# I Help understanding internal energy

1. Aug 6, 2016

### nashed

Edit: @Dale managed to do a far better job in stating the problem, essentially the question is why do we get the same internal energy for different microstates corresponding to a single thermodynamic state

Original Post:

So I'm self studying a course about thermodynamics and statistical physics ( due to personal issues I could not attend the lectures ) using Mehran Kardar's Statistical Physics of Particles, the book starts with thermodynamics, it started with the zeroth law which after some struggle I think I've got a grip on.

Now I'm tackling the first law and and I just cannot understand the reasoning behind it, it states that the work required to change an adiabatically insulated system from one state to another is a function of only the initial and final states, in other words it implies the existence of and energy function for the states, which is called the internal energy.

After some digging around the internet I saw that it was stated in multiple places that the internal energy is a measure of the energy of the constitute particles (kinetic, potential, bond, etc..), this explains the conservation of energy ( because essentially when talking about heat transfer we're talking about the mechanical transfer of energy at the microscopic level), what it doesn't explain is how come that the internal energy can be a function of state seeing that there are multiple microscopic configurations to produce the same state, that implies that every single configuration has the same energy which I doubt is true.

I guess my question is can anyone explain internal energy to me? because the way I understand it, it should not exist in the first place.

Last edited: Aug 6, 2016
2. Aug 6, 2016

### QuantumQuest

As you're trying to understand the 1st law of thermodynamics, it is more appropriate to focus on the change of the internal energy of a thermodynamic system.

As is well known, this equation states: $\Delta U = Q - W$ where $\Delta U$ is the change in internal energy, $Q$ is heat added to the system and $W$ is the work done by the system. Now, in words, this states that the change in internal energy of a system is the heat added minus the work done by the system. So, the focus is on the changes of internal energy: what we give minus what the system does as work. As an example take the case of having a gas. We heat it and this expands. This expansion can be used to push a piston for instance and this is the work done by the system.

I think that the above example combined with constant pressure explains this.

The system has some internal energy and we're dealing with its changes as we give heat and get some work done by the system, as the expression of 1st law states.

3. Aug 6, 2016

### Staff: Mentor

@nashed if I understand you correctly then I think you are asking why the change in internal energy is 0 for changes between different microstates having the same macrostate. Is that correct?

4. Aug 6, 2016

### nashed

Yes that would be a much better way to state it

5. Aug 6, 2016

### Staff: Mentor

I have not thought about that before. I would be interested to hear from others about it.

I guess maybe by definition a macrostate is a set of microstates that are indistinguishable from each other, and if you could get energy out then they would be distinguishable. That doesn't answer why the distinguishable features are pressure, volume, and temperature.

6. Aug 7, 2016

### nashed

After giving it some thought I had an idea, if all the microstates corresponding to the same macrostate are the result of a series of elastic collisions of each other ( that is assume a microstate A, a series of elastic collisions occurs during some time frame t, then after that time frame we are in microstate B) then conservation of energy applies and indeed all microstates have the same energy, but this rests on the assumption that all microstates are obatinable through a series of elastic collision from some reference state.

7. Aug 7, 2016

### Staff: Mentor

For an ideal gas that certainly is the case, since there are no internal degrees of freedom within the gas molecules. But as you get real gasses with molecular internal degrees of freedom the collisions become inelastic. I'm not sure where to go from there.