48
$\begingroup$

By the end of the 19th century all gasses had been liquefied apart from helium (He). What is it about helium that makes it so hard to liquefy compared to the other gases? And why does it need to be pre cooled in the Joule-Kelvin expansion?

$\endgroup$
49
$\begingroup$

The next approximation beyond the ideal gas is given by the Van der Waals fluid equation. It is a phenomenological law which takes into account the finite size of the molecules and their interactions with themselves.

When you plot several Van der Vaals isotherms for a given substance, you observe that some of them show a phase transition from gas to liquid while others do not. The ones which do not show a phase transition are above a so called critical temperature $T_c$. Graph of pressure against volume showing critical temperature Above this temperature you can decrease the volume or increase the pressure of the gas and it will not liquefy.

Actually, the isotherms below the critical temperature need a correction given by Maxwell. To avoid instability (lower pressure giving lower volume giving lower pressure...) the actual path in the $PV$ diagram must avoid the "bumps" and follow the dashed line, as in the figure below Graph of P against V with some peaks/troughs bypassed by a dashed line The dashed line is the phase transition region. To see this, notice that if you keep decreasing volume further below $V_L$ you will need a huge amount of pressure. This means we got a liquid. Also notice that if the substance is above the critical temperature there is no need to apply that Maxwell correction. So there is no phase transition. The phase transition prediction by Van der Waals gave him the 1910 Nobel prize in Physics.

Examples of critical temperatures are (in degrees Celsius): \begin{align} T_c(H_2O)&=+374,35,\\ T_c(O_2)&=-118,55,\\ T_c(N_2)&=-147,15,\\ T_c(H_2)&=-240,17,\\ T_c(He^4)&=-267,96. \end{align} As you can see, we are only able to liquefy Helium when it is below $-267,96^oC$. For a long time chemists called the gases $O_2$, $N_2$, $H_2$ and $He^4$ as permanent gases, since they were not able to drop the temperature enough to turn them liquid.

Edit: I basically said that the great difficulty in liquefying helium is due to its extremely low critical temperature. The next question would be: Why is the helium critical temperature so low? Let me try to answer to that question too.

The van der Waals equation for one mol of gas reads $$\left(P+\frac{a}{v^2}\right)(v-b)=RT.$$ The parameter $a$ characterizes the strength of the attractive intermolecular interaction while $b$ is related to the effective volume occupied by the molecules. The critical temperature can be calculated in terms of these parameters (remember the temperatures are always given in Kelvin), $$T_c=\frac{8a}{27bR}.$$ So a small $T_c$ means either small $a$ (weak interaction) or high $b$ (big molecules) or a combination of both. For the gases above mentioned we have, \begin{array}{|c|c|c|} \hline & a(Pa\cdot m^3/mol^2) & b(m^3/mol) \\ \hline H_2O & 554\cdot 10^{-3} & 3.05\cdot 10^{-5} \\ \hline O_2 & 138\cdot 10^{-3} & 3.19\cdot 10^{-5} \\ \hline N_2 & 137\cdot 10^{-3} &3.87\cdot 10^{-5} \\ \hline H_2 & 24.8\cdot 10^{-3}& 2.66\cdot 10^{-5} \\ \hline He^4 & 3.46\cdot 10^{-3} & 2.38\cdot 10^{-5} \\ \hline \end{array} These data suggest that the extremely weak (compared to the other) intermolecular interaction is the reason it has such a low critical temperature.

$\endgroup$
  • 2
    $\begingroup$ If the critical temperature of water is 374 c, why does it boil at 100 c? $\endgroup$ – John Dvorak May 24 '16 at 17:32
  • 12
    $\begingroup$ @Jan Dvorak: The critical temperature is the temperature that a substance can't be condensed to a liquid at any pressure. $\endgroup$ – Dan May 24 '16 at 17:41
  • 13
    $\begingroup$ @JanDvorak Critical temperature is not the same thing as boiling point. Imagine you have water vapor at constant temperature and you change pressure to find the point where it liquefies. At $100\, C$ you will find the pressure is around $1\, atm$. Now if the temperature is above $374\, C$ you won't find any pressure that liquefies water. $\endgroup$ – Diracology May 24 '16 at 17:46
  • 1
    $\begingroup$ @Luaan Please have another look at the answer. I tried to explain (using van der Waals equation) why helium critical temperature is so low. $\endgroup$ – Diracology May 26 '16 at 1:20
  • $\begingroup$ So... the larger the molecules, the lower the boiling point? That seems counterintuitive. Wouldn't larger molecules have a) more Van der Vaals forces and b) occupy more space as a liquid, resulting in a higher gas transition point? I am wondering. $\endgroup$ – Ber May 26 '16 at 6:37
25
$\begingroup$

Getting from gas to liquid is a matter of interparticle interaction winning over thermal agitation.

There are several reasons why interparticle interactions are very weak in the case of helium atoms. On one hand, it is a noble gas and thus cannot form covalent bonds. On the other hand, it is very light hence highly non-polarizable: its Van der Waals interactions are weak.

$\endgroup$
6
$\begingroup$

Throttling the gas (Joule-Kelvin expansion) only lowers the temperature of the gas when the Joule–Thomson coefficient is positive. For Helium, that point (the "J–T inversion temperature") is reached at 43°K (source: Cryogenic Society of America; the wikipedia article gives an incorrect value of 51°K). Above that temperature, Joule-Kelvin expansion will increase the temperature of the gas instead of lowering it, that's why pre-cooling is required.

Throttling is an isenthalpic process; definition and formula for the Joule–Thomson coefficient (see the link for more details):

$\mu_{\mathrm{JT}} = \left( {\partial T \over \partial P} \right)_H = \frac{V}{C_{\mathrm{p}}}\left(\alpha T - 1\right)\,$

V is the gas volume, $C_p$ the heat capacity at constant pressure and $\alpha$ the coefficient of thermal expansion. $\mu_{\mathrm{JT}}$ gives the temperature drop in °K per bar.

Only helium, hydrogen and neon have an inversion temperature below ambient (neon: 250°K) and require pre-cooling.

$\endgroup$
  • 1
    $\begingroup$ I'm happy to see that my images are used, but you have to attribute them to the author, like the majority of images on Wikipedia. $\endgroup$ – Han-Kwang Nienhuys May 26 '16 at 6:48
  • $\begingroup$ What do you mean ? I thought that linking to Wikipedia was enough, as your name appears in caption of the image. $\endgroup$ – Dimitri May 26 '16 at 8:02
  • $\begingroup$ I've removed the image, given the unclear copyright status. Does "reproduced, stored in a retrieval system or transmitted, in any form or by any means..." (The Standard Reference Data Act, nist.gov/srd/publiclaw.cfm) include the representation of data in a graph? $\endgroup$ – Previous May 26 '16 at 8:56
3
$\begingroup$

For the first question, it is the low boiling temperature, 4.21K for Helium-4 and 3.19K for Helium-3, that makes helium difficult to be liquefied. Hydrogen's boiling temperature at 1 atm is 20.27K, or about 4-5 times higher.

For pre-cooling, one can take a look of entropy $$\delta S = \frac {dQ}{T}$$ We can see that, because T is very small, a slight changing in heat will increases the entropy a lot, which makes liquefation difficult. Thus pre-cooling and creating a extremely cold environment is critical in the process.

$\endgroup$
3
$\begingroup$

Simply Said:

It doesn't attract itself enough to be solid. So much resistance to attraction that it is very hard to liquefy. The electrons prefer to repel because the electrons repel like the protons. Since it has electrons, sometimes it becomes polar for a very short amount of time. This help pull together the molecule, letting it liquefy but at very low temperature. This is call the London Dispersion Force. Since helium has a few amount of electrons and a full shell, it has a very low probability of an atom becoming polar for a second and thus reduces the boiling point even more.

$\endgroup$
  • $\begingroup$ I don't think so. How would you explain superliquid statoe of helium near $0 K$ temperatures? $\endgroup$ – Crowley Jul 1 '16 at 14:40
  • $\begingroup$ Then it would rely on quantum mechanics afterward $\endgroup$ – Arvin Singh May 23 '18 at 11:42

protected by Qmechanic May 25 '16 at 13:15

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.