# Energy within gas and liquid

1. Dec 27, 2009

Hi all, everyone know that molecules hold more energy in their gasous phase than their liquid phrase, i would like to ask why is that so?

If i look at bernoulli's equation and the ideal gas equation, let's say both phases are filled in their own containers which has the same volume and pressure, from ideal gas equation P*V=mRT, PV will give me the energy, whereas in bernoulli's equation, if i rearrange the equation to => P + v^2*rho/2 +rho*g*h = Total pressure, then if i times the volume of the cylinder, i would get energy within the system, then if you look at the equation, wouldn't liquid phase has more energy than the gasous phase?

Thanks!

2. Dec 27, 2009

### Astronuc

Staff Emeritus
In a two phase system, the molecules in the gaseous phase have received sufficient energy to overcome the intermolecular forces, which is why they are in the gaseous phase.

In a closed system, there certainly could be more energy in the liquid phase than in the gaseous phase depending on the number of molecules in each phase and the energy per molecule.

3. Dec 27, 2009

Thanks for the answer, but why is it that vessels containing gas are considered more danagerous than vessel with liquid?

Thanks!

4. Dec 27, 2009

### Astronuc

Staff Emeritus
Well that depends in the liquid or gas. Does one refer to flammable (fuel) gases or liquids. Gases mix with air and are much more quickly dispersed than liquids. Liquids may be confined locally, or they flow on the ground, if they are not absorbed by the ground. Gases simply flow in the atmosphere.

5. Dec 28, 2009

I think i didnt clarify enough, let me make it more clear.
If there are two vessels of equal sizes, one filled with steam and one filled with water, the common preception is that the vessel containing steam is more danagerous, in the sense that gasous phase hold much more energy and that it will explode if you were to poke a hole in the vessel.

The best comparison i can think of is a balloon filled with air and a ballon filled with water, if you poke a hole in the balloon with air, it will go 'pop', but if you do the same on the water balloon, it will simply leak water. And if i am correct, that would be because of the higher energy state in gasous molecules.

But from the equations i posted in my first post, it looks like liquid phase actually contains more energy, and this is confusing me?

Thanks!

6. Dec 28, 2009

### Astronuc

Staff Emeritus
What are the conditions of the liquid and vapor in the two vessels. If they are at the same pressure then there is not difference on the stress in the vessel wall. If they contain the same mass, then the pressure of the vapor will be extremely high and the vessel would explode if a flaw or hole were introduced.

Given the same mass, the vapor phase will have more energy than the liquid phase. At the saturation temperature, the difference in energy would be the heat of fusion, which takes the fluid from liquid to gas/vapor.

In the balloon analogy, if the balloon is under sufficient tension, it will rupture if pierced, whether it contains liquid or gas, although it is likely that there is a range of tension where the balloon would leak.

7. Dec 30, 2009

Thanks for your input, but how do you calculate the energy contained in a gas (assume it is just steam or something)? you would use the ideal gas law PV=mRT; PV or mRT will give you the energy contain in the gas under the specific parameters (pressure, temperature etc) in this case right?

Let us go back to the equations and use our vessel example (one contain steam, one contain water), assuming the pressure applied, volume in both vessel are equal, now let us calculate the energy in each of this vessel.

We can get the energy in the vessel containing steam with PV (for argument sake, let us assume the same mass in both steam and water too, and since PV is already the energy, the mass of steam don't come into the equation). And we can get the energy in the water vessel by Bernoulli's equation (in pressure instead of head): P + v^2*rho/2 + rho*g*h = Total Pressure
Then to get energy, we times it by the Volume, thus equation becomes: PV + V*v^2*rho/2 +rho*g*h*V = Total energy
since the water is not moving out of the vessel, velocity term is zero, and we get left with:
PV+rho*g*h*V = Total energy

So it brings me to my question: As we have said we will assume the same mass of both steam and water to be the same (however, it doesn't seem from the above equations that you need the mass when you have pressure and volume). And i do agree with you 100% that the energy should be higher for the gas phase if they are contain in the same volume of space and pressure. But what i dont get is that from the equations above, liquid phase will always have more energy because it has an extra static head energy in the equation when compare to the ideal gas law? So where did i go wrong?

Thanks!

8. Dec 30, 2009

### Astronuc

Staff Emeritus
Typcially for steam one uses 'steam tables' or correlations derived from them. There are other tables for other gases.