Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

In compression of gas, which is first kinetic energy or heat?

  1. Jun 5, 2014 #1

    mdn

    User Avatar

    Hello,
    when gas compressed it get heated, proper reasons i don't know.
    If we consider first kinetic energy increases due to external work done but hows that happens?
    and if we consider heat is first, where from it came?

    my guess is that when molecules come closer in confined wall( due to low volume), frequency of collision occurs rapidly between confined molecules means they excited and radiate heat energy.
    Heat energy first?
     
  2. jcsd
  3. Jun 5, 2014 #2

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    Gas can be compressed adiabatically (no heat transfer). This leads to an increase in temperature due to work done on the system. How we know this happens? Experiment tells us this happens. Alternatively, the first law of thermodynamics (which is verified by experiment) tells us this happens. The first law: ΔU=pdV+dQ and dQ=0 leads to ΔU=pdV.

    If you want a microscopic reason, as you compress the walls of the container, the atoms are taking the energy of the motion of the wall and moving faster and hitting the wall more often. This leads to an increase in both temperature and pressure. (Assuming heat doesn't leave the system)
     
  4. Jun 5, 2014 #3

    maajdl

    User Avatar
    Gold Member

    Heat cannot be radiated. It can be the result of radiations, like infra-red. Heat is essentially random motion of molecules.
     
  5. Jun 5, 2014 #4

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    ? Heat is not the random motion of molecules, maybe you're thinking of temperature? Heat is not a state-function so it's not really a property of a system at all. It denotes a kind of transfer of energy. Radiative transfer is a perfectly valid way of transferring heat. The other two types of heat transfer are convection and conduction.
     
  6. Jun 5, 2014 #5

    WannabeNewton

    User Avatar
    Science Advisor

    The gas will necessarily only increase its temperature if the compression is adiabatic i.e. ##Q = 0##. Otherwise I can compress the gas while at the same time cooling it by having heat flow out of the system which will keep the gas at the same temperature i.e. I can perform an isothermal process, which is in general not adiabatic. So if you want the gas to necessarily increase in temperature (which is what I presume you mean by "heat up") then the compression needs to have no heat flow ##Q## into or out of the system, as noted above.

    Now why does the adiabatic compression necessarily increase the gas temperature? Well ##\Delta E = W > 0## so the average energy of the gas increases. Since the average energy is proportional to the temperature for a classical gas, the temperature will also increase (recall ##S \propto \ln E## so ##\frac{1}{T} = \frac{\partial S}{\partial E}|_{V,N} \propto \frac{1}{E}##). Microscopically, for any system in classical statistical mechanics, the average energy is proportional to the temperature through the equipartition theorem.
     
    Last edited: Jun 5, 2014
  7. Jun 5, 2014 #6

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    If you make heat flow into the system, while doing work on it, the gas will heat up even more. You just can't take out more heat than the work you're putting in. In other words for dQ<0 (heat out), we must require |dQ|<|pdV| for the gas to heat up. If |dQ|=|pdV| then you have an isothermal process, and if |dQ|>|pdV| then you're actually cooling down the gas as you're compressing it...(Assuming no phase changes are occurring! Let's not worry about such things as evaporative cooling for this problem.)
     
  8. Jun 5, 2014 #7

    WannabeNewton

    User Avatar
    Science Advisor

    I said that the heat flow has to be zero to necessitate the desired result so I'm not sure if you're replying to me specifically or making a general remark based on what I said. For non-adiabatic processes we must do things case by case, in the manner you have described, so we can't construct a generalized process.
     
  9. Jun 5, 2014 #8

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    I'm saying that even if the heat flow isn't 0, you can still get the desired result (heating up the gas), you just need to put some conditions on the heat flow. (I.e. you can't remove heat faster than you're putting in work is all.)

    It sounded like you were saying that even if you put heat INTO the system, and do work on it, for some reason this might not imply the system heating up, and that the only way to get the system to heat up is by requiring no heat flow period.
     
  10. Jun 5, 2014 #9

    WannabeNewton

    User Avatar
    Science Advisor

    Ah no I was just saying that if you take an arbitrary process involving compression and we know ##Q = 0## then we know that the temperature must necessarily increase. Otherwise we have to do things case by case. AFAIK there is no general classification of processes for which ##W + Q \geq 0## or for which ##W + Q \leq 0## but I might have just never seen the terms for them if they do exist. Twas' a poor choice of words on my part.

    Regardless it's probably good to make light of the fact that the temperature rising need not have anything to do with heat flux, as the OP kept referring to heat flux with regards to the rising temperature during compression, as there can be processes with no heat flux for which compression raises temperature as well as processes for which compression occurs but temperature still decreases due to sufficient heat flux out of the system. I wanted to make that clear.
     
  11. Jun 5, 2014 #10

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    Ok, no arguments here. I just think you worded it really weird haha. :tongue:
     
  12. Jun 5, 2014 #11

    WannabeNewton

    User Avatar
    Science Advisor

    Oh no doubt, I'm rereading it now and realize it sounds totally off the mark :)
     
  13. Jun 5, 2014 #12

    mdn

    User Avatar

    so what is conclusion
     
  14. Jun 5, 2014 #13

    WannabeNewton

    User Avatar
    Science Advisor

    There need not be any heat flux involved in order to raise the temperature of the gas. If you do work by compressing the gas, enough work to have the change in average energy be positive between initiating and ending the compression, then the temperature of the gas will rise because a positive change in average energy means a positive change in temperature. So, if it helps, you can loosely think of the order as: first sufficient work is done so as to increase average energy and as a result temperature increases.
     
  15. Jun 5, 2014 #14

    mdn

    User Avatar

    does it means mechanical movement is sufficient to increase the kinetic energy of molecule, i mean would slow moving piston increase kinetic energy of gas molecules?
     
  16. Jun 5, 2014 #15

    mdn

    User Avatar

    and one thing, my concept behind this stuff is, the warm we perceive is in the form of radiation and not kinetic energy or vibration of particles (measure of temperature).
     
  17. Jun 5, 2014 #16

    WannabeNewton

    User Avatar
    Science Advisor

    Yes.

    I'm not sure what you mean by this. What "perceived warmth" are you referring to? If I touch a warm stove then the warm feeling I get in my hand is due to the flow of heat upon touching the stove, not because of radiation. Do you mean the warmth you feel when walking into a particularly hot room? This also need not have anything to do with radiation. Have you ever walked into your room late at night, when no lights are on and the windows are closed, and felt a sudden pang of warmth? It's just because of the hot air in the room; there's no radiation around. Sure the air would have heated up from the radiation it interacted with during the day but that's besides the point.
     
  18. Jun 5, 2014 #17

    mdn

    User Avatar

    do you mean warmth (of course in room at night) we feel is due to vibration of particle and not due to energy radiated by the particles in the form of heat?.
    if answer is yes, than vibration of particle itself is the form of energy that body will perceive as warm.
     
    Last edited: Jun 5, 2014
  19. Jun 5, 2014 #18

    WannabeNewton

    User Avatar
    Science Advisor

    You seem to have some misconceptions about heat. Heat isn't necessarily transported by radiation. We feel the heat flowing into our bodies because of the temperature difference between the hotter gas and our colder selves, during the process of evolving towards thermal equilibrium, but, again, the heat transport does not have to be through radiation; in the situation described above in post #16 it is certainly not through radiation. Matterwave already addressed this, see post #4.
     
  20. Jun 5, 2014 #19

    mdn

    User Avatar

    Temperature difference is responsible for transfer of heat (radiant energy), heat itself is physical energy as like current, more kinetic energy of particles produces more temperature our surrounding, and if there is temperature difference, heat has to flow, but where from this heat came?
     
  21. Jun 5, 2014 #20

    WannabeNewton

    User Avatar
    Science Advisor

    Heat is just a spontaneous transfer of energy from hotter systems to colder systems, with the heat flux being proportional to the temperature gradient and consequently solving a transport equation. The heat is drawn from the energy of the hotter system and imparts this energy to the colder system. This could be the translational kinetic energy of monoatomic gases, the rotational kinetic energy of diatomic gases, the energy of a gas of photons etc.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: In compression of gas, which is first kinetic energy or heat?
Loading...