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

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

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" $$\begin{gathered} \mathrm{惟}鈮�\frac{V^N(4m蟺U{)}^{\frac{3N}{2}}}{h^{3N}N!\left(\frac{3N}{2}\right)!} \\ \mathrm{惟}鈮�\frac{V^N(4m蟺U{)}^{\frac{3N}{2}}}{h^{3N}\left(\frac{3N}{2}\right)!} \end{gathered}$$

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

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

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

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

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

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

The monatomic gas of internal energy is

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

Helium is monatomic gas so $f=3$.

By degree of freedom,

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

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

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

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