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Does compressing a gas mean the temperature has to change?

  1. Oct 19, 2011 #1
    I'm taking a thermodynamics class and am curious about how changes in volume of a gas affect temperature.

    I have the relations

    Δu= ∫c(T)dT

    If we take we take Q, ΔKE and ΔPE to be 0 this leaves us with.

    -W= ∫c(T)dT

    So does this mean any time you compress a gas (do work on it) there has to be a temperature change? At least with a well insulated system?
  2. jcsd
  3. Oct 19, 2011 #2
    Temperature is directly related to the average kinetic energy of a substance. This kinetic energy can harvest itself in translations, vibrations, and rotations. If you compress a bunch of particles you are most definitely increasing the number of collisions. If you imagine millions of collisions per second you will see a picture in your head of extremely fast vibrations. With that, you can see that the average speed of particles is increased because of how rapidly they are hitting each other, thus temperature increases.

    Disclaimer: I may have misconceptions based on this, this is just my intuition.
  4. Oct 19, 2011 #3
    PV = nrt so yes temperature usually rises as compression pressure increases.

    There is a discussion I have seen in these forums and someone knew of an exception
    which was explained, but as a general rule its true.
  5. Oct 19, 2011 #4
    In this scenario volume is decreasing as well.
  6. Oct 19, 2011 #5


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    Gold Member

    One funny thing about sudden compression is that the pressure does not double when you compress to half the volume, but more than double. As the temperature rise under sudden compression, the pressure will also rise in addidion to the compression rate found by the change in volume.

  7. Oct 19, 2011 #6
    And if pressure rises due to the temperature will the temperature rise again due to the pressure?
  8. Oct 19, 2011 #7
    Depends on the path the gas follows in phase space. Adiabatic compression, where everything is well insulated and you don't add heat to the system, does result in an increase in temperature following PV^γ = constant together with the equation of state, where γ is the heat capacity ratio that depends on the gas and is 5/3 for an ideal gas. If you extract heat during compression, you have isothermal compression where T does not change, but in general and classically at least, T will rise.
  9. Oct 19, 2011 #8

    Andrew Mason

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    The first law is:

    [tex]\Delta Q = \Delta U + W[/tex]

    where W is the work done by the gas. The ΔKE and ΔPE are part of ΔU.

    So if ΔQ = 0, then ΔU = - W. So if work is done ON the gas (as in a compression), ΔU will be positive. If U is proportional to T then ΔT will necessarily be positive. This would be the case for an ideal gas.

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