lecture 17 - For a system that has three phases...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: For a system that has three phases (solid, liquid, and gas), the complete phase diagram in the (P,T) plane looks like this: Each of these curves is described by the Clapeyron Eq. Since the molar volume of liquid can be lower than that of the solid (e.g., for water), dP/dT for the liquid ­solid coexistence can be either positive or negative. The above picture corresponds to the case of dP/dT<0. T is called the triple point and C is the critical point. For water, the triple point is at P~ 611 Pa and T ~ 0.010C, and the critical point is at 647K and ~220 bar = 220 x 105 Pa A gas can be converted to liquid either by lowering T or via compression (i.e. by raising P). It is possible to go from gas to liquid without experiencing a phase transition. If we fix P and raise T our process is s →l→g (melting then vaporization) if P > PT or s →g (sublimation) if P < PT Question: why no critical point on the liquid ­solid coexistence curve? The phase diagram of He. Helium has a very unique phase diagram: This shows that, unless at very high pressure, helium will stay liquid all the way down to zero temperature. To understand this, you will need to know quantum mechanics. Even a more interesting phenomenon is the superfluid phase of helium. Superfluid liquid can creep up the walls of a container and escape it. It can also flow without any friction. To understand superfluid behavior, you will need to know quite a bit of quantum mechanics and condensed matter physics. Unusual states of matter (supercooling, superheating, and the glassy state) If we lower the temperature of a gas, condensation will take place. Often condensation begins through the formation of droplets of liquid within the gas. Consider a spherical droplet of liquid in vapor. When can these be in equilibrium? What’s missing in our previous analysis is the excess energy due to having a surface, an interface between the liquid and the gas (a molecule at a surface has a higher energy than one inside because it has fewer neighbors). This excess energy is proportional to the surface area. If the droplet has radius r, the Gibbs energy will have the form: G = ng µ g + nl µl + 4πσ r 2 , where σ is the surface tension coefficient. We have nl = 4π r 3 / 3Vl and ng = n − 4π r 3 / 3Vl so that G = (n − 4π r 3 / 3Vl )µ g + 4µlπ r 3 / 3Vl + 4πσ r 2 View this as a function of r and try to find the equilibrium value of r: The condition is dG/dr = 0: G′ = (4π r 2 / V )l µ g + 4µlπ r 2 / Vl + 8πσ r = 0 This equation has two solutions: One described buy the equation µ g = µl + (2σ / r )Vl and the other is r=0. Consider the first solution. One can interpret this equation by saying the chemical potential inside the droplet became redefined, µl′ = µl + ΔPVl where ΔP=2σ/r is the excess pressure inside the droplet. Using this interpretation, the equilibrium condition is the same as before: the chemical potential should be the same in the liquid and in the gas. Note that the pressure difference inside and outside the droplet leads to all kinds of interesting phenomena (capillary effects). Consider the equilibrium condition with nonzero r: if µg < µl it cannot be satisfied as the radius must be positive. However below the boiling point µg > µl and our equilibrium condition is satisfied for r = r* = 2σ Vl /(µ g − µl ) We still have to check if what we found is a real equilibrium condition. For r* to be a minimum, the second derivative of G must be positive. G ''(r*) = −8π r * µ g / Vl + 8π r * µl / V + 8πσ = −8πσ < 0! What we found is a maximum not a minimum. The free energy G(r) would tend to decrease. That is, if I start with a big droplet, r>r*, it will grow (move away from the maximum) and if r<r* it will disappear. This means that below the condensation point there’s a certain critical size of a droplet that has to form to result in condensation. The gas will remain stable with respect to the formation of small droplets, smaller than r*. As the temperature goes down, it gets easier to create such a critical droplet, as r* decreases. Such a state of the gas is called supercooled vapor. It is a metastable state. In reality, it’s hard to supercool vapor because some impurities, dust, may be condensation centers. In addition, if there are walls, condensation may occur at the walls, which does not require critical droplets (a wall is flat, r is infinite!). ...
View Full Document

Ask a homework question - tutors are online