During a phase transition, what is the nature of the material?

[richardc]

As a material is heated, it increases in temperature until it enters a phase transition. During the transition, added heat does not increase the temperature. As this heat is being added, what is the substance? Is it a mixture of liquid and gas perhaps? If so, what are the proportions?

In other words, what exactly is the substance during one of the flat parts of the temperature vs. heat graph?

http://www.physicstutorials.org/images/stories/phaseofchangeofwater.png

 

[Drakkith]

If we look at water we would see molecules being given enough energy to overcome the bonds between them to turn into a gas. The proportions of gas and liquid is hard to determine, as this depends on how even something is heated, how hot the source is, if the gas can be lost from the container, and other factors. Heating water evenly in a sealed container is obviously different than boiling water on my stove at home.

But, assuming a perfectly sealed container I would expect the amount of gas to liquid ratio to increase as the amount of heat absorbed by the water increases. At a certain point practically all of the water would be a gas. However, as temperature is a measure of the average energy of the particles in a gas, I would expect there to be some liquid present at all times.

 

[Chestermiller]

As a material is heated, it increases in temperature until it enters a phase transition. During the transition, added heat does not increase the temperature. As this heat is being added, what is the substance? Is it a mixture of liquid and gas perhaps? If so, what are the proportions?

In other words, what exactly is the substance during one of the flat parts of the temperature vs. heat graph?

http://www.physicstutorials.org/images/stories/phaseofchangeofwater.png

The flat part indicates a mixture of liquid and gas. At a point along the flat part, the fraction of liquid is proportional to the portion to the right, and the fraction of gas is proportional to the portion to the left. This is called the "lever rule."

 

[mikeph]

The graph shouldn't have sharp turns strictly. Even at 50 degrees there will be some water vapour in thermodynamic equilibrium, it's just a case of that gas/liquid ratio increasing very sharply as you approach 100 degrees. You'd visually see some molecules breaking free from the van der waals molecular interactions, then as the temperature increases, far more gain the kinetic energy necessary to break free.

 

[Chestermiller]

The graph shouldn't have sharp turns strictly. Even at 50 degrees there will be some water vapour in thermodynamic equilibrium, it's just a case of that gas/liquid ratio increasing very sharply as you approach 100 degrees. You'd visually see some molecules breaking free from the van der waals molecular interactions, then as the temperature increases, far more gain the kinetic energy necessary to break free.

You can have all saturated liquid, all saturated vapor, or a combination of saturated liquid and saturated vapor at 50C. So at 50C, the ratio of saturated vapor to saturated liquid can vary anywhere from zero to infinity. The same can be said for the case of any temperature below the critical temperature. There is nothing particularly special about 100 C.

 

[0xDEADBEEF]

Your answers are highly confusing Chester. Of course 100°C is a special point for water (at atmospheric pressure), this is where the phase transition of water from liquid to gas takes place under equilibrium conditions. Yes overheating is possible up to the point where the p(V) curve changes slope, but this is not an equilibrium state and there are strong thermodynamic reasons why substances have a boiling point. Just because there is always some vapour pressure doesn't change that fact. The way you phrase it is sounds more like the behaviour behind the critical point, where there is indeed no difference any more between liquid and solid.

 

[Chestermiller]

Your answers are highly confusing Chester. Of course 100°C is a special point for water (at atmospheric pressure), this is where the phase transition of water from liquid to gas takes place under equilibrium conditions. Yes overheating is possible up to the point where the p(V) curve changes slope, but this is not an equilibrium state and there are strong thermodynamic reasons why substances have a boiling point. Just because there is always some vapour pressure doesn't change that fact. The way you phrase it is sounds more like the behaviour behind the critical point, where there is indeed no difference any more between liquid and solid.

A phase transition between liquid and gas does not only take place under equilibrium conditions at 100 C. It takes place at all temperatures between the freezing point and the critical temperature. Here is a link to the Temperature-Enthalpy diagram for water which illustrates and confirms everything I said:

http://www.google.com/imgres?imgurl...UAxUZXoIsS20QGSvYGYAg&ved=0CDkQ9QEwAA&dur=557

Displayed on the figure the various percentages of saturated liquid and saturated vapor present as a function enthalpy of the mixture at each equilibrium temperature (the green lines). Note also that there is no special significance ascribed to 100 C, 1 atm. on the figure.

 

[0xDEADBEEF]

Well I don't see how that graph is supposed to prove your point. On the right you have the constant pressure lines. The lowest one is for atmospheric pressure. If you are coming from the right you start with gas. When you remove Energy/Enthalpy by cooling, you move down the line and the temperature is dropping until the temperature drops to 100°C. When you remove more energy you move on the horizontal red line to the left, so the temperature stays at 100°C while the gas/liquid fraction drops (which is indicated by crossing the green lines) until all the water is liquid. All the rest of the diagram is for different pressures. So 100°C is the very special temperature at atmospheric pressure, where the constant pressure line for 1 atm touches the dome. At other pressures the phase transition happens at other temperatures and above the critical point there is no phase transition.

Vapour pressure has more to do with the Maxwell tail of the Boltzmann distribution so you always have some part of the water in the energetically less favourable state, but there is a sharp phase transition none the less. That is the whole idea of a phase transition.

 

[Chestermiller]

Well I don't see how that graph is supposed to prove your point. On the right you have the constant pressure lines. The lowest one is for atmospheric pressure. If you are coming from the right you start with gas. When you remove Energy/Enthalpy by cooling, you move down the line and the temperature is dropping until the temperature drops to 100°C. When you remove more energy you move on the horizontal red line to the left, so the temperature stays at 100°C while the gas/liquid fraction drops (which is indicated by crossing the green lines) until all the water is liquid.

No. There are other lines at lower pressure that are just not shown on the figure. For example, you could also draw another line in at 50 C to the right of the lowest pressure line you referred to. It would be analogous to the lines at 200 C and 300 C, except below the 100 C line. The person who drew this graph just happened to show contours with pressure increments corresponding to saturation temperatures of 100C, 200C, and 300C. If the graph had been drawn with finer resolution, you would see the contours at lower pressures.

Here is a link to a pressure-enthalpy diagram for water which illustrates this with finer resolution.
http://www.google.com/imgres?imgurl...EyUa_bDsrs0gGB64DABQ&ved=0CEIQ9QEwBA&dur=8216

Also shown in the figure are the specific fractions of saturated vapor and liquid during the transition.

All the rest of the diagram is for different pressures. So 100°C is the very special temperature at atmospheric pressure, where the constant pressure line for 1 atm touches the dome. At other pressures the phase transition happens at other temperatures and above the critical point there is no phase transition.

In terms of the fundamental thermodynamic behavior of water, there is nothing really unique about the point 100 C, 1 atm. On days when the atmospheric pressure is lower than 1 atm, say 740 mm Hg, water would boil at a lower temperature, and if you went to mile high Denver CO to boil your water, it would also boil at a lower temperature. A pressure of 1 atm is merely the time average value at sea level. So at best, 1 atm and 100C is an interesting reference point, but it bears no fundamental relevance to the thermodynamic behavior of water. The equilibrium properties of a substance don't depend on where the measurements are made.

 

[0xDEADBEEF]

No. There are other lines at lower pressure that are just not shown on the figure. [...] In terms of the fundamental thermodynamic behaviour of water, there is nothing really unique about the point 100 C, 1 atm.

I completely agree with that.

 

[robphippen]

This is a very interesting question.

When a substance is at a phase transition temperature it means that two phases are in thermal equilibrium. e.g. at boiling point: where , what this means is that the gas phase is in perfect equilibrium with the liquid phase.

In molecular terms: this means that the number of molecules escaping the surface of the liquid and becoming gas ('evaporating') is exactly matched by the number of molecules colliding with the liquid surface and 'joining' the liquid phase ('condensing').