Question

I'm uncertain about what defines a system and a handful of things, so here's a thought experiment to try to get at some of my doubts.

Answer 1

If we conduct a reaction in a beaker open to the atmosphere that occurs at approximately a constant temperature without heat exchange, I'd like to know the change in internal energy based on the amount of gas that is released. So in calculating the work done,

\[\Delta U = w= -\int_{V_i}^{V_f} p_{ex} dV = -p_{ex} \Delta V\]

Now I believe that the total amount of escaping gas is represented by \(\Delta V\), so if the gas is ideal, we can rewrite the energy change as

\[\Delta U = - (\Delta n)RT\]

So now we can look at the difference in mass of the system and the temperature at which the reaction occurred in order to predict the change in internal energy. The point being, the gas escaped and I couldn't easily measure the volume change. Also, as a side note, it appears that I can use this to measure equilibrium constants.

SO! Did I make too many assumptions? Did I leave anything out? Is this reasoning sound? When does it break? Since internal energy is a state function, it shouldn't matter what the change in heat transfer is over the course of the reaction as long as I bring it to the same temperature at the end, right or wrong?

Answer 2

@Frostbite Hey! I'm creeping into thermo stuff if you have time I'd appreciate it!

Answer 3

It all checks out perfectly for me. We are doing an expansion against a constant external pressure so the law you wrote holds. And your rewriting assuming your gas behave ideally also checks out.

And yes the internal energy is a state function meaning, it is only how the work was done that can change the state function.

Answer 4

e.g. reversible and non-reversible work.

Answer 5

Awesome thanks, I find it hard to confirm my understanding since it's hard to come up with examples that I can confirm plausibly or not. I suppose I would have to conduct some experiment in order to really verify this but theoretically at least I think I have the concepts down on what should happen.

I am still pretty fuzzy on what all reversible vs irreversible entails. I think being a state function means it doesn't matter how you got somewhere, but there's still some kind of "excess" "something" going "somewhere" if you do the process irreversibly so I'm not sure what that is.

In this case, we did work irreversibly, so if we were to do this reversibly I don't know what would be different since the internal energy change should be the same since it's a state function. Hmmm.

Answer 6

Well.... http://assets.openstudy.com/updates/attachments/55daf37ce4b05a670c277d75-frostbite-1440412923962-thermodynamics_fundamentals_and_internal_energy.pdf

page 10

Answer 7

I'll read it, I had forgotten about this, I won't waste your time any more on stuff you've already answered!

Answer 8

Regarding state functions I am having a hell of a time to explain it to people so usually I just want to say "two systems with the same state variables will have the same state functions. But when you do a change to the state, it depends on how the change is made "

:P

Answer 9

And you don't waist my time, right be contrary. :D

Answer 10

Aha, ok right this is sorta the concept of an ensemble, sorta seeing now I should look more carefully back at this!

I think I am finally getting to some point in my life where I understand a thermodynamic system. I'll just go ahead and run down what I think I know and maybe you can guide a direction for me to point towards next.

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The first law is cool because the changes in internal energy are what we can observe and heat and work change the internal energy so we don't have to know "the absolute energy" of the system.

A system in equilibrium has its states Boltzmann distributed by the partition function.

The internal energy of an ideal gas is solely dependent on sum of only the translational motion of all the particles contained within it and not their interactions, the size of the box, electronic, rotational, or vibrational motions, is there anything else I left out / am wrong about here? (Also curious, I'd like to know practically speaking how far this approximation stretches past simple cases, like what about higher pressures of non inert molecules? I guess this is why you use van der waals gas and virial equation but I have not ever really thought about them with the intent to really use them like I am now.)

My big goal is to next really look into chemical potential and understand that as well as possible. Also, I realize I said nothing about entropy but it's just an odd concept to me, and I think I'll get some more info outta that over time the more I get comfortable with using Boltzmann distribution stuff.

:O Haha

Answer 11

Uh what the hell, it says I wasn't your fan, weird. Fixed that haha.

Answer 12

lol for the first time I'm of absolutely no use here frostbite made a great tutorial, I haven't had the time to look at it though.

Answer 13

Overall I want to say you are correct yes @Kainui

In this this part:
"The internal energy of an ideal gas is solely dependent on sum of only the translational motion of all the particles contained within it and not their interactions, the size of the box, electronic, rotational, or vibrational motions"

I fast need to check what terms is in included as I honestly can't remember and kinda embarrassing as I did that lecture a couple of weeks ago, but you can derive the partition function into 2 parts. The ideal part and excess part... the ideal part is like the zero point energy.... the system always present in the system while the excess comes from interactions and stuff.
\[Q=Q_{ideal} \times Q_{excess}\]

Answer 14

the energy always present*

Answer 15

Yes you are right. It is the translational term that survive as, it is the only term surviving as the configuration integral ends up equal to 1.

Answer 16

@Photon336 Those words makes me heart soft :'D.... embarrassing however I haven't written the other tutorials yet.... I have the manuscripts but I am to lazy translating it to English xD

Answer 17

Ok awesome, I guess the energy always present comes from quantum mechanics which is exactly the kind of thing I'm really starting to want to bridge the gap. Although I guess since you can only observe changes in internal energy the zero point energy is not really observable is it?

Also, I know a system can be a thought boundary, so I'm really trying to test how far that goes. Can you make a system that isn't connected? I am imagining a beaker with a racemic mix of some chiral molecule, and making two systems, each containing all of the left handed or all of the right handed isomers. So these would be two disconnected and open systems, however particle number is is conserved between the systems, so if you convert one isomer into the other, it changes from being in one system to the other.

Answer 18

Yes it is mostly an analogy :D

If you want to do a direct link to quantum mechanics and the internal energy, we can look at the most useless equation in thermodynamics:
\[E=\sum_{i}^{}N_i \epsilon_i\]
And here is \(\epsilon\) the energy of the states, derived from the Schrodinger equation.

Answer 19

The issue is size. In quantum you work on 1 molecule. in thermodynamics you work on 1 mol of molecules.

Answer 20

The argument is a little vague, but in principle if we could know how many molecules is found in each state, then we could calculate the internal energy from just that.

Answer 21

Ok, that's more or less how I understand it so far, we separate the different energy states which have their relative populations determined by the Boltzmann distribution which in turn only depends on the temperature. The energies of the molecular systems are determined by Schrodinger's equation, and then we can multiply all the partition functions for the translational, rotational, vibrational and electronic wave functions together to get the total partition function which we can then get the total energy of our macroscopic system out of... I think? I will have to go back to look now because I think I read this almost a year ago out of Atkins.

Anyways I think that's all for now, I think this has given me a lot of peace of mind about what these models are saying haha.

Answer 22

Thanks again :D

Answer 23

http://cavemanchemistry.com/cavedemo.pdf