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Homework Help: Kinetic energy of a gas

  1. Feb 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A stationary gas has a kinetic energy of 500 J at 30 C. What is its kinetic energy at 60 C?

    2. Relevant equations
    KE = 3/2 RT

    3. The attempt at a solution
    I first converted T to Kelvin = 333K. I substituted all the values in to find KE but my answer is around 40K, which is not even close to the right answer. KE = 3/2RT = 3/2 * 0.08*333 = 40 K. What am I doing wrong?
  2. jcsd
  3. Feb 19, 2015 #2
    Isn't ##R \approx 8.31## , not 0.08?

    Also isn't the formula you need
    ##\Delta K_{energy} = \frac{3}{2} R\ \Delta T##, where ##\Delta## means the difference between final and initial.
  4. Feb 19, 2015 #3
    Right. For these problems, I meant to use R = 8.314 J/moleK. Even if I used that one, I end up with a number huge = 4146 J. The correct answer is 550J
  5. Feb 19, 2015 #4
    Read my edit. You need the changes, not the absolute in kelvin.
  6. Feb 19, 2015 #5
    I never seen kinetic energy in that way before but yes, that makes sense. I set up the equation as:
    500-x = 3/2 (8.31)(-30)
    However my final KE = 873.
  7. Feb 19, 2015 #6


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    Staff: Mentor

    No idea what you are doing here. How come the answer to a question asking for energy is in K?
  8. Feb 19, 2015 #7


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    Homework Helper
    Gold Member

    For some reason we might discuss elsewhere students who (I presume) grasped simple proportion in early years at school can't do it any more later when it is called chemistry.

    It is essential to know and understand whence comes the formula KE = (3/2) RT . If you had been asked to calculate KE from scratch you would need to be given R (or to be able to calculate it from something else you know, like a mole occupies 22.4 l at STP). You can calculate R also from the information given in the excercise.

    But here you don't need to even consider R. R is independent of temperature. (3/2) is also the same at all temperatures(!). KE is proportional to absolute temperature. You have a ratio of temperatures, what is the ratio of KEs?

    (Added - and this cancelling of things that are constant is already there hidden away in the problem because they haven't told you how much of the gas there is - if it doesn't change you don't need to know that either.)

    (A separate point is that the temperature scale, practical origins understandable, is in theoretically rather irrational units, degrees, that depend on theoretically complicated and I believe incompletely understood or calculable properties of particulat substance, water . As Feynman pointed out, we might have defined temperature in joules/mole and in that case R would have been 1).
    Last edited: Feb 19, 2015
  9. Feb 19, 2015 #8
    I understand that this question is asking for KE at 60 C. I also understand that R is a constant. I also understand that KE is proportional to temperature. Yes, I understand these realtionships. The answer should be in J not K that was my mistake.

    However, my question is why can't we use KE = 3/2RT to calculate kinetic energy of a gas at 60 C? There has to be a formula to use if the answer is 550J. I mean if we were to look at KE = 3/2RT we would know that if KE goes up then T goes up but the question is by how much?
  10. Feb 19, 2015 #9


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    Staff: Mentor

    Because the formula is wrong. It holds for exactly 1 mole of gas, and you don't know how much gas there is. Sure, you can calculate number of moles of gas remembering that the correct formula is [itex]KE = \frac 3 2 nRT[/itex] (where n is number of moles), and then plug it back with anew temperature into the same formula, but as epenguin have explained - that's completely unnecessary.
  11. Feb 19, 2015 #10
    Oh I see the subtlety in that equation. This makes much more sense. When I take KE = 3/2nRT I get n = 0.13. Then substituting in to the equation again to find KE under 60 C I get approx. 540 J which is close to 550 J. Thank you for explaining that! My book did not specify that KE = 3/2RT for one mole. That's where my confusion was.
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