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Thermodynamics; find the thermal energy

  1. Aug 26, 2013 #1
    question:Calculate the total thermal energy in a liter of helium at room temperature and atmospheric pressure. Then repeat the calculation for a liter of air.



    I'm just confused because i thought thermal energy only depended on the translational kinetic energy of the particles. So why do i need all the pressure if the temperature is already given?

    the only equation that comes to mind is E(kinetic)=3/2kT

    and ...maybe the gas law?..

    where do i go from here?
     
  2. jcsd
  3. Aug 26, 2013 #2
    Maybe this way:

    U(thermal) = N*1,5kT

    and the ideal gas law: pV = NkT
     
  4. Aug 26, 2013 #3
    so E(thermal)=(3/2)NkT -----> E(thermal)=(3/2)PV? since N=PV/kT? and then i just plug and chug?
     
  5. Aug 26, 2013 #4
    Yes, I believe so...
     
  6. Aug 26, 2013 #5
    isn't a "PV" term dynamically associated with the pressure with respect to a change in volume? ie calculating the work done on a system from a PV diagram? (ie the area under the PV curve)
     
  7. Aug 26, 2013 #6

    DrClaude

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    Staff: Mentor

    You're missing something here. The correct formula for an ideal gas is
    $$
    U = \frac{f}{2} N k T
    $$
    where ##f## is the number of (quadratic) degrees of freedom. That is why you get a different answer for helium and air.
     
  8. Aug 26, 2013 #7

    DrClaude

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    Staff: Mentor

    Yes, expansion/contraction work done by/on a gas is obtained from
    $$
    W = - \int_{V_i}^{V_f} P \, dV
    $$
    but ##PV## by itself is just the product of the pressure and the volume.
     
  9. Aug 26, 2013 #8
    well what i was getting at was i thought that that quantity (PV) was the case where the P is constant (isobaric) but still a dynamic case where the Volume is changing. so basically i don't understand why the quantity PV is used for a static case.
     
  10. Aug 27, 2013 #9

    DrClaude

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    Staff: Mentor

    Pressure doesn't have to be constant. The formula for work is valid even when ##P## varies, although this might make it complicated to calculate the integral (unless ##P## can be expressed as a simple function of ##V##).

    Equations of state are equations that relate the different macroscopic observables of a system. In the case of a gas, these observables are ##P##, ##V##, and ##T## (for a fixed quantity of gas). For an ideal gas, the relation is exactly
    $$
    PV = N k T
    $$
    or
    $$
    PV = n R T
    $$
    Such equations of state also exist for more realistic gases: they are slightly more complicated, but again relate ##P##, ##V##, and ##T##, such that if you fix two of them you can know the value of the third.

    As an example, if you measure the pressure inside a bicycle tire and know what the temperature is, then you can calculate the volume inside the inner tube. So you see, this has nothing to do with "dynamics."
     
  11. Apr 14, 2015 #10
    This is the first response to this thread in over a year and a half. I am closing this thread.

    Chet
     
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