Question

For either a monatomic ideal gas or a high-temperature Einstein solid, the entropy is given by times some logarithm. The logarithm is never large, so if all you want is an order-of-magnitude estimate, you can neglect it and just say . That is, the entropy in fundamental units is of the order of the number of particles in the system. This conclusion turns out to be true for most systems (with some important exceptions at low temperatures where the particles are behaving in an orderly way). So just for fun, make a very rough estimate of the entropy of each of the following: this book (a kilogram of carbon compounds); a moose of water ; the sun of ionized hydrogen .

Short answer

The rough estimate of the entropy of each of the following:

For$1\mathrm{kg}$book is $S~692.484\mathrm{J}\times {\mathrm{K}}^{-1}$;

For $400kg$moose is $S~184644\mathrm{J}\times {\mathrm{K}}^{-1}$;

For sun is$S~1.662\times {10}^{34}\mathrm{J}\times {\mathrm{K}}^{-1}$.

Step-by-step solution

Step: 1 Finding entropy of book:

We can see that the entropy of an ideal gas is $Nk$times a logarithm, and that the entropy of an Einstein solid is likewise $Nk$times a logarithm. Because $N$is a high number for any macroscopic item and the logarithm is considerably smaller, we may ignore the log part and choose $S~Nk$for an approximate order-of-magnitude estimate of the entropy. Here are a few examples of such estimates:

Taking $1kg$of carbon with molar mass of $12\times {10}^{-3}\mathrm{kg}\times {\mathrm{mol}}^{-1}$for $1kg$of book,the number of molecule is

$$\begin{gathered} N=Numberofatomsinonemole\times Numberofmoles \\ N=Numberofatomsinonemole\times \frac{Mass}{Molarmass} \\ N=6.022\times {10}^{23}\times \frac{1}{12\times {10}^{-3}} \\ N=5.018\times {10}^{25}. \end{gathered}$$

Entropy as

$S~Nk$

Substituting $k=1.38\times {10}^{-23}\mathrm{J}\times {\mathrm{K}}^{-1}$,so the entropy of book as

$\begin{array}{l}S~5.018\times {10}^{25}\times 1.38\times {10}^{-23}\\ S~692.484\mathrm{J}\times {\mathrm{K}}^{-1}\end{array}$

Step: 2 Finding the entropy of moose:

For a $400kg$of mooses,the approximate of $400kg$of water with molar mass of $18\times {10}^{-3}\mathrm{kg}\times {\mathrm{mol}}^{-1}$,the number of molecules is

$$\begin{gathered} N=6.022\times {10}^{23}\times \frac{400}{18\times {10}^{-3}} \\ N=1.338\times {10}^{28} \end{gathered}$$

The entropy mooses is

role="math" localid="1650273675935" $$\begin{gathered} S~Nk \\ S~1.338\times {10}^{28}\times 1.38\times {10}^{-23} \\ S=184644\mathrm{J}\times {\mathrm{K}}^{-1}. \end{gathered}$$

Step: 3 Finding entropy of sun:

For sun,the ionized hydrogen of $2\times {10}^{30}\mathrm{kg}$with molar mass is${10}^{-3}\mathrm{kg}\times {\mathrm{mol}}^{-1}$ ,the number of molecules is

$$\begin{gathered} N=6.022\times {10}^{23}\times \frac{2\times {10}^{30}}{{10}^{-3}} \\ N=1.2044\times {10}^{57} \end{gathered}$$

The entropy is

$$\begin{gathered} S~1.2044\times {10}^{57}\times 1.38\times {10}^{-23} \\ S=1.662\times {10}^{34}\mathrm{J}\times {\mathrm{K}}^{-1}. \end{gathered}$$