Chapter 22, Problem 81PQ

(a)

To determine

The number of microstates available to the combined system.

Answer to Problem 81PQ

The number of microstates available to the combined system is $2\times {10}^{20}$.

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.

  $w=w_1{w}_2$

Here, w is the microstates of the combined system, $w_1$ is the microstates of the system $A_1$ and $w_2$ is the microstates of the system $A_2$.

Conclusion:

Substitute ${10}^{10}$ for $w_1$ and $2\times {10}^{10}$ for $w_2$ in the above equation to calculate w.

  $\begin{array}{c}w=\left({10}^{10}\right)\left(2\times {10}^{10}\right)\\ =2\times {10}^{20}\end{array}$

Therefore, the number of microstates available to the combined system is $2\times {10}^{20}$.

(b)

To determine

The entropies of the systems A, $A_1$ and $A_2$.

Answer to Problem 81PQ

The entropies of the systems A, $A_1$ and $A_2$ respectively $6.45\times {10}^{-22}\text{J}/\text{K}$, and $3.27\times {10}^{-22}\text{J}/\text{K}$ $3.18\times {10}^{-22}\text{J}/\text{K}$.

Explanation of Solution

Write the expression to calculate the entropy of the system $A_1$.

  $s_1=k\ln w_1$                                                                                                                 (I)

Here, $s_1$ is the entropy of the system $A_1$ and k is the Boltzmann constant.

Write the expression to calculate the entropy of the system $A_2$.

  $s_2=k\ln w_2$                                                                                                               (II)

Here, $s_2$ is the entropy of the system $A_2$.

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

  $s=k\ln w$                                                                                                                 (III)

Here, $s$ is the entropy of the system $A$.

Conclusion:

Substitute ${10}^{10}$ for $w_1$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for k in the above equation (I) to calculate $s_1$.

  $\begin{array}{c}{s}_1=\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\ln {10}^{10}\\ =\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)10\ln 10\\ =3.18\times {10}^{-22}\text{J}/\text{K}\end{array}$

Substitute $2\times {10}^{10}$ for $w_2$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for k in the above equation (II) to calculate $s_2$.

  $\begin{array}{c}{s}_2=k\ln \left(2\times {10}^{10}\right)\\ =\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(\ln 2+10\ln 10\right)\\ =3.27\times {10}^{-22}\text{J}/\text{K}\end{array}$

Substitute $2\times {10}^{20}$ for $w$ and $1.38\times {10}^{-23}\text{J}/\text{K}$ for k in the above equation (III) to calculate $s$.

  $\begin{array}{c}s=\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\ln \left(2\times {10}^{20}\right)\\ =\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(\ln 2+20\ln 10\right)\\ =6.45\times {10}^{-22}\text{J}/\text{K}\end{array}$

Therefore, the entropies of the systems A, $A_1$ and $A_2$ respectively $6.45\times {10}^{-22}\text{J}/\text{K}$, and $3.27\times {10}^{-22}\text{J}/\text{K}$ $3.18\times {10}^{-22}\text{J}/\text{K}$.