# I Heat and Adiabatic Confusion

1. Dec 17, 2016

### Ehden

I have a dilemma, which has been nagging at me for a week.

So first off, can anyone verify my definitions are correct?
Heat is the flow of energy, and that flow is caused by the collisions of the atoms in a system. The collisions cause a transfer of KE, hence heat will flow from a substance that has higher KE to a substance with lower KE, in other words, heat diffuses from higher temp to lower temp due to this microscopic view.

The internal energy of a system is the total KE and PE of the atoms. And in an adiabatic process, the internal energy change is only determined by the work being done on or by the system.

So if my previous definitions are correct, this is where my confusion comes in. How is an adiabatic process possible? Isn't the ability to do work essentially the transfer of energy? In the microscopic view, the atoms in an adiabatic process does work on its surrounding by expanding against it, so the atoms collides and transfers its KE. And by earlier definition, heat is this type of transfer, so isn't there also a loss of heat?

My textbooks says, "... is defined as one with no heat transfer into or out of a system; Q = 0. ....When a system expands adiabatically, W is positive (the system does work on its surroundings), so ∆U is negative and the internal energy decreases. When a system is compressed adiabatically, W is negative (work is done on the system by its surroundings) and U increases. In many (but not all) systems an increase of internal energy is accompanied by a rise in temperature, and a decrease in internal energy by a drop in temperature."

Isn't a drop in internal energy/temperature a loss of heat?

2. Dec 17, 2016

### Andrew Mason

No. Heat, or heat flow, has a specific meaning in thermodynamics. Heat, or heat flow, is the transfer of energy that occurs by means other than mechanical work. Heat flow occurs due to a temperature difference between two bodies that are in thermal contact with each other. If the body is not in thermal contact with any other body while work is being done on it, or while it is doing work on its surroundings (ie. an adiabatic process) its internal energy (and usually its temperature) will change but no heat flow will occur (Q=0).

AM

3. Dec 18, 2016

### Ehden

Thank you for the clarification, so heat is just another method of energy transfer besides mechanical work.

Can you explain why does this occur? On the microscopic level, what exactly is happening to the atoms when there is a heat transfer? I always suspected the reason why heat flows from higher temp to a lower temp was because of the kinetic collisions between the atoms.

4. Dec 18, 2016

### Ehden

To add on top of that, if heat is the transfer of KE through the collisions of atoms, why isn't it considered mechanical work? Aren't the atoms essentially doing work on other atoms?

5. Dec 18, 2016

### Andrew Mason

Temperature of a system is a measure of the average translational kinetic energies of its molecules (in thermodynamic equilibrium). Heat flow occurs between two such systems if there is a difference in their temperatures and if there is thermal contact between them so that the molecules of each system can physically interact at some interface between them. Eventually the two systems will reach thermodynamic equilibrium (same temperature) due to the heat flow from the higher temperature system to the lower.

When heat flow occurs between systems, the molecules of the higher temperature system do work on the molecules of the cooler system. But that work causes random changes in translational kinetic energy of the molecules within each system so that the transfer of energy does not change the translational kinetic energy of the centre of mass of either system. Mechanical work is work that can change the centres of mass of systems or macroscopic parts of systems.

AM

6. Dec 18, 2016

### Ehden

Neat, thanks again for that comprehensive explanation.

So to sum it up, temperature is the measure of average translational KE, but, this is excluding the KE due to vibrations and rotations. In an adiabatic process, there is no gain in or loss in heat, then by definition there is no transfer of energy? However, the system can have a lower/higher temperature, so isn't that drop/gain associated with the heat flow? Why are the molecules losing/gaining translational KE during a adiabatic process without losing/gaining heat?

7. Dec 18, 2016

### Andrew Mason

Their energy has increased/decreased due to mechanical work done on/by the system. For example, when a gas is compressed, the moving walls of the container cause an increase in speed of the gas molecules, which increases the temperature of the gas. But this is not the result of random molecular transfers of translational kinetic energy (ie. heat flow). It is the result of the macroscopic organized transfer of energy from the wall (ie. mechanical work being done on the gas by an external source pushing on the walls).

AM

Last edited: Dec 19, 2016
8. Dec 18, 2016

### Ehden

Cool this is starting to come together. So one last question. The reason why engines aren't 100% efficient is because of this random factor? For example, the random movement of these atoms can't all be used to expand a piston, since some of that "work" will be put into changing the KEs of other atoms?

9. Dec 18, 2016

### Cutter Ketch

I read this topic from the top, and while you seem to be satisfied it looks to me like your understanding is getting further and further from correct. I thought I would try.

"Heat" gets used as a property of a system, the "heat content" or similar, but officially heat is the transfer of that energy. So we should talk about internal energy and temperature. The internal energy has two parts: the kinetic energy and the chemical potential energy. The kinetic energy part is the sum of the kinetic energy of all the microscopic motions including rotations and vibrations excluding the macroscopic kinetic energy. That is to say if the whole block of material is flying across the room or spinning as a whole, that kinetic energy is not part of the internal energy. Think of it as all the microscopic motions in the center of mass system. For understanding let's ignore the chemical potential part for now. This is appropriate in many thermodynamics problems for example if they say the fluid in question is an ideal gas. The chemical potential will be extremely important in other cases, for example phase changes, but let's ignore it for now. The kinetic part is directly related to temperature and can be thought of as thermal energy. The thermal energy is referenced to some state and the logic doesn't depend on the choice of reference state. However, for general understanding let's refer to absolute 0 as the reference state. If it weren't for quantum mechanics we could say this is the temperature where all motion stops and the kinetic energy is zero. Quantum mechanics requires a nonzero minimum kinetic energy, but the zero point is very close to zero.

So ignoring chemical potential and not worrying about the small zero point energy required by quantum mechanics, internal energy is thermal energy. It is the total kinetic energy of all the microscopic motions in the center of mass system. It is a total, so it does depend on how much stuff you have. It is for all motions, so it includes vibrations and rotations. It is directly related to temperature. However temperature doesn't depend on how much stuff you have and so is more like an average rather than a sum. The temperature is proportional to the average kinetic energy per degree of freedom so per atom or molecule and per all the different ways they can move including vibrations and rotations.

In classical mechanics we already learned how to keep track of the macroscopic forces and kinetic energies. In thermodynamics we are trying to keep track of these microscopic energies. The microscopic and the macroscopic are distinct and treated separately. Heat flow is the transfer of internal energy of one body to the INTERNAL energy of another body. Any macroscopic motion imparted to an external body is not heat flow.

In an adiabatic process there is no heat transfer. This is a very specific statement. It means that no object outside the system being considered had its INTERNAL energy changed. This does not mean that no thermal energy was gained or lost by the system or working fluid in question. It does mean that any change in internal energy of the system went to or came from macroscopic motions of the outside world. If a gas expands and pushes a piston it does work on the piston. The work comes from the internal energy of the gas. However the internal energy of the piston does not change. No heat flowed between the gas and the piston. For the purpose of keeping track of the microscopic thermodynamic energy you don't need to look at the piston. You do note the work done, the motion of the piston, the conservation of energy, but the macroscopic energy is tallied separately and doesn't show up in the accounting of the thermodynamic energy. That is what's meant by adiabatic. All the microscopic internal thermodynamic energy can be accounted for by considering only the system in question. Macroscopic quantities are accounted for in the usual nonthermodynamic way.

10. Dec 18, 2016

### Andrew Mason

You are on the right track. A heat engine uses heat (Q, or heat flow) to produce useful mechanical work. Energy flows from a hot to a cold reservoir in the form of heat flow and only some of that energy produces useful mechanical work. To get a better understanding of why not all the heat flow can produce useful mechanical work, you will have to study the second law of thermodynamics and the concept of Entropy.

AM

11. Dec 18, 2016

### ehild

The piston does work on the surroundings, and it moves as the colliding molecules transfer momentum to it. In an adiabatic process, these collisions are elastic, the piston and the walls are insulating. Consider the process of collision. A molecule with average KE approaches the piston with velocity vi and the piston moves with velocity V, in the same direction. vi>>V. The relative velocity of the molecule with respect the piston is vi-V. As the mass of the piston is much larger than that of the molecule, the molecule reflects with the same relative speed, so its velocity is -(vi-V) with respect to the piston,and vf =-(vi-V)+V=2V-vi with respect to the gas, The speed after collision is vi-2V , less than before the collision, so the KE of the molecule is less than the average. It collides with other molecules, and the energy distribution sets back to the normal one, but with less average energy than before. That means, the internal energy decreases, without transferring heat.