Question

Compute the entropy of a mole of helium at room temperature and atmospheric pressure, pretending that all the atoms are distinguishable. Compare to the actual entropy, for indistinguishable atoms, computed in the text.

Short answer

The actual entropy indistinguishable atoms is$S_{\text{dist}}=572.816\mathrm{J}\times {\mathrm{K}}^{-1}.$

Step-by-step solution

Step: 1 

The Sackur-Tetrode formula by $3-d$ideal gas is

$S=Nk\left[\ln \left(\frac{V}{N}{\left(\frac{4m\pi U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{5}{2}\right]$

Where,$V$represents volume, $U$represents energy, $N$represents the number of molecules, $m$represents the mass of a single molecule, and $h$represents Planck's constant.These are some of the assumptions used to generate this formula is that the molecules are indistinguishable, therefore altering any of the molecules makes no change in any arrangement of the molecules in position and momentum space. This assumption inserts the $N!$ component into the multiplicity function's denominator.

role="math" localid="1650281260492" $$\begin{gathered} \mathrm{\Omega }\approx \frac{V^N(4m\pi U{)}^{\frac{3N}{2}}}{h^{3N}N!\left(\frac{3N}{2}\right)!} \\ \mathrm{\Omega }\approx \frac{V^N(4m\pi U{)}^{\frac{3N}{2}}}{h^{3N}\left(\frac{3N}{2}\right)!} \end{gathered}$$

The logarithm factor $\frac{V}{N}$loses its $N$, we get

$S_{dist}=Nk\left[\ln \left(\frac{V}{N}{\left(\frac{4m\pi U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{3}{2}\right]$

Step: 2 Finding degree of freedom:

The mole mass of helium is $4.0026g$,the mass of helium molecule is

$\begin{array}{l}m=\frac{\text{Mass of one mole}}{\text{Number of atoms one mole}{N}_A}\\ m=\frac{4.0026\times {10}^{-3}}{6.022\times {10}^{23}}\\ m=6.646\times {10}^{-27}\mathrm{kg}.\end{array}$

From ideal gas law, the pressure of $1atm=101325Pa$and temperature of ${300}^{\circ }K$one mole occupies a volume of

$$\begin{gathered} V=\frac{nRT}{P} \\ V=\frac{8.31\times 300}{101325} \\ V=0.0246{\mathrm{m}}^3. \end{gathered}$$

The monatomic gas of internal energy is

$U=\frac{f}{2}NkT$

Helium is monatomic gas so $f=3$.

Step: 3 

By degree of freedom,

$$\begin{gathered} U=\frac{3}{2}NkT \\ U=\frac{3}{2}nRT \\ U=\frac{3}{2}\times 8.31\times 300 \\ U=3739.5\mathrm{J}. \end{gathered}$$

Substituting the values of$k=1.38\times {10}^{-23}\mathrm{J}\times {\mathrm{K}}^{-1};h=6.626\times {10}^{-34}\mathrm{J}\times \mathrm{s}$, we get

$\begin{array}{l}{S}_{\text{dist}}=Nk\left[\ln \left(V{\left(\frac{4m\pi U}{3N{h}^2}\right)}^{\frac{3}{2}}\right)+\frac{3}{2}\right]\\ S_{\text{dist}}=Nk\left[\ln \left(0.0246{\left(\frac{4\left(6.646\times {10}^{-27}\right)\pi \times 3739.5}{3\left(6.022\times {10}^{23}\right){\left(6.626\times {10}^{-34}\right)}^2}\right)}^{\frac{3}{2}}\right)+\frac{3}{2}\right]\\ S_{\text{dist}}=\left(6.022\times {10}^{23}\right)\left(1.38\times {10}^{-23}\right)\left[68.928\right]\\ S_{\text{dist}}=572.816\mathrm{J}\times {\mathrm{K}}^{-1}\end{array}$

Because there are many more molecular orbitals accessible to the system if the molecules are distinct, the entropy is substantially larger.