Question

According to the Sackur-Tetrode equation, the entropy of a monatomic ideal gas can become negative when its temperature (and hence its energy) is sufficiently low. Of course this is absurd, so the Sackur-Tetrode equation must be invalid at very low temperatures. Suppose you start with a sample of helium at room temperature and atmospheric pressure, then lower the temperature holding the density fixed. Pretend that the helium remains a gas and does not liquefy. Below what temperature would the Sackur-Tetrode equation predict that $S$is negative? (The behavior of gases at very low temperatures is the main subject of Chapter $7$.)

Short answer

The critical temperature of helium at room temperature and atmospheric pressure is$T_{\text{crit}}=0.01213.$

Step-by-step solution

Step: 1 Getting degree of freedom:

By Sackur-Tetrode formulua for $3-d$ideal gas is

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

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. The logarithm term can go below$-5/2$if the energy $U$falls low enough, making $S$negative. Because this is not conceivable, the Sackur-Tetrode equation must fail at low energies.

The monatomic gas of inernal energy is

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

Where, $f$is degree of freedom,the monatomic gas has localid="1650268768225" $f=3.$

localid="1650268764922" $U=\frac{3}{2}NkT$

Step: 2 Equating critical temperature:

Having an mole of helium and it cools.

The critical temperaature form as

$-\frac{5}{2}=\ln \left(\frac{V}{N}{\left(\frac{4m\pi U_{crit}}{3N{h}^2}\right)}^{\frac{3}{2}}\right)$

substituting equation we get,

$\begin{array}{l}-\frac{5}{2}=\ln \left(\frac{V}{N}{\left(\frac{4m\pi \left(\frac{3}{2}Nk{T}_{\text{crit}}\right)}{3N{h}^2}\right)}^{\frac{3}{2}}\right)\\ -\frac{5}{2}=\ln \left(\frac{V}{N}{\left(\frac{2m\pi k{T}_{\text{crit}}}{h^2}\right)}^{\frac{3}{2}}\right)\end{array}$

Taking exponential on both sides,

As $e^{\ln \left(a\right)}=a$we get,

$\begin{array}{l}{e}^{-\frac{5}{2}}=\frac{V}{N}{\left(\frac{2m\pi k{T}_{crit}}{h^2}\right)}^{\frac{3}{2}}\\ \frac{2m\pi k{T}_{crit}}{h^2}={\left(\frac{N}{V}{e}^{-\frac{5}{2}}\right)}^{\frac{2}{3}}\\ T_{crit}=\frac{h^2}{2m\pi k}{\left(\frac{N}{V}{e}^{-\frac{5}{2}}\right)}^{\frac{2}{3}}\end{array}$

Step: 3 Finding critical temperature value:

The mole mass of helium is $4.0026g$so, the molecule of mass helium 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 $1atm=101325\mathrm{Pa}$and temperature of ${300}^{\circ }K.$

The volume occupies one one mole is

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

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

$\begin{array}{l}{T}_{\text{crit}}=\frac{{\left(6.626\times {10}^{-34}\right)}^2}{2\pi \left(6.646\times {10}^{-27}\right)\left(1.38\times {10}^{-23}\right)}\times {\left(\frac{6.022\times {10}^{23}}{0.0246}{e}^{-\frac{5}{2}}\right)}^{\frac{2}{3}}\\ \to T_{\text{crit}}=0.01213\end{array}$

In reality, because helium liquefies at roughly $4^{\circ }K$, formula appears to be valid for the region where helium is still a gas.