Q:

A cylinder with piston contains only water and steam in equilibrium. The cylinder is insulated and the piston too. So the process inside it may be thought as adiabatic. We start to compress further the steam inside the cylinder.
What happens exactly ?

- George Kourtis (age 48)

Athens, Greece

- George Kourtis (age 48)

Athens, Greece

A:

Here's the phase diagram of water, to help picture the answer.

https://cnx.org/contents/RTmuIxzM@9.17:oFoO44pW

cnx.org/contents/RTmuIxzM@9.17:oFoO44pW

As long as there's some liquid and some vapor, the material has to stay on the liquid-vapor coexistence line. As you squeeze on the cylinder, you're doing work, adding energy, and raising the temperature. The material will move up that line, toward the critical point. If there's initially just the right ratio of liquid to gas, the critical point will be reached. After that you lose the phase separation and just continue toward higher p and T in the highly non-ideal supercritical fluid.

Mike W.

*(published on 08/01/2013)*

Q:

The answer tells me what I already knew.The phase state diagram doesn't tells us if water becomes steam or steam becomes water in that adiabatic compression.
What I don't know is it the water becomes steam or the steam water and that isn't shown there because that diagram doens't shows what happens exactly.

- George Kourtis (age 48)

Athens, Greece

- George Kourtis (age 48)

Athens, Greece

A:

This is a great question, which I'd evaded in the first answer. (That answer is now slightly modified after further thought.)

Let's look at the reverse process, adiabatic expansion. Start at the critical point. The adiabatic expansion will cool the material and reduce the pressure (p), giving phase separation. This adiabatic process will have a *unique outcome*, so at any particular temperature (T), the ratio of gas molecules to liquid molecules is determined. Reverse the process and of course you return to the critical point, as described in the previous answer.

Here's another way to see why there's only one special liquid/gas ratio at some T that will get to the critical point on adiabatic compression. For a given number of molecules, there's some particular entropy at the critical point. Reversible adiabatic processes don't change entropy. At any lower T, the liquid and gas have different entropies per molecule. So at each T there will only be one liquid/gas ratio that will give the correct entropy to match the critical entropy.

Here's something more interesting. What if you start at some sub-critical T but not at the right ratio of gas to liquid? What I realized as I started to calculate was that if there's more liquid than the special ratio, adiabatic compression won't heat things up much because of the large heat capacity of the liquid. So the increase of vapor pressure due to heating won't be enough to compensate for the loss of volume. Thus the number of gas molecules will decrease further on compression, falling farther below the special ratio for the new T. Before reaching the critical point, all the gas will be lost. Then you'll leave the coexistence line and just get plain adiabatic compression (with just a little heating) of the liquid.

What if there's *less* liquid than the special ratio at that T? Then you get a lot of heating on compression, so the increase of vapor pressure increases the fraction of gas more than the volume loss decreases it, more than it would have starting at the special ratio. That means that on further compression, all the liquid will convert to gas before you get to the critical point. Once that happens, you leave the coexistence line and get simple compressive adiabatic heating of the (somewhat non-ideal) gas.

Bottom line: whether the compression makes more liquid or more gas depends not just on p and T but also on the starting liquid/gas ratio. There's a special liquid/gas ratio at each T, and whichever phase beats that ratio gains further, compared to the special ratio, as the cylinder is compressed. It's sort of like economics.

You may be particularly interested in water around room temperature, for which some approximations are good. The heat capacities of the vapor and liquid are fairly constant, c_{G} and c_{L}, as is the latent heat, L. The gas is nearly ideal, much less dense than the liquid, which is nearly incompressible. I believe that if we call the fraction of the molecules in the gas "x", we get:

dx/dln(T)=x-(kT/L)(xc_{G} +(1-x)c_{L}), where k is Boltzmann's constant. Near room temperature for water we have roughly c_{G} =4k and c_{L}=9k and L/kT = 16, each per molecule. So if more than about 40% of the molecules are in the gas, compression makes more gas. If less than that are in the gas, compression makes more liquid. I'm not sure exactly what percent would be right to stay on the critical adiabat.

Mike W.

p.s. You may also want to know at some particular T< T_{C} whether this compression increases or decreases the fraction that's in the gas phase, assuming it's right on the adiabat that leads to the critical point. That's a hard problem but I can give a crude estimate of the answer. Say T is well below T_{C}. What fraction N_{G}/N of the material needs to be gas to make the entropy, S, the same as the critical entropy? Let's say that the liquid, not very compressible, has S dependent mostly on T, not p. Let's also say that its heat capacity per molecule is fairly constant, c. Then if all the material stayed liquid the entropy would drop by an amount Nc*ln(T_{C}/T) on cooling from T_{C}. (There will be some shift from that because the first cooling from T_{C} will be in a range where the approximations aren't close.) To make up for that, you need to take N_{G} molecules from the liquid to the higher-S gas. The S gained per molecule of gas is L/T, where L is the latent heat per molecule at T. So we get roughly N_{G}/N= (cT/L)*(ln(T_{C}/T)+some constant shift). This expression is not necessarily monotonic in T. For low enough T, it increases with T, but for water that region corresponds to the solid-gas equilibrium, not liquid-gas. For higher T, but still less than T_{c}, it can decrease with T, depending on the constant. Then L becomes a strong function of T, approaching zero at T_{c}, and things will get much more complicated.

So although I'm sure that at low T pressing the piston will make more gas, I'm not sure if there is a regime where pressing the piston makes more liquid.

p.p.s. I found a paper ( Melosh, Meteoritics & Planetary Science 42, 2079–2098 (2007)) describing adiabats near a liquid vapor critical point. It's for SiO_{2} rather than H_{2}O, but the ideas should be similar. Figure 2 shows the curves we're interested in, with the p-T axes switched from the figure we show above. Notice that curve "15" follows the coexistence curve, then peels off into the vapor. Curve "5" peels off into the liquid. In between there are several curves that appear, to a crude visual inspection, to go through the critical point. Only one of them can do so exactly.

8/19/2013 Using a simple model, mean-field van der Waals, I can figure out what the liquid/gas ratio must be just below the critical point. Just below the critical point, it looks like there's a leading term in the volume change from the critical value that is proportional to

+/- sqrt(T_{c}-T), the plus solution being the gas and the minus one being the liquid. The lowest order term in the volume change vs. T is linear, so these sqrt terms must cancel. That requires that the gas and liquid fractions be equal just below the critical point.

*(published on 08/01/2013)*

Q:

I am wondering what would happen during adiabatic compression of a boiling cryogen in a piston cylinder? If I consider compression of the bulk liquid and vaporizer to also be adiabatic (ie no heat transfer between liquid and vaporizer) I would expect the vapor to increase at a different rate than the liquid and when I reach my ultimate pressure the molecules that started as a vapor are at a higher temperature than the liquid. If I let that concoction sit and equalize, will the entropy be the same as when I started?

- Ken Kratschmar (age 52)

Vancouver Canada

- Ken Kratschmar (age 52)

Vancouver Canada

A:

Interesting question! I've marked it as a follow-up to a closely related question, on which I worked off and on for several weeks.

Your question differs in that you want to adiabatically compress the gas quickly enough for it to fall out of equilibrium with the liquid. The adiabatically compressed gas will then heat up, without changing its entropy, as for any adiabatically compressed material. Then as the gas and the cooler liquid equilibrate, the net entropy will increase, as must happen in any non-equilibrium process in an isolated system.

As in the process described above in the thread, whether you end up with more or less liquid than you started with depends on the initial ratio of liquid to gas.

Mike W.

*(published on 04/08/2020)*