up:

S2=kBlnΩ2

Here, S2 is the entropy when 100 coins are arranged with 50 heads up and Ω2 is the number of ways in which 100 coins can be arranged with 50 coins facing headup.

According to the problem, 100 coins can be arranged with 50 coins having their heads up in 1×1029 ways.

Substitute 1.38×1023 J/K for kB and 1×1029 for Ω1

S2=(1.38×1023 J/K)[ln(1×1029)]=92.1×1023 J/K 

Calculate the change in entropy when the box is shaken till there are 50 coins with heads up:

ΔS=S2S1

Here, ΔS is the change in entropy during the process.

Substitute 92.1×1023 J/K  for S2 and 6.36×1023 J/K for S1

ΔS=(92.1×1023 J/K )(6.36×1023 J/K)8.6×1022 J/K

Conclusion:

The change in entropy is 8.6×1022 J/K.

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