EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684651
Author: Katz
Publisher: VST
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Chapter 22, Problem 81PQ

(a)

To determine

The number of microstates available to the combined system.

(a)

Expert Solution
Check Mark

Answer to Problem 81PQ

The number of microstates available to the combined system is 2×1020.

Explanation of Solution

The number of microstates for the combined system is the product of microstates of individual systems.

Write the expression to calculate the microstates of the combined system.

  w=w1w2

Here, w is the microstates of the combined system, w1 is the microstates of the system A1 and w2 is the microstates of the system A2.

Conclusion:

Substitute 1010 for w1 and 2×1010 for w2 in the above equation to calculate w.

  w=(1010)(2×1010)=2×1020

Therefore, the number of microstates available to the combined system is 2×1020.

(b)

To determine

The entropies of the systems A, A1 and A2.

(b)

Expert Solution
Check Mark

Answer to Problem 81PQ

The entropies of the systems A, A1 and A2 respectively 6.45×1022J/K, and 3.27×1022J/K 3.18×1022J/K.

Explanation of Solution

Write the expression to calculate the entropy of the system A1.

  s1=klnw1                                                                                                                 (I)

Here, s1 is the entropy of the system A1 and k is the Boltzmann constant.

Write the expression to calculate the entropy of the system A2.

  s2=klnw2                                                                                                               (II)

Here, s2 is the entropy of the system A2.

Write the expression to calculate the entropy of the system A.

  s=klnw                                                                                                                 (III)

Here, s is the entropy of the system A.

Conclusion:

Substitute 1010 for w1 and 1.38×1023J/K for k in the above equation (I) to calculate s1.

  s1=(1.38×1023J/K)ln1010=(1.38×1023J/K)10ln10=3.18×1022J/K

Substitute 2×1010 for w2 and 1.38×1023J/K for k in the above equation (II) to calculate s2.

  s2=kln(2×1010)=(1.38×1023J/K)(ln2+10ln10)=3.27×1022J/K

Substitute 2×1020 for w and 1.38×1023J/K for k in the above equation (III) to calculate s.

  s=(1.38×1023J/K)ln(2×1020)=(1.38×1023J/K)(ln2+20ln10)=6.45×1022J/K

Therefore, the entropies of the systems A, A1 and A2 respectively 6.45×1022J/K, and 3.27×1022J/K 3.18×1022J/K.

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Example Two charges, one with +10 μC of charge, and another with - 7.0 μC of charge are placed in line with each other and held at a fixed distance of 0.45 m. Where can you put a 3rd charge of +5 μC, so that the net force on the 3rd charge is zero?
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Chapter 22 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

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