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I'm having a bit of difficulty grasping the concept of heat flow in isothermal expansion -- most likely because I don't fully grasp the concepts yet. We were given a problem that the professor answered using basic algebra.

When an ideal gas undergoes isothermal expansion a) heat flows into the gas b) heat flows out of the gas c) there is no heat flow in or out d) any of these

From the algebra, it was determined that heat flowed into the gas. I'm convinced that this is the case. However, I was wondering if (D) could be an answer as well. Could heat be flowing into the system and simultaneously out of it? Could this process occur with no heat flow in or out of the system?

I was thinking that (C) could be the reason why (D) isn't the answer, but I think this highlights a misconception that I have about heat flow.

Could someone help me try to conceptually grasp this?

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  • $\begingroup$ Are you paraphrasing the question? When an ideal gas undergoes isothermal expansion into a vacuum, the answer is (C): there is no heat flow in or out. For (A) to be correct, the question must specify that the ideal gas is expanding by doing work against its surroundings. (B) can never be correct because the temperature would drop, so (D) cannot be correct. $\endgroup$
    – Chemomechanics
    Feb 21, 2019 at 2:54
  • $\begingroup$ I wasn't paraphrasing the question. I directly quoted it. Interesting thought though! $\endgroup$
    – ghste
    Feb 22, 2019 at 4:59

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In an isothermal expansion, the temperature of the gas remains unchanged (by definition). Now, if the gas is expanding, it must be doing work on e.g. the walls of the container. But total energy must be conserved by the first law, and a system "doing work" means energy is leaving the gas due to expansion. (Think of pushing a box—you're doing work on the box, and energy is flowing to the box, in kinetic form). To conserve energy, then, "thermal" energy must flow in—that's heat.

Another way to look at it is to consider what one might intuitively ``expect." For me, I imagine that if a gas is expanding, then it is getting colder (because the opposite is generally true, meaning if you compress a gas, it gets warmer). But we've constrained the expansion to be isothermal, so to maintain the temperature, you have to put energy in, which is a heat inflow.

You see that, if heat didn't flow in, the gas would cool, or, more generally, energy would not be conserved, so (C) doesn't obey the first law. Neither does (B). "Heat flow" is simply a transfer of energy, just like work; they are only defined by what they "change." "Heat" generally means thermal energy, i.e. kinetic energy of particles, so that temperature changes; while "work" generally means changes in volume.

Note that the first law is really a statement about conservation of energy for an isolated system... generally in these types of problems, we make the assumption of isolation.

EDIT: For slight clarification at the end of the penultimate paragraph.

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  • $\begingroup$ Ahhhh, okay. So, generally, even though these are types of energy with the same metric units, we do not view "Heat" as equivalent to "Work." I was getting caught up by the fact, as you stated, that if volume expanded, the system would lose "Heat" in the form of "Work." However, I viewed Heat as being equivalent to Work. And was thinking too much about the ideal situation as you suggested. Am I correct with these statements? Thanks for the comprehensive answer! I really think you hit on everything that was causing my head to spin! $\endgroup$
    – ghste
    Feb 22, 2019 at 5:02
  • $\begingroup$ @ghste You may be interested in Loverude et al.'s “Student understanding of the first law of thermodynamics: Relating work to the adiabatic compression of an ideal gas,” Am J Phys 70(2) (2002), which discusses common misconceptions of this type. $\endgroup$
    – Chemomechanics
    Feb 22, 2019 at 20:44
  • $\begingroup$ @ghste Yes, I think you've got it! I had a bit of trouble identifying your exact misconception, so I'm glad this helped. Yes, even though "heat" and "work" are both just energy transfers at the end of the day, it's the fact that one is thermal energy and the other relates to volume changes that makes them different. $\endgroup$
    – flevinBombastus
    Feb 23, 2019 at 13:37
  • $\begingroup$ @ghste Also, my apologies, I made a slight error—I'd typed that heat "generally means thermal energy, i.e. kinetic energy of particles, so that volume changes," but I meant "so that temperature changes." Hope that didn't throw you off; it's edited now. $\endgroup$
    – flevinBombastus
    Feb 23, 2019 at 13:38
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    $\begingroup$ Ah, this makes much more sense! That was a good entry, Chemo, good find! And flevin you really helped me get down to the basics for understanding it all. Thanks for helping me out, you two. $\endgroup$
    – ghste
    Feb 27, 2019 at 1:59

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