Chapter 6: Q. 6.43 (page 249)

Some advances textbooks define entropy by the formula
$S = - k \sum_{s} P (s) \ln P (s)$
where the sum runs over all microstates accessible to the system and $P (s)$ is the probability of the system being in microstate $s$ .
(a) For an isolated system, role="math" localid="1647056883940" $P (s) = \frac{1}{Ω}$ for all accessible states $s$ . Show that in this case the preceding formula reduces to our familiar definition of entropy.
(b) For a system in thermal equilibrium with a reservoir at temperature $T$ ,role="math" localid="1647057328146" $P (s) = \frac{e^{- \frac{E (s)}{k T}}}{Z}$ . Show that in this case as well, the preceding formula agrees with what we already know about entropy.

Short Answer

Expert verified

Part (a): $S = k \ln Ω$

Part (b): $S = \frac{E - F}{T}$

Step by step solution

Part (a): Step 1. Given information

The entropy of the system is defined as

$S = - k \sum_{s} P (s) \ln P (s) . . . . . . . . . . . . . . . . . . . . . . . . (1)$

and the probability of finding the system in a microstate $s$ is given by

$P (s) = \frac{1}{Ω} . . . . . . . . . . . . . . . . . . . . . . . . (2)$

Part (a): Step 2. Calculation

Substitute the value of $P (s)$ from equation (2) into equation (1) and simplify to obtain the entropy of the system.

$\begin{array}{rcl} S & = & - k \sum_{s} \frac{1}{Ω} \ln (\frac{1}{Ω}) \\ = & \frac{k \ln Ω}{Ω} \sum_{s} (1) \\ = & \frac{k \ln Ω}{Ω} Ω \\ = & k \ln Ω \end{array}$

Part (b): Step 1. Given information

The probability is given by

$P (s) = \frac{e^{- \frac{E (s)}{k T}}}{Z} . . . . . . . . . . . . . . . . . . (3)$

Part (b): Step 2. Calculation

Take logarithm from both sides of equation (3).

$\begin{array}{rcl} \ln P (s) & = & \ln [\frac{e^{- \frac{E (s)}{k T}}}{Z}] \\ = & - β E (s) - \ln Z \\ = & \frac{1}{k T} (- E (s) + F) . . . . . . . . . . . . . . . . . (4) \end{array}$

Substitute the values of the parameters from equation (3) and equation (4) into equation (1) and simplify to obtain the required entropy of the system.

$\begin{array}{rcl} S & = & - k \sum_{s} [\frac{e^{- \frac{E (s)}{k T}}}{Z}] \frac{1}{k T} (- E (s) + F) \\ = & \frac{1}{T} \sum_{s} E (s) \frac{e^{- β E (s)}}{Z} - \frac{F}{T} \sum_{s} \frac{e^{- β E (s)}}{Z} \\ = & \frac{E - F}{T} \end{array}$

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Short Answer

Step by step solution

Part (a): Step 1. Given information

Part (a): Step 2. Calculation

Part (b): Step 1. Given information

Part (b): Step 2. Calculation

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