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Answers without the blur. Sign up and see all textbooks for free! Q. 3.9

Expert-verified Found in: Page 95 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # In solid carbon monoxide, each CO molecule has two possible orientations: CO or OC. Assuming that these orientations are completely random (not quite true but close), calculate the residual entropy of a mole of carbon monoxide.

The residual entropy for a carbon monoxide is calculated to be $5.76J{K}^{-1}$.

See the step by step solution

## Step 1: Given

Carbon monoxide is the given solid. CO is its chemical formula. Each molecule can be arranged in two different ways: CO or OC.

## Step 2: Calculation

Multiplicity for $N$ molecules is given as:

$\Omega ={2}^{N}\dots \dots \dots ..\left(1\right)$

And Residual entropy is given as:

${S}_{\text{residual}}=k\mathrm{ln}\Omega \dots \dots \dots \text{(2)}$

Where, $k$ is Boltzmann constant.

In the given case, $N=6.022×{10}^{23}$

By substituting this value in equation (1), we get,

$\Omega ={2}^{6.022×{10}^{23}}$

Now, by substituting this value of $\Omega$ and $k=1.38×{10}^{-23}J{K}^{-1}$ in equation (2), we get,

${S}_{\text{residual}}=\left(1.38×{10}^{-23}\right)\mathrm{ln}\left({2}^{6.022×{10}^{23}}\right)\phantom{\rule{0ex}{0ex}}{S}_{\text{residual}}=\left(1.38×{10}^{-23}\right)\left(6.022×{10}^{23}\right)\mathrm{ln}2\phantom{\rule{0ex}{0ex}}{S}_{\text{residual}}=5.76{\mathrm{JK}}^{-1}$

Hence, the residual entropy is calculated to be $5.76J{K}^{-1}$. ### Want to see more solutions like these? 