# What is Internal Energy + Pressure * Volume

1. Nov 6, 2006

### kmarinas86

Can anyone tell me if $Internal\ Energy + Pressure * Volume$ equals heat, total energy, energy in transit?

What if all energy is assumed to be in transit (heat all the way down)? Would that mean all energy is heat? I tend to think so for the following reason:

In a "complete" system (with no inputs and outputs):

$Total\ (E)nergy=Internal\ Energy + Pressure * Volume$

Would imply that $Internal\ Energy + Pressure * Volume$ is constant. It would therefore mean any decrease in internal energy would be matched by a increase in $pressure * volume$; this implies that entropy must increase if pressure * volume increases. We see this happening with the cores of planetary nebula. However, if pressure * volume decreases (i.e. internal energy increases), this implies an entropy decrease.

If volume became

Pressure*Volume would amount to "un-internal energy".

But am I wrong?

I'm trying to figure this out:

http://web.mit.edu/16.unified/www/FALL/thermodynamics/chapter_4.htm

http://en.wikipedia.org/wiki/Internal_energy#Expressions_for_the_internal_energy

Oh wait.... i think this is it:

http://64.233.187.104/search?q=cache:tnyYIhREbdMJ:www.physicsforums.com/archive/index.php/t-98358.html+%22total+energy%22%22internal+energy%22+%22pressure+times+volume%22&hl=en&gl=us&ct=clnk&cd=2&client=firefox-a [Broken]

Also, if radiation is emitted, at the finest particles and/or at the largest scales, does pressure * volume increase?

Is $Internal\ Energy + Binding\ Energy = constant$ a correct statement for closed systems?

Last edited by a moderator: May 2, 2017
2. Nov 7, 2006

### lightarrow

Why don't you call it Enthalpy (H)?
H is a state function (it doesn't depend on the path, like heat or work) and is commonly used in thermodinamics.
H is the heat transferred at constant pressure. This is the only physical meaning I know, but certainly there are others (not so simple, however).

Last edited: Nov 7, 2006
3. Nov 9, 2006

### Epicurus

In planetary nebulae I would say that it is right to say that pressure is constant through the process.

dQ=dU+pdV.

This is a reversible process and as such i.e. no heat is dissipated or absorbed, dQ=0 and as such

S=dQ/T does not change, i.e. there is no change in entropy. This is a simple thermodynamic argument, so we assume no emission of photons or radiation of any sort and that we have a closed system.

4. May 26, 2008

### jimRH9

Hmm, I was thinking about this today. we did an experiment where we put a gas under pressure and measured the volume to verify boyle's law? But when we removed the pressure, it did not return to it's original volume. I am assuming this is because of frictional effects with the piston and the cylinder - if it wasn't the case, and the piston returned to it's original position, then the heat produced due to friction would have come from nowhere, violating the law of conservation of energy.
I was thinking though, the idea of the volume of a (fixed mass of) gas changing without the pressure or temprature changing, seems very odd to me. I was thinking, if (change in)internal energy = heat + work, and since there is no change in temp. (it's at room temprature the whole time), (change in) internal energy = work.
So what i need, i suppose, is some correlation between internal energy and volume or pressure, Are the two related? does anyone know a formula that relates the two?

5. May 26, 2008

### Redbelly98

Staff Emeritus
The gas heats up when compressed, and cools down when it expands. Are you allowing enough time for the gas to reach room temperature again, after the volume changes each time?

6. May 26, 2008

### jimRH9

well, that's what i thought it might be. However, the lab supervisor mentioned an effect called hysterisis, I had a quick look on wikipedia for it, but i haven't found very much useful stuff on it other than what's in the wiki article. I'm thinking of saying that, Once all the pressure has been unloaded from the piston, and the volume stops increasing before it has returned to what it was originally because at the point where it stops, the difference in pressure between the gas inside the cylinder and outside is less than what it was for that volume as the load was increasing. The smaller difference in pressure is not enough to overcome static friction on the piston so it does not move, despite the fact that the volume is less than it was originally... Is this correct?

7. May 26, 2008

### Redbelly98

Staff Emeritus
It's definitely possible that friction is causing what you see. To test this you could force the piston back to its original volume, and see if it stays at that volume.

Other possible causes are that some gas might have leaked out (if the piston's seal leaks) or, as I mentioned before, the gas temperature has changed.

Those are just 3 possibilities that come to mind. I'm not sure what hysteresis would have to do with this, perhaps someone else has an idea?

8. May 26, 2008

### Mapes

This sounds like a reasonable explanation and it would constitute hysteresis. As a mechanical engineer, I associate hysteresis with energy dissipation, the "opening up" of a linear relationship between pressure and volume to enclose some finite area inside the curve, the need to apply a little extra force to get the system back to its original condition. These are the characteristics of hysteresis in magnetization, polarization, and spring motion with friction, which is one way to model your system.

9. May 27, 2008

### jimRH9

Exellent, I think i'll include both temprature and hysterisis in the discussion. As well as the obvious and most probable cause, that the gas leaked out.