VincentG wrote:I will however argue that in real gas, density and volume does matter. And that the same gas molecules when spread out can have the same kinetic motion (internal energy), even if the apparent "temperature" is lower.
There are formulas that more accurately predict real gas behavior. They can be found in the following link:
https://en.m.wikipedia.org/wiki/Real_gas
And for internal energy:
https://en.m.wikipedia.org/wiki/Internal_energy
Wikipedia wrote:Therefore, the internal energy of an ideal gas depends solely on its temperature (and the number of gas particles): 𝑈=𝑈(𝑁,𝑇). It is not dependent on other thermodynamic quantities such as pressure or density.
It seems to me that temperature and internal energy run hand in hand. This is unlike "heat" in that the same amount can be transfered between any two temperature differences. Or any amount between fixed temperature differences. Temperature is the measurement of internal energy.
There appears to only be internal energy per mass/mole, not density, or volume mole, just the total isolated mass.
I was a little confused on this years ago, as I thought number of impacts dictated temperature too. I now think, it's just speed of impact. Number just dictates Wattage of heat flow. The more dense, the faster the thermometer will heat up. The faster the molecules, the higher the temperature reading will eventually be.
That is one of the reasons highly pressurized engines are more powerful, better conduction of heat into the gas. It's not because the compressed gas appears hotter.
The problem I see with the thoughts of this thread is, are we describing heat or internal energy being converted?
Internal energy is converted to work in an adiabatic with work process. Many tests have shown this. The diesel engine, or fire Piston, come to mind. Work of compression makes the internal gas hotter, for a higher temperature and higher internal energy. Single stroke. Compressing it slowly, so energy can leave, keeps the temperature rise lower.
Heat going into an engine does one of to things, it either raises the temperature of the working gas, or keeps it from going down or down as fast. It is, IMHO, misleading ones self, to describe heat being converted to work, as it really isn't something that exists anyway. For that mater neither does work. Internal energy gets converted to potential or kinetic energy.
Those links provided, and others, take some serious studying time to understand. That is why I try to avoid the confusion between, heat, work, internal energy, enthalpy, entropy, kinetic, potential, electrical, chemical, etc..., and all the measurable energy effects.
One more confusing point, heat entering, or exiting, an isothermal process, stroke, equals the work coming out, or in, during that process. The heat and work are still mathematical concepts, so the real conversion is between the internal energy's of the hot, cold, and height masses. If you want to avoid the confusion, think of those values, rather than heat and work.
Tom and my disagreement appears to be how much energy possibly gets converted to work verses how much gets cooled down to Tc and rejected as heat. No arguments about whether any come out as work, therefore less as heat. The reason for the disagreement is washed out buy eliminating the confusing terms. He is beginning to see the difference a little at a time, as he's using the terms more and more and in more correct ways. He's learning this by studying links like the ones I post. Doesn't take my word for it, as it should be.
My beef now is that when I think I've put out an obvious point, he slips back into single stroke process and ignores the completion of the cycle.
If a compressed tank of gas at room temperature is released in an isolated room, the room will become hotter. PV=MRT. The pressure of the room will be higher. The tank will cool. Eventually the two temperatures will equalize as slightly higher. Energy has been released into the room. The room pressure will be higher. Overall entropy will be higher. Any energy saved by lifting a weight will leave the room cooler by the same amount of potential energy saved. T=Q/CvM. Density unimportant, only the total mass isolated.
If the tank is compressed in the room temperature will be higher, by the amount of electrical energy put into the room minus the energy stored in the tank.