Use the Sackur-Tetrode equation to calculate the entropy of a mole of argon gas at room temperature and atmospheric pressure. Why is the entropy greater than that of a mole of helium under the same conditions?

The entropy of sackur-tetrode formula by

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

From ideal gas law,

At pressure of $1\mathrm{atm}=101325\mathrm{Pa}$and temperature of ${300}^{\circ }K$.

The 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

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

where $f$is the degree of freedom.

Argon is monatomic gas,so it has degree of freedom as $f=3.$

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

The mass of argon mole is $39.948g.$So the Argon mass molecule is

$\begin{array}{l}m=\frac{\text{Mass of one mole}}{\text{Number of atoms one mole}{N}_A}\\ m=\frac{39.948\times {10}^{-3}}{6.022\times {10}^{23}}\\ m=6.634\times {10}^{-26}\mathrm{kg}.\end{array}$

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

we get

$\begin{array}{l}S=Nk\left[\ln \left(\frac{0.0246}{6.022\times {10}^{23}}\right){\left(\frac{4\left(6.626\times {10}^{-26}\right)\pi \times 3739.5}{3\left(6.022\times {10}^{23}\right){\left(6.626\times {10}^{-34}\right)}^2}\right)}^{\frac{3}{2}}+\frac{5}{2}\right]\\ S=Nk\left[\ln \left(\frac{0.0246\times 2.4596\times {10}^{32}}{6.022\times {10}^{23}}\right)+\frac{5}{2}\right]\\ S=Nk\left[\ln \left(10047519\right)+\frac{5}{2}\right]\\ S=\left(6.022\times {10}^{23}\right)\left(1.38\times {10}^{-23}\right)\left[18.62\right]\\ S=154.76\mathrm{J}\times {\mathrm{K}}^{-1}.\end{array}$

Because argon has a larger mass than helium, this figure is somewhat higher.