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Hydrogen Properties for Energy Research (HYPER) Lab ME 527

ME 527 Lesson 39: Fluid Friday – Helium

We’ve now come to an end of our list of cryogenic fluids arriving at helium:

  1. Krypton (119.73 K)
  2. Methane (111.67 K)
  3. Oxygen (90.188 K)
  4. Argon (87.302 K)
  5. Fluorine (85.037 K)
  6. Carbon Monoxide (81.64 K)
  7. Nitrogen (77.355 K)
  8. Neon (27.104 K)
  9. Deuterium (23.31 K)
  10. Hydrogen (20.369 K)
  11. Helium (4.222 K)

Discovery, origins, and current usage

Evidence of helium was first discovered in 1868 while looking at spectral lines of our sun. It was later liquefied by Kammerlingh-Onnes, who discovered it’s supercritical properties and won the Nobel prize.


Fixed Properties

Helium is unique among the cryogenic fluids due to a change of phase from classical to quantum fluid behavior known as a superfluid. This occurs at what is known as the Lambda line, or lambda point (2.17 K at 1 atm):

In the superfluid state the liquid is quantized with no viscosity, extremely fast thermal conductivity and sound speed, etc.

Obviously below the lambda line our classical “macroscale” thermodynamics ceases to remain valid. That’s why we’re restricting ourselves to helium I.

Helium EOS fixed points

Surface of State

Helium Pressure versus Temperature

Helium Pressure versus density

Helium isochoric heat capacity versus temperature

Helium sound speed versus temperature

Helium Gruneisen versus Temperature

Helium phase identification parameter versus temperature


ME 527 Lesson 36: Fluid Friday – Deuterium

Now that we are familiar with hydrogen, we will cover it’s only stable isotope, Deuterium on our list of cryogenic:

  1. Krypton (119.73 K)
  2. Methane (111.67 K)
  3. Oxygen (90.188 K)
  4. Argon (87.302 K)
  5. Fluorine (85.037 K)
  6. Carbon Monoxide (81.64 K)
  7. Nitrogen (77.355 K)
  8. Neon (27.104 K)
  9. Deuterium (23.31 K)
  10. Hydrogen (20.369 K)
  11. Helium (4.222 K)

Introduction and Discovery

Deuterium, also known as heavy hydrogen, is one of two stable isotope of hydrogen. Deuterium comes from the Greek deuteros meaning “second” referring to the two particles (proton + neutron) in the nucleus.  Deuterium accounts for 0.0156% of the hydrogen in the oceans or one in 6420 hydrogen atoms.  Deuterium was created during the Big Bang and accounts for 0.0013% of the Universe.

Deuterium was discovered by Harold Urey in 1931.  He predicted that their would be a difference between the vapor pressure of pure hydrogen and it’s heavier isotopes. He developed a method to isolate the heavier isotope through the distillation of liquid hydrogen. This lead to Urey receiving the Nobel Prize in Chemistry in 1934 “for his discovery of heavy hydrogen”.

Similarly to hydrogen, deuterium also has 2 spin isomers denoted as orthodeuterium and paradeuterium. Contrary to hydrogen and tritium, the lower-energy, even-states are denoted “ortho” for deuterium, while the higher-energy, odd-J states are denoted “para.” This is due to the ortho-para compositions at room temperature are different for deuterium than they are for hydrogen because deuterium has nuclear spin of +1 whereas hydrogen has a nuclear spin of +1/2.

Uses and Production

Deuterium is used as fuel in nuclear fusion reactors like the National Ignition Facility (NIF) and ITER (in development). Deuterium is also used in NMR spectroscopy, as a stable isotope tracer, nuclear weapons, and medicine for deuterated drugs.  Research grade deuterium gas costs $1,112 for 500 standard liters which equates to $12.38 per gram!  This price can be reduced to $1/L for larger orders of research grade deuterium.  Due to the military applications, the deuterium market is very secretive and little information is available to the public. After a discussion with a Linde representative, I was able to learn that the deuterium market has rapidly grown over the last 10 years because of it’s use in the manufacturing of semiconductors and fiber optics. Linde is the largest producer of deuterium in the world producing millions of liters a years.

Due to the small differences in molecular mass, it is energy intensive and expensive to separate the isotopes of hydrogen.

H2 Combinations

Deuterium can be separate from the other hydrogen isotopes through several processes including chemical exchange, thermal diffusion, cryogenic distillation, electrolysis, and permeation.The most common method for producing deuterium is through electrolysis of heavy water, D2O, which is widely used in heavy water reactors. Until it’s closure in 1997, the Bruce Heavy Water plant in Ontario, Canada was the largest producer of D2O utilizing the Girdler Sulfide process.  This is an isotopic exchange process between H2S and H2O:  H2O + HDS ⇌ HDO + H2S. After several iteration this process is able to enrich the water to 15-20% D2O where it can then be purified using distillation or electrolysis.  This process takes 340,000 kg of feed water to produce 1 kg of heavy water.

D2O Production Overview

Canada is the largest producer and consumer of heavy water where it is used to cool their CANDU reactors. The price of heavy water is $300 per kg (2001).

D2O Production



Fixed Properties

D2 Fixed Properties

Surface of State

Only the equation of state for normal deuterium is included in REFPROP.  The variation in thermophysical properties between orthodeuterium and paradeuterium were not significant enough to warrant separate equation to be include in REFPROP.

Critical Isotherm

Though the rectilinear diameter has a slight change in the curvature, the critical isotherm exhibits correct thermodynamic behavior.

D2 Extrapolation

Extrapolating to extreme pressures and densities shows the equations behaves theoretically correct at the high extremes.


The Cv vs. T plot for deuterium behaves similar to traditional fluids with saturated vapor line crossing the saturated liquid line.


The deuterium equation was complete in 2013 and is a 21 term, Helmholtz explicit equation of state.




Tritium is the heaviest of the hydrogen isotopes having 2 neutrons.  Tritium is highly radioactive having a half-life of 12.3 years.  Tritium is very rare in nature accounting for only 3 in 10^18 atoms of hydrogen.  The radioactive decay causes phosphors to glow making it useful for self-power lighting in devices like watches, exit signs, firearm sights, etc.  Tritium is also used in nuclear weapons and as fuel for fusion reactors like ITER and NIF.  Tritium costs a staggering $30,000 per gram.  The commercial demand is 400 grams per year.  Pure tritium is produced  by cryogenically distilling enriched deuterium gas streams that have been electrolyzed from the heavy water CANDU reactors.

Cryo Distillation





There are only a handful of vapor pressure and density experimental measurements for tritium. The available information on the critical and triple point properties are discussed below:

Tritium Properties

















A Helmholtz explicit equation of state does not exist for tritium.

ME 527 Lesson 35: Binary mixture phase diagrams and property estimation

In the 1960’s Konynenburg and Scott extensively analyzed the qualitative behavior of the Van der Waals EOS to identify distinct classifications of fluid mixtures to draw similar phase envelopes. The Gibbs energy is utilized to draw the critical loci. Specifically when:

(∂²Gm/∂y²)|T,P = 0 and (∂³Gm/∂y³)|T,P = 0, where Gm is the molar Gibbs energy of the binary mixture. When applying this to the Van der Waals equation Koynenberg needed to relate the parameters (a and b) of the Van der Waals EOS in order to generalize this behavior:

ξ = (b22-b11)/(b11+b22) expresses the differences in sizes between the two molecules of the fluids

ζ = (a22/b22²-a11/b11²)/(a11/b11²+a22/b22²) is related to the difference in critical pressures and temperatures between the fluids

Λ = (a11/b11²-2a12/b11b22+a22/b22²)/(a11/b11²+a22/b22²) is related to the low temperature enthalpy of mixing.

The enthalpy of mixing is key to the complexity of fluid behavior and strongly related to the change in Gibbs Energy (g=h-Ts). If Λ is too positive or negative, the system will form an azeotrope. By converting the Van der Waals equation into Helmholtz energy (close to Gibbs and simpler) Konynenburg was able to come up with the following classifications of fluids that should be observed:

Class 1: Continuous Critical Line– Mixtures of two components with similar gas-liquid critical temperatures. In such mixtures the critical points of the pure components are continuously connected by a critical line.

Type I. One critical line: C1 to C2 (G-L)

Type I-A. The same as I, with the addition of a negative azeotrope

Type II. Two critical lines: C1 to C2 (G-L); Cm to upper critical end points (UCEP) (L-L).

Type II-A. The same as II, with the addition of a positive azeotrope.

Class 2: No Continuous Critical Line: Mixtures of two components with very different gas-liquid critical temperatures. In these mixtures there is no continuous critical line joining the critical points of the pure components.

Type III. Two critical lines: C1 to upper critical end points (G-L); Cm to C2 (L-L to G-L. A three phase line runs from a upper critical end points to P=0.

Type IV. Three critical lines: C1 to upper critical end points Two critical lines: C1 to upper critical end points (G-L); Cm to C2 (L-L to G-L. A three phase line runs from a upper critical end points to P=0, T0 (G-L); lower critical end point to C2 (L-L to G-L); Cm to upper critical end points (L-L).

Type V. Two critical lines: C1 to upper critical end points (G-L); lower critical end point (LCEP) to C2 (L-L) to G-L).

Type V-A. The same as V with the addition of a negative azeotrope.

Class 3. Very complex mixtures exhibiting such phenomena as low critical solution temperatures (LCSTs). In this class the low-temperature phase behavior results from strong specific interactions between the two components. These interactions lead to a high degree of order in the liquid mixture and thus cannot be represented by a ‘one-fluid’ equation of state such as that of Van der Waals. Carbon dioxide and butane are good examples of this type of mixtures.

These classes and types are show qualitatively:


Konynenberg Binary Phase diagrams

The three generalizing parameters (ξ, ζ, and Λ) can, in principle, be applied to all possible fluid mixtures. Clearly this is too complex for visualization. Konynenburg chose the simplified case of equal-size molecules ξ=0 to construct maps of how just the differences in critical points and heats of mixing relate to azeotropes of binary fluid mixtures and the Van der Waal’s equation. He was then able to further break down the classification of different fluids further into the graph on the left. The graph on the right has unequal, ξ=0.33, sized molecules.


Konynenberg Binary Phase diagrams

From the map we can see that fluids of equal sizing molecules conforming to more ideal fluid mixing fall along the Λ=0 line and have little heat of mixing and no azeotrope formation. Positive azeotropes occur with positive values of Λ and vice versa. As the differences in critical parameters (ζ) increases, the higher types of mixtures emerge as the critical regions have difficulty, and in many cases cannot overlap. As can be seen in the right hand figure, the behavior becomes even more complex as the molecule sizes begin to deviate ξ≠0.

What should be taken away from this is that mixtures, even just binary, are incredibly complex! From here, we could begin with one class of fluid mixture EOS, identify the mixture we are using, find the mixing rules/parameters from a database, and begin estimating mixture properties. Which classification you end up using is likely dependent on the culture of the organization you work with. Here’s a few general databases to get started:

  1. Predictive Soave-Redlich-Kwong
  2. UNIversal QUAsi-Chemical (UNIQUAC)

Next time we’ll dive into how REFPROP (NIST standard) works through mixtures and the available models it employs.


ME 527 Lesson 34: Multi-component Phase Equilibrium and Azeotropes

Last week we discussed analysis of a single-phase binary gas mixture using both Kay’s Law and cubic EOS. It’s actually fairly straight forward to derive a phase envelope for a fluid mixture (see page 773 in your text). As we learned earlier in the semester in the discussion of Gibbs energy, g, is equivalent to the chemical potential, μ, under certain conditions of constant temperature and pressure through changing molar composition.The chemical potential of each substance in the mixture must be in equilibrium during all coexisting phases. If a gas phase obeys the ideal gas law then μl=μv and

xiPsat,i=yiP and a phase envelope can be drawn for two fluids.

phase football

The points at either end correspond to the saturated vapor temperature of the pure componet at the 2 atm pressure. This shape is commonly known as the ‘football’ of simple fluid mixtures. It is able to predict, for very similar fluids, when one fluid will begin to condense or boil from the mixture at given temperatures or pressures and compositions. The key deviation from this type of behavior is the existence of azeotropes:



The name azeotrope literally means ‘constant boiling’ and applies to mixtures that maintain constant composition of liquid and vapor during boiling. A well known azeotrope is in the ethanol-water mixture where water cannot be further removed from a 96% ethanol mixture through distillation. These appear as ‘kinks’ in out traditional football phase diagrams:

Whether an azeotrope is referred to as positive or negative is based on whether the vapor pressure increase (positive) or decreases (negative) below the vapor pressures of both pure components.

Next time we’ll dive into how Koynenenberg used this information to generalize classification of binary mixtures.





ME 527 Lesson 33: Fluid Fridays — Hydrogen

Continuing down our list of cryogenic fluids we arrive at Hydrogen, the lightest stable atom or molecule on our list of cryogens:

  1. Krypton (119.73 K)
  2. Methane (111.67 K)
  3. Oxygen (90.188 K)
  4. Argon (87.302 K)
  5. Fluorine (85.037 K)
  6. Carbon Monoxide (81.64 K)
  7. Nitrogen (77.355 K)
  8. Neon (27.104 K)
  9. Deuterium (23.31 K)
  10. Hydrogen (20.369 K)
  11. Helium (4.222 K)

Discovery, origins, and current usage

Discovery of hydrogen is attributed to Robert Boyle in 1671 after reacting iron filings with acid, although Henry Cavendish is often given credit for the discovery in 1766. Antoine Lavoisier named the element hydrogen in 1783 from hydro “water” gen “creator” as he verified the gas produces water when combusted.

Today we know that hydrogen is 75% of all baryonic mass in the universe:

Here on earth the proportions change and hydrogen is the fifth most abundant:

Hydrogen comprises just over 10% of the mass of earth’s oceans.

James Dewar first liquified hydrogen in 1898 after replecating the experiments of Cailletet, Pictet, Wróblewski, and Olszewski. The prior experimentalists had difficult preserving the liquified samples and many of the experiments were considered ‘dynamic’ instead of static due to the lack of accumulated liquid. Dewar invented the silvered vacuum flask to minimize radiation heating of the liquid. This allowed for bulk storage. As legend has it, Dewar benefited from an insight that occurred when a family friend, whom recently given birth, was needed to keep milk warm. Dewar used one of his silvered vacuum flasks to store the milk, which was the invention of the Thermos. However, Dewar did not patent the invention and lost the rights when the Thermos corporation went into production.

Cryogas recently published an analysis by PNNL researcher Daryl Brown of the 2014 US and World hydrogen market. Hydrogen production is divided into the following categories: 1) Captive: 5.86 Million Metric Tonnes (MMT) (55.39 World), 2) Byproduct: 5.68 MMT (40.93 World), 3) Merchant: 3.83 MMT (6.16 World), 4) Totals: 15.36 MMT (102.48 World). The consumption by industry breakdown includes 1) Oil refining (10.10 MMT, 32.10 World), Ammonia (2.03 MMT, 31.81 World,), Coke/Iron (0.38 MMT, 15.63 World), Methanol (0.43 MMT, 9.43 World).

Fixed Properties

Hydrogen fixed points


Surface of State

Here I’m showing the property plots for Parahydrogen. I fit the parahydrogen EOS first, because the best data were available in the parahydrogen form. Normal hydrogen and then orthohydrogen were fit as derivative EOS to the parahydrogen EOS.

Hydrogen Pressure versus Density

Hydrogen Heat capacity versus Temperature

In the CV plot you can clearly start seeing problems in the low temperature liquid phase near the triple point. I did not have the ability to look at the melting line behavior until the entire EOS was finalized.

Hydrogen Sound Speed versus Temperature

Although the sound speed behaves well, the Gruneisen below has problems in the same region as the heat capacity data.

Hydrogen Gruneisen versus Temperature

Although the Gruneisen had problems, the PIP is much better. The slight increase in slope on the liquid side near the triple point may be an issue.

Hydrogen PIP versus temperature











ME 527 Lesson 32: cubic-EOS mixing rules

What we saw early on in class is that decidedly different schools of thought emerged early on for fluid equations of state. The EOS’s ranged from virial, 2-parameter a.k.a. Cubic EOS of Van der Waals, Peng-Robinson (PR), and Redlich-Kwong-Soave (RKS), to the multi-parameter EOS of Benedict-Webb-Rubin and reduced Helmholtz versions. Each class of EOS has different mixing rules. While we’ve been covering generalized mixing models and ideal mixing models, we need to handle phase change with mixtures. The simplest class of EOS for this are the 2-parameter ‘cubic’ equations.

Van der Waals EOS

Van der Waals was one of the first to attempt prediction of fluid mixtures. Van der Waals defined his fluid mixture equation of state as a simple modification of his usual EoS:

(P +a_x/V²)(V-b_x) = RT

where a_x and b_x are the two-parameter coefficients characterizing the fluid mixture. Van der Waals had to then decide what “mixing rules” to apply to deduce a_x from a_1 and a_2, and b_x from b_1 and b_2 (for a two component mixture). Since these should assume pure-fluid behavior at x=0 and x=1, he initially chose a quadratic function to scale between.

a_x=(1-x)²a_1+2x(1-x)a_12+x²a_2 and b_x=(1-x)²b_1+2x(1-x)b_12+x²b_2

Although Van der Waals did not commit to a specific mixing rule for a_12, a simple geometric mean can be assumed a_12=(a_1*a_2)^0.5 (See Sengers 2002).


The Redlich-Kwong EOS was originally developed in 1949 and was subsequently modified by Soave in 1972. The RKS equation is cubic similar to Van der Waals’:

P=RT/(v-b)-a/(v(v+b)) where a=0.42747αR²Tc²/Pc and b=0.08664RTc/Pc

The parameter α is defined by

α^0.5=1+m(1-Tr^0.5) where m=0.48+1.574ω-0.176ω²

This parameter ω is called the acentric factor. It was originally defined by Pitzer in 1955 as an additional empirically determined parameter to account for deviations from Corresponding States. It is defined by the curvature of the saturation line approaching the critical point:


In order to mix fluids with RKS the mixing rules for the coefficients include:

a=(∑yiai^0.5)² and b=∑yibi.


ME 527 Lesson 31: General Mixing Rules

Up until now, our treatment of mixtures has assumed that they were either: 1) ideal, or 2) pseudo-ideal, i.e. air or moist gas with phase change (psychrometrics). Both are relatively simple from an applications standpoint and we did not need a rigorous set of rules to scaffold the analysis. As we dive further into real fluid mixtures though, the complexity increases quickly. We need to lay out a set of simple rules to follow for approaching our mixing analysis.

Rule #1: The Gibbs Phase Rule

J. Willard Gibbs introduced one of the most simple, yet powerful, rules of thermodynamics almost in passing. The Gibbs Phase rule states that:

f = c – p + 2

where f is the degrees of freedom (the total number of independent variables needed to fix a fluid state), c the number of components (species) in the mixture, and p the number of phases. For most tradition thermodynamic analysis of real fluids just one fluid (c=1) is used with one phase (p=1), so the number of independent pieces of thermodynamic information is two (f=2). For a real fluid in a saturated state c=1, and p=2 so f=1. At the triple point c=1, and p=3 so f=0. The critical point is special in that the two criticality conditions must be obeyed, dP/dv=0, and d²P/dv²=0 and two degrees of freedom are lost so f=0.

Getting into binary mixtures c=2 but the number of phases p adds addition complexity. A homogeneous phase (vapor+vapor, liquid+liquid) has f=3 degrees of freedom. A two-phase system (liquid+vapor) has f=2 degrees of freedom, and a three-phase system has one (vapor-liquid+vapor).

Getting into ternary and higher mixtures the degeneracy increases.

Rule #2: Dalton’s Rule

Approximates the total pressure by the sum of the partial pressures:

P=P_1+P_2+P_3…..  P=y_i*P_i

where y_i is the mole fraction. We covered this in class last week.

Rule #3: Amagat’s Rule

One of the things I absolutely stress in undergraduate thermodynamics is that VOLUME IS NOT CONSERVED to prevent folks from adding volumetric flow rates without considering differences in density of the flows. However, in mixture land knowing whether the volume of two substances sums to the combined volume is actually important. Enter Amagat’s Rule:


This is also known as an ideal solution- the substances being mixed do not notice the presence of eachother. Of course this works well for ideal gas mixtures and is how we solved the CO2 sequestration problem we went through earlier in class.

Rule #4: Kay’s Rule

This is the first mixing rule that begins to account for real-fluid effects on density changes. Remember our discussion from Corresponding States, for most light-gases and petroleum components the compressibility factor Z causes the fluids to follow general compressibility lines. If all of these fluids conform to these universal lines, then by reason, we should be able to approximate a mixture using similar techniques. Kay’s rule creates pseudo-critical points for mixtures for precisely this purpose:

T’c=∑yiTci,  P’c=∑yiPci, and V’c=∑yiVci

Let’s take an example of a 50-50 nitrogen + argon mixture at T=150 K and P=8.5 MPa. You’ll need the critical point information for N2: Tc= 126.19 K, Pc= 3395800 Pa, Vc= 1/313.3 m³/kg, molarmass = 28 kg/kmol, and Ar= Tc=150.69 K, Pc= 4.863 MPa, Vc=535.6 kg/m³, molarmass = 40 kg/kmol.

T’c=0.5*(126.19)+0.5*(150.69) = 138.4 K  this gives a T’r= T/T’c = 150/138.4 = T’r = 1.084

P’c=0.5*(3395800)+0.5*(4863000) = 4,129,000 Pa this gives a P’r= P/P’c = 8.5/4.129 = P’r = 2.059

Going back to our plot of the compressiblity reduced isotherms we find that z’=0.38. Since Z=Pv/RT we just need the gas constant for the mixture. The apparent molecular weight of the mixture is: mw=0.5(28)+0.5(40) = 34 kg/kmol.

This gives v=0.00164 m³/kg. REFPROP gives the solution of 0.001834 m³/kg, 11.8% off. Not bad.

How far off would we have been calculating the volumes independently and averaging them?

From the ideal gas law v_n2= 0.005238 m³/kg, and v_ar=.003673 m³/kg, for v_ave=0.004456 m³/kg, way off!

Using the corresponding states curves with z_n2=0.4, v_n2=0.005761 m³/kg, and z_ar=0.55, v_ar=0.00404 m³/kg. So v_ave=0.004905 m³/kg.

What is clear from this, it’s important to consider the fluid as a mixture, and not simply a combination or weighted average of the separate components. Considering how precise corresponding states was for nitrogen and argon, even when we use Kay’s law the deviation from the standard is significant and shows the inadequacies of Kay’s law when applied to mixtures.



ME 527 Lesson 30: Fluid Fridays — Neon

Continuing down our list of cryogenic fluids we arrive at Neon, the first of the “quantum” fluids on our list of cryogens:

  1. Krypton (119.73 K)
  2. Methane (111.67 K)
  3. Oxygen (90.188 K)
  4. Argon (87.302 K)
  5. Fluorine (85.037 K)
  6. Carbon Monoxide (81.64 K)
  7. Nitrogen (77.355 K)
  8. Neon (27.104 K)
  9. Deuterium (23.31 K)
  10. Hydrogen (20.369 K)
  11. Helium (4.222 K)

Discovery, origins, and current usage

Neon was discovered in 1898 by William Ramsay during the process of isolating the gases contained in air. As we discussed earlier during our Fluid Friday discussions of Krypton and Argon. Ramsay liquefied a large quantity of air and through carefully distillation discovered the gases in order of their approximate abundance in the atmosphere (Argon, Neon, Krypton, Xenon), for which he won the 1904 Nobel Prize in Physics:

The Nobel Prize in Physics 1904 was awarded to Lord Rayleigh “for his investigations of the densities of the most important gases and for his discovery of argon in connection with these studies”.


Today, neon is produced at air separation plants that have an additional distillation column for neon separation. As Richard Betzendahl explained last December in Cryogas magazine, the former Soviet Union installed these columns on all air separation plants in the late 1970’s through 1990’s during the “SpaceWar” years. This created a glut of crude neon on the market that forced many eastern European air separation plants to vent or mothball their crude neon systems in the late 1990’s and 2000’s. At the same time, increased use of excimer laser systems for the production of micro-chips has increased steadily at 6-10% for the last 4-5 years. Recent conflicts in the Ukraine and Crimea have caused the price of this crude neon to fluctuate highly. The price is currently 100x what it was in 2012. We recently purchased a bottle of pure neon for ~$2/standard liter, that’s about $2.40/gram! As Betzendahl reports, most gas shortages don’t last longer than 2 years and the price is supposed to reduce in 2016.


Fixed Properties

The most recent equation of state for neon was developed in 1986 at the Center for Applied Thermodynamic Studies (CATS) at the University of Idaho. The equation is fundamental (Helmholtz explicit) with 29 terms.

Neon Fixed Point Properties

Surface of State

Remember from our Quantum Law of Corresponding States discussion that neon is the first fluid that begins to differentiate itself from normal fluids as measured by the quantum parameter and virial behavior. With a quantum parameter near ~0.6, neon is still more classical than quantum from an interaction standpoint. So let’s get into the surface.

Looking at the standard P-rho plot we see that liquid neon is very dense, even denser than water. The rectilinear diameter also shows that the critical point density may be out of place.


Where the standard pressure density plot looks fine, more complex thermodynamic properties reveal another story. Although the isochoric heat capacity of the saturated vapor crosses the saturated liquid line like most other fluids, the saturated vapor line diverges as the triple point is approached. This is clearly not physical. The behavior of the surface at low temperatures and near the melting line is also suspect. Only severe phase changes could cause this type of behavior.

Neon Heat capacity versus temperature

Moving onto speed of sound versus temperature we see the similar problem on the liquid side near the triple point temperature where the surface “turns over” on itself. Sound speed should not decrease with increasing pressure at fixed temperature. What also begins to be noticed is how condensed the critical region appears relative to other fluids that have more expansive regions.

Neon Sound Speed versus Temperature

As we recall, the Gruneisen is a higher level property, but similar to the sound speed. Here we again see the problem on the liquid side near the triple point, but also on the vapor side at low temperatures.

Neon Gruneisen versus Temperature

Finally the Phase Identification Parameter (PIP) which we introduced during our discussion of thermodynamic property trees shows serious issues on the liquid side of the surface, in addition to the vapor line at low temperatures.


What’s important to remember about all of this is that in 1986 we barely had the use of personal computers or the ability to generate these types of surfaces that are indicative of equation of state behavior. Looking at this now we know that we could quickly improve a surface like this. Your classmate, Ian Richardson, is currently taking liquid neon measurements upstairs which could serve as the impetus to go back and update this equation.













ME 527 Lesson 26: Quantum Law of Corresponding States

Last time we introduce the Theory of Corresponding States through the historical context via Van der Waals’ equation of state. At the end of class we tried a simple exercise to estimate the density of nitrogen and hydrogen via classical corresponding states. Although nitrogen was accurate to 0.8%, hydrogen was off by 12%. Here’s a plot over our cryo fluids of interest this semester:


So clearly, something weird is happening with the very low temperature fluids neon, deuterium, hydrogen, and helium. The deviations are primarily due to the classical corresponding states curves being developed for hydrocarbon and refrigerant blends. The heavy gases like krypton, and the very light gases are clearly not well modeled by the technique. Why?

Early in the semester we introduced the Kammerlingh-Onnes’ virial equation:


If the corresponding states in compressibility (Z) space are not working then something must be happening to the virial coefficients of these light gases. Plotting the virials we see that is indeed the case where a classical curve exists and the quantum fluids depart from:

Virial coefficients versus reduced temperature

But how can this be if the particles are acting as billiard balls? As early as 1922 Byk suggested that all deviations from classical corresponding states should be due to Quantum Mechanics. That is indeed a very early suggestion. Although Wave-particle duality was first postulated by Einstein in 1905:

“It seems as though we must use sometimes the one theory and sometimes the other, while at times we may use either. We are faced with a new kind of difficulty. We have two contradictory pictures of reality; separately neither of them fully explains the phenomena of light, but together they do

it wasn’t until 1924 when Louis de Broglie extended the theory to all matter. De Broglie’s premise was simple, if light has mass, which it does, a very, very small amount, then all mass should also have wave-particle duality. When he submitted his Ph.D. thesis the committee sent it to Einstein and inquired whether he should receive a diploma. Einstein famously replied something along the lines of: “The matter of the diploma is up to the committee. But if it’s up to me, I’d award him the Nobel Prize.” Which he did receive in 1929:

The Nobel Prize in Physics 1929 was awarded to Louis de Broglie for his discovery of the wave nature of electrons.

Mathematically speaking the reasoning went like this:

E = mc² where m is mass and c is the speed of light, has to equal E = hf where h is Plank’s constant and f is the frequency of a light wave.

The wavelength is defined as speed over frequency: λ = c/f and the momentum defined as energy over speed: i = E/c. By substituting these into above we arrive at the definition of a de Broglie wavelength:

λ = h/i  the de Broglie wavelength is equal to Plank’s constant divided by the momentum of the particle, atom, or molecule in question. But what does this mean? As the mass of molecules decreases, so does the momentum, since Plank’s constant is a constant, the wavelength increases. Eventually the particles will behave more wave-like than billiard ball-like.

For reference a heavy baseball hit at 110 miles/hour has a de Broglie wavelength λ = 9.5 x 10^-35 meters. A hydrogen molecule moving at 1 x 10^6 m/s has a wavelength λ = 2 x 10^-13. The wavelength of hydrogen is 21 orders of magnitude larger than the baseball because the mass is so small!

Now the key challenge is relating relevant properties of these light gases to the de Broglie wavelength. Going all the way back to statistical mechanics we’ll find a direct relationship for calculating the 2nd virial coefficient:

B = 2πNa ∫[1-exp(-U/kbT)]r²dr

where Na is Avogadro’s number, U is the intermolecular potential, kb is Boltzmann’s constant and r is the distance between molecules. The intermolecular potential is the attraction and repulsive forces between molecules. The Lennard-Jones 12-6 potential is one of the simplest forms:

lennard-jones potential

From this, σ is the radius (distance between molecules) at the maximum well depth (ε) which provides a measure of the molecule speed. Combining these with the definition of the de Broglie wavelength above allows estimation of an approximate wavelength for the molecules:

λ = Nah/(Mε)^0.5 where M, the molecular weight, times the well depth and taking the square root gives momentum.

In 1947 de Boer proprosed a dimensionless comparison between the de Broglie wavelength for a molecule above and the average radius of interaction σ to see if on a scale relevant to the molecules waves or particles dominated. Here’s a plot of the de Boer, aka Quantum Parameter calculated for the cryogenic fluids:

Quantum Parameter

From this we can see that tritium is the balance point where wave and particle effects balance and as we reduce in mass, the molecules become more wavelike than particle like. For example, the hydrogen wavelength is nearly double the average radius of particle interaction! Moreover, we can now use this Quantum Parameter to scale (similar to Z) and map the fluids relative to one another. First we have to non-dimensionalize the temperature and pressure by the lennard-jones parameters (similar to how Van der Waals used the critical point as reducing parameters): T*=RT/ε and P*=Naσ³P/ε Now plot versus quantum parameter. First reduces temperature versus quantum parameter:

reduced temperature versus quantum parameter

Now the pressure:

reduced pressure versus quantum parameter

So if we know the difference in intermolecular potential between these quantum fluids, we can estimate properties using the slope of the fit lines above and the equations below. I then used the equations to transform parahydrogen vapor pressures into normal hydrogen vapor pressures for developing the equations of state we talked about previously.

Vapor Pressure Transformation

Another application was during my Ph.D. dissertation when we were trying to correlate the shear strengths of solid cryogens. I had a hunch that the shear stress would reduce in a similar way as the pressure. When I tried it, I almost fell out of my chair when I made the plot:

Reduced shear stress versus quantum parameter

Which then meant we were able to estimate shear strength data for tritium with a high degree of confidence:

transformed shear strength data



ME 527 Lesson 25: Corresponding States

We’re now shifting gears, and deviating from the syllabus, to touch on a classic topic that will set the stage for our Fluid Friday discussions this Friday.

A fantastic book I’ve recently discovered covering the last half of class is “How fluids unmix: Discoveries by the School of Van der Waals and Kamerlingh Onnes” by Johanna Levelt-Sengers. The book covers the primary historical context from which we can understand the status quo. Levelt-Sengers studies in Kamerlingh Onnes’ lab under Michels and is very familiar with the historical background.

The real triumph of the Van der Waals equation started from the very beginning. Van der Waals wrote his equation


and realized he could find the critical point from derivatives: (∂P/∂V)|T=0, (∂²P/∂V²)|T=0:

By solving for the critical parameters directly we get:

Pc=a/27b², Vc=3b, RTc=8a/(27b), and the combined critical ratio of PcVc/RTc=3/8.

This then allowed his equation to be generalized to other fluids if the critical parameters themselves are known. Know Pc, Vc, and Tc? Then you can find a and b for the fluid. Plug a and b into the top equation and you can P, V, or T.

Van der Waals propsed directly using the critical parameters as units of measurement in his equation. Mathematically:

(P*+3/V*²)(3V*-1)=8T* with P*=P/Pc, V*=V/Vc, T*=T/Tc.

Note that this equations is generalized, or universal. The reduces pressures of two fluids are the same if the fluids are in corresponding states, that is, at the same reduced volume and temperature. Kamerlingh Onnes used this technique to estimate the amount of liquid hydrogen he would need to liquefy helium. He wrote Van der Waals a letter stating:

“In what I described to you, your theory has been my guide. To calculate the critical temperature of a permanent gas from the [P-V] isotherm brings your dissertation to memory in a new way. The calculations were performed entirely on the basis of the law of corresponding states. Guided by that law, I estimated – even though I did not put that on paper – to need 20 liters [of hydrogen]. Had I estimated a few liters fewer, the experiment would not have succeeded – had I estimated much more, then I would have judged it unwise to proceed, in view of the available resources.” ~ Letter from Kamerlingh Onnes to Van der Waals on his retirement

As demonstrated, Van der Waals’ equation is good enough to get in the ballpark, he used saturated vapor pressure (P,T) data of water to estimate the critical temperature of 390°C, not far from 373.9°C for an extrapolation of 150°C. However the inaccuracies of his cubic equation became clear, estimation of the critical ratio for common fluids ranged from 0.23 for water to 0.29 for nobel gases, far away from the 3/8 his equation predicted. As you know, we developed better cubic equations (Peng-Robinson and Redlich-Kwong) and corresponding states can be applied in much the same with these as Van der Waals originally intended. So in the end, the principle — of corresponding states — was established and is still used today.

Let’s apply this in an example. Remember back to our derivation of the virial coefficients when the compressibility (Z) originated? Compressibility, Z=Pv/(RT), is similar to the critical ratio above as Van der Waals originally defined. Now, 150 years later we have many experimentally determined corresponding fluid states to test the theory with. When taking most normal fluids it’s then possible to overlay different fluid measurements on the same compressibility plot similar to the following:

Reference: Cengel, Y. A. and Boles, M. A., “Thermodynamics: An Engineering Approach”, 7th ed., WCB/McGraw-Hill, Boston, 2011.

Let’s estimate the specific volume of nitrogen at T=150 K and P=8.5 MPa using the above chart. You’ll need the critical point information for N2: Tc= 126.19 K, Pc= 3395800 Pa, Vc= 1/313.3 m³/kg.

Now let’s do the same for hydrogen at T=40 K and P=3.2 MPa using the above chart. You’ll need the critical point information for H2: Tc=33.145 K, Pc=1296400 Pa, Vc=1/31.362 m³/kg.


How’d you do?

Nitrogen: v_act=0.002834 [m^3/kg], Hydrogen: v_act=0.02441 [m^3/kg]

Washington State University