Why exactly does high pressure (72 bar) nitrous oxide's pseudocritical¹ temperature drop so far when oxygen is added to make 50%v/v mixture? (Entonox)

Question

This answer to Can precipitation occur in states of matter other than liquids? mentions the unusual case of Entonox.

The effects of temperature on nitrous oxide and oxygen mixture homogeneity and stability (Litwin P. D. 2010 BMC Anesthesiol. 2010; 10: 19. doi: 10.1186/1471-2253-10-19) says:

Poynting found that the critical temperature and pressure of a vapour may be affected when it is mixed with another gas. The critical point of a substance is the temperature at above which a gas cannot be liquefied, no matter how much pressure is applied. Conversely, it is the temperature and pressure point at which condensation of the gas into liquid will commence. For nitrous oxide this critical point occurs at a temperature of +36.4°C and at a pressure of 72.45 bar (1050 psi) [12]. For a cylinder of a 50% nitrous oxide and 50% oxygen v/v mixture filled to 138 bar (2000 psi), the new critical temperature of the nitrous oxide (known as the pseudocritical temperature) decreases from +36.4°C to -6°C.

Answers to Can precipitation occur in states of matter other than liquids? tend to guide us away from thinking of the Poyting effect as being related to dissolution or solubility but my concern is that "Poynting effect" is simply a name for an observation, and by itself not a well developed scientific concept or explanation.

That Wikipedia article says:

In thermodynamics, the Poynting effect generally refers to the change in the fugacity of a liquid when a non-condensable gas is mixed with the vapor at saturated conditions.

and that linked subsection ends with:

This equation allows the fugacity to be calculated using tabulated values for saturated vapor pressure. Often the pressure is low enough for the vapor phase to be considered an ideal gas, so the fugacity coefficient is approximately equal to 1.

So if we start with tabulated basic measurements of vapor pressure and calculating fugacity f and then see an unexpected, significant increase in vapor pressure of A when introducing gas B to the vapor, we say "Oh, that's the Poynting effect!"

We might then quickly change the subject or excuse ourselves before "What's that?" or "Why does that happen exactly?" gets ask.

Question: This change has a name, but what is happening? Why does the high pressure (~72 bar) nitrous oxide's pseudocritical¹ temperature drop so far below it's pure substance critical temperature (from +36.4°C to -6°C) when oxygen is add to achieve a 50% v/v mixture at ~138 bar? Is the mechanism at least analogous to dissolution in some ways? Or could it be considered a "liquification point depression" analogous to a freezing point depression?

The citation of a readily available reference or published work explaining "exactly why" this is thought to happen will be most helpful, and if in what ways the process is and is not like dissolution can be added, then that's a complete answer.

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¹As pointed out in comments, critical temperature is defined for a pure substance so "pseudocritical temperature" is used in the block quoted work accordingly. See Real gases; Real gas law or Interaction Parameters for Kay's Pseudocritical Temperature for some examples of the term in use.

Answer 1

Dissolution(solvation) is solvation of solute molecules by molecules of solvents. This decreases chemical potential $\mu={\left(\frac{\partial G}{\partial n}\right)}_{T,p}$ of the solute, comparing it (often hypothetically ) to the chemical potential of the solute at the same concentration in gaseous phase.

Different solvents cause different chemical potential drop, leading to different solubility. As dissolution equilibrium means the equal solute chemical potentials in the solid phase and the solution.

Chemical potential and fugacity

It [fugacity] is equal to the pressure of an ideal gas which has the same temperature and molar Gibbs free energy[=chemical potential] as the real gas.

are linked by relation $\mu=\mu_0 + RT \ln{\frac{f}{f_0}} \tag{1}$

Molecules in gaseous phase move independently on each other, unless there is a strong chemical affinity, what is not this case. Pointing effect does not depend(*) on the used gas, as there is no molecular wrapping in gas.

Pointing effect does not directly affect chemical potential of vapor. But pressure affects the chemical potential of liquid instead, leading to higher saturated vapor pressure at the same temperature. The vapor pressure therefore increases to balance chemical potentials of the substance in both phases.

For liquids near their critical conditions, like for the case of nitrous oxide with or without oxygen, there are big changes of state variables like molar volume or density with small change of conditions like pressure and temperature. This leads to relatively big change of saturated vapor pressure with the change of pressure of the system.

Note that for the liquid at its critical temperature $T_\mathrm{c}$ and near its critical pressure $p_\mathrm{c}$, $$\lim_{p \to p_c}{\ \left(\frac {\partial V}{\partial p}\right)}_{T }= -\infty \tag{2} $$

to get a picture what these changes are like.

As we know from the thermodynamics, for stable composition and zero non-volume work:

$$\mathrm{d}G=-S \cdot \mathrm{d}T + V. \mathrm{d}p \tag{3}$$

Therefore $${\ \left(\frac {\partial G}{\partial p}\right)}_{T }=V \tag{4}$$

$${\ \left(\frac {\partial \mu_\mathrm{liq}}{\partial p}\right)}_{T }=V_\mathrm{m, liq} \tag{5}$$

If we assume ideal vapor behavior, we can consider $p_\mathrm{vap} = f_\mathrm{vap}$ and

$$RT \cdot {\left(\frac{ d \ln{p_\mathrm{vap}}}{\mathrm{d}p}\right)}_T = V_\mathrm{m, liq} \tag{6}$$

$$RT \cdot \ln {\frac{p_\mathrm{vap}}{p^{\circ}_\mathrm{vap}}} = V_\mathrm{m, liq} \cdot (p - p^{\circ}_\mathrm{vap}) \tag{7}$$

$$ p_\mathrm{vap} = p^{\circ}_\mathrm{vap} \cdot \exp {\left(\frac {V_\mathrm{m, liq}}{RT} \cdot (p - p^{\circ}_\mathrm{vap})\right)} \tag{7}$$

(7) is a simplified saturated vapor pressure dependency rather for "normal" liquids with $T_\mathrm{c} \gg T$, when the liquid compressibility can be neglected. Otherwise, $V_\mathrm{m,liq}=f(p)$ would have to be included to the integration, as the liquid molar volume and compressibility start steeply grow toward the critical point. E.g. water vapour and liquid densities converge to about $\rho = \pu{0.21 g/cm}$ at the water critical point.

For illustration, we can take water at the boiling point with $V_\mathrm{m} = M/\rho = (\pu{18.02 g/mol})/(\pu{0.9584 g/cm3})=\pu{18.8 cm3/mol}$, $T=\pu{373 K}$ a $p^{\circ}_\mathrm{vap}=\pu{e5 Pa}$. It can be seen the pressure effect can be neglected for small and medium pressure.

$\begin{array}{cc}\small \\ \text{pressure [MPa]} & \text{H2O vapor pressure [kPa]} \\ 0.1 & 100.0 \\ 0.2 & 100.1 \\ 0.5 & 100.2 \\ 1.0 & 100.5 \\ 2.0 & 101.2 \\ 5.0 & 103 \\ 10 & 106 \\ 20 & 113 \\ 50 & 135 \\ 100 & 183 \\ \end{array}$

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(*) Special cases are systems with significant chemical interactions, including gas solubility in liquid. E.g. well soluble gases cause additional effect of the opposite sign than the Pointing effect. They decrease the saturated vapour pressure by decreasing solvent molar fraction and eventually its chemical potential in both phases by chemical interaction. E.g. $\ce{NH3(g + aq),H2O(l + g)}$.