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Q. 6.43

Expert-verified
Found in: Page 249

### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

# Some advances textbooks define entropy by the formula$S=-k\sum _{s}P\left(s\right)\mathrm{ln}P\left(s\right)$where the sum runs over all microstates accessible to the system and $P\left(s\right)$ is the probability of the system being in microstate $s$.(a) For an isolated system, role="math" localid="1647056883940" $P\left(s\right)=1}{\Omega }$ 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" . Show that in this case as well, the preceding formula agrees with what we already know about entropy.

Part (a): $S=k\mathrm{ln}\Omega$

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

See the step by step solution

## Part (a): Step 1. Given information

The entropy of the system is defined as

$S=-k\sum _{s}P\left(s\right)\mathrm{ln}P\left(s\right)........................\left(1\right)$

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

$P\left(s\right)=1}{\Omega }........................\left(2\right)$

## Part (a): Step 2. Calculation

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

$\begin{array}{rcl}S& =& -k\sum _{s}\frac{1}{\Omega }\mathrm{ln}\left(\frac{1}{\Omega }\right)\\ & =& \frac{k\mathrm{ln}\Omega }{\Omega }\sum _{s}\left(1\right)\\ & =& \frac{k\mathrm{ln}\Omega }{\Omega }\Omega \\ & =& k\mathrm{ln}\Omega \end{array}$

## Part (b): Step 1. Given information

The probability is given by

## Part (b): Step 2. Calculation

Take logarithm from both sides of equation (3).

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.