0
$\begingroup$

I understood that in the adiabatic stroke, the internal energy is being used up to do work

Again, assuming that the internal energy of all the gas molecules of the system are the same (can we?)

Why would the internal energy flow?(since it flows only when there is a difference), since the system pushes the piston only using it's kinetic energy, i.e, pressure being applied.

What I mean is, is the internal energy used in pushing the piston? If yes, how and why?

Improve this question
$\endgroup$
6
  • $\begingroup$ I think my answer to this question may also cover the comment you made on my answer to your previous related question. If not, let me know $\endgroup$
    – Bob D
    Commented Dec 15, 2019 at 11:36
  • $\begingroup$ Yes, it has. The main misunderstanding was about considering the kinetic energy of the system and it's internal energy to be two different things. $\endgroup$
    – Swaroop Joshi
    Commented Dec 15, 2019 at 14:47
  • 1
    $\begingroup$ Glad it cleared things up. But a word of caution. It only applies to ideal gas behavior. For real gases, liquids and solids the internal energy consists of both kinetic and potential energy. $\endgroup$
    – Bob D
    Commented Dec 15, 2019 at 14:54
  • $\begingroup$ @BobD again, may I ask, how is it assured that the gas will expand till it reached T2, how will it know that it's temperature has matched that of the temperature outside the system? Is it that, once the force matches that of the surrounding matches it, it will stop? Or how? $\endgroup$
    – Swaroop Joshi
    Commented Dec 20, 2019 at 18:46
  • 1
    $\begingroup$ I have answered this follow up question in a revision to my answer. Hope it helps. $\endgroup$
    – Bob D
    Commented Dec 20, 2019 at 19:57

2 Answers 2

Reset to default
3
$\begingroup$

The internal energy, or more correctly kinetic energy, of the individual gas molecules is not the same.

For an ideal gas, the internal energy is the sum of the kinetic energies of the individual gas molecules. The individual kinetic energies of the molecules are not the same because their velocities are not the same. For an ideal gas the velocities follow the Maxwell-Boltzmann distribution.

Rather than thinking about internal energy “flow”, which could be confused with the movement of a fluid, think about it as energy transfer in the form of heat and/or work. For an adiabatic process the only means of transfer is work.

When the collisions of the molecules against the piston cause it to move, the molecules are collectively doing $pdV$ work causing the molecules to slow down giving up kinetic energy. That reduces the internal energy of the gas as a whole.

So yes the internal energy is used to push the piston.

Again, may I ask, how is it assured that the gas will expand till it reached T2, how will it know that it's temperature has matched that of the temperature outside the system? Is it that, once the force matches that of the surrounding matches it, it will stop? Or how?

The gas doesn't "know" anything. The idea is to gradually reduce the external pressure allowing the gas to slowly expand adiabatically while monitoring the gas temperature (e.g., with a thermometer) until the gas temperature reaches the temperature of the low temperature reservoir and then stop reducing the pressure. Then you follow that process with the reversible isothermal compression rejecting heat to the low temperature reservoir.

Hope this helps.

Improve this answer
$\endgroup$
7
  • $\begingroup$ So the kinetic energy of the molecules are a part of the internal energy? $\endgroup$
    – Swaroop Joshi
    Commented Dec 15, 2019 at 13:24
  • 3
    $\begingroup$ @Swaroop Joshua For an ideal gas the kinetic energy of the molecules is ALL of the internal energy. $\endgroup$
    – Bob D
    Commented Dec 15, 2019 at 13:27
  • $\begingroup$ What about VdP, can it be considered as a change in the internal energy? $\endgroup$
    – Swaroop Joshi
    Commented Dec 16, 2019 at 19:53
  • 1
    $\begingroup$ Exactly right. The external pressure is intentionally slowly reduced by some external means. $\endgroup$
    – Bob D
    Commented Dec 20, 2019 at 20:04
  • 1
    $\begingroup$ @SwaroopJoshi No problem. did I answer your question? $\endgroup$
    – Bob D
    Commented Dec 20, 2019 at 22:05
1
$\begingroup$

Molecules bounce against the piston. That is pressure, that is what makes the piston move.

The molecules bouncing back from an outward moving piston will have a lower speed. That is the microscopic view of why the internal energy decreases.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.