(a) Interpretation: Among four coins in which two are heads and two are tails and four coins in which one is head and three are tails, the one thathas higher entropy is to be stated. The reason for the answer is to be explained. Concept introduction: Entropy is defined as the randomness of molecules in a system. It is an extensive property. Entropy is directly proportional to natural logarithm to probability. It is represented as shown below. S = k B ln W

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Answer 1

Question

Chapter 8, Problem 8.18P

Interpretation Introduction

(a)

Interpretation:

Among four coins in which two are heads and two are tails and four coins in which one is head and three are tails, the one thathas higher entropy is to be stated. The reason for the answer is to be explained.

Concept introduction:

Entropy is defined as the randomness of molecules in a system. It is an extensive property. Entropy is directly proportional to natural logarithm to probability. It is represented as shown below.

$S=kBlnW$

Expert Solution

Answer to Problem 8.18P

Four coins in which two are heads and two are tails has a higher entropy. The entropy for the first case $2.47×10−23 J/K$ and second case is $1.91×10−23 J/K$ .

Explanation of Solution

Entropy is directly proportional to natural logarithm to probability. It is represented as shown below.

$S=kBlnW$ …(1)

Where,

• $S$ is entropy.

• $kB$ is Boltzmann constant is a proportionality constant $(1.38×10−23 m2kg s−2K−1)$ .

• $lnW$ is natural logarithm to the number of distributions.

When four coins are flipped, there are six possible cases in which two are heads and two are tails which are shown below.

$HHTT TTHHHTHT THTHHTTH THHT$

Similarly, when four coins are flipped, there are three possible cases in which one is head and three are tails which are shown below.

$HTTT THTTTTHT TTTH$

Substitute the above values in equation (1). Therefore, the entropy for the first case is calculated as shown below.

$S=kBlnW=(1.38×10−23 m2kg s−2K−1)ln6=(1.38×10−23 m2kg s−2K−1)1.79=2.47×10−23 J/K$

Similarly, the entropy for the second case is calculated as shown below.

$S=kBlnW=(1.38×10−23 m2kg s−2K−1)ln4=(1.38×10−23 m2kg s−2K−1)1.38=1.91×10−23 J/K$

Therefore, the first case has a higher entropy.

Conclusion

The entropy for the first case $2.47×10−23 J/K$ and the second case is $1.91×10−23 J/K$ . The first case has a higher entropy.

Interpretation Introduction

(b)

Interpretation:

Among six coins in which two are heads and four are tails and six coins in which two are tails and four are heads, the one that has higher entropy is to be stated. The reason for the answer is to be explained.

Concept introduction:

Entropy is defined as the randomness of molecules in a system. It is an extensive property. Entropy is directly proportional to natural logarithm to probability. It is represented as shown below.

$S=kBlnW$

Expert Solution

Answer to Problem 8.18P

The entropy for both the cases is same and has a value of $3.72×10−23 J/K$ .,

Explanation of Solution

Entropy is directly proportional to natural logarithm to probability. It is represented as shown below.

$S=kBlnW$ …(1)

Where,

• $S$ is entropy.

• $kB$ is Boltzmann constant is a proportionality constant $(1.38×10−23 m2kg s−2K−1)$ .

• $lnW$ is natural logarithm to the number of distributions.

When six coins are flipped, there are $15$ possible cases in which two are heads and four are tails which are shown below.

$HHTTTT HTHTTTHTTHTT HTTTHTHTTTTH THTTTHTTHTTH TTTHTHTTTTHH THHTTTTTHHTT TTTHHTTHTHTT TTHTHTTHTTHT$

Similarly, when six coins are flipped, there are $15$ possible cases in which two are tails or four are heads which are shown below.

$TTHHHH THTHHHTHHTHH THHHTHTHHHHT HTHHHTHHTHHT HHHTHTHHHHTT HTTHHHHHTTHH HHHTTHHTHTHH HHTHTHHTHHTH$

Substituting the above values in equation (1). Therefore, the entropy for the first case is calculated as shown below.

$S=kBlnW=(1.38×10−23 m2kg s−2K−1)ln15=(1.38×10−23 m2kg s−2K−1)2.70=3.72×10−23 J/K$

Similarly, the entropy for the second case is calculated as shown below.

$S=kBlnW=(1.38×10−23 m2kg s−2K−1)ln15=(1.38×10−23 m2kg s−2K−1)2.70=3.72×10−23 J/K$

Therefore, entropy for both cases is equal.

Conclusion

The entropy for both the cases is same.

Interpretation Introduction

(c)

Interpretation:

Among $1 mole$ of solute in $1 L$ of water and $5 mole$ of solute in $1 L$ , the one that has higher entropy is to be stated. The reason for the answer is to be explained.

Concept introduction:

Entropy is defined as the randomness of molecules in a system. It is an extensive property. Entropy is directly proportional to natural logarithm to probability. It is represented as shown below.

$S=kBlnW$

Expert Solution

Answer to Problem 8.18P

The number of moles is higher in the second case which means its entropy is also higher. Due to the directly proportional relation between the number of moles and entropy.

Explanation of Solution

Entropy and the number of moles are directly proportional to each other. If the number of moles increases on the product side. Then the entropy of the reaction increases. In the given cases, the number of moles is higher in the second case that is $5 moles$ in $1 L$ . Due to this, the entropy is also higher in the second case.

Conclusion

Entropy and the number of moles are directly proportional to each other. The number of moles is higher in the second case which means its entropy is also higher.

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