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Compressing ideal gas, show no heat exchange

  1. May 4, 2015 #1
    1. The problem statement, all variables and given/known data
    An ideal monatomic gas is contained in a cubic container of size ##L^3##. When ##L## is halved by reversibly applying pressure, the root mean square ##x##-component of the velocity is doubled. Show that no heat enters of leaves the system.


    2. Relevant equations
    ##dU = dQ -pdV##
    ##\langle {v_x}^2 \rangle = kT/m##
    ##pV = nRT##
    ##U = \frac{3}{2}nRT##

    3. The attempt at a solution

    Since ##2\langle {v_x}^2 \rangle = k2T/m##, the temperature has doubled.

    The change in internal energy is ##\Delta U = \frac{3}{2}R(2T -T) = \frac{3}{2}RT##.

    The final pressure is ##p' = \frac{nR2T}{V'} = \frac{16nRT}{L^3}##.

    I must somehow show that $$\int -pdV = \Delta U,$$

    but I seem to know nothing about the path.
     
  2. jcsd
  3. May 4, 2015 #2

    BiGyElLoWhAt

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    Gold Member

    What path do you mean? If youre integrating over a volume, youre integrating from a beginning volume to an end volume. Have you tried comparing the state variables? What are constants and what are variables? Compare them before and after the compression.
     
  4. May 4, 2015 #3
    Well isn't p a function of V? I'm not sure what the constraints are to be honest. I mean you can write p = nRT/V, but then T also varies.
     
    Last edited: May 4, 2015
  5. May 4, 2015 #4
    You know something about the path. The problem statement says reversible.
     
  6. May 4, 2015 #5
    At first I thought that reversibility was contained in the equation ##dW = -pdV##.
     
  7. May 4, 2015 #6
    It is implicitly contained . Otherwise it would read - pIntdV, where pInt is the pressure at the interface with the surroundings. See my recent PF Insights article.

    Chet
     
  8. May 4, 2015 #7
    For a reversible path, dU = -pdV. You need to integrate.
     
  9. May 4, 2015 #8
    Then I'm confused. I thought I must show that dQ = 0? I wanted to show that the integral of -pdV was equal to the change in internal energy.
     
  10. May 4, 2015 #9
    You want to write CvdT=(RT/V)dV, divide both sides of the equation by T, and then integrate. See what you obtain. You should end up with an identity.

    Chet
     
  11. May 4, 2015 #10
    But isn't that assuming dQ = 0? Also, I know the relation you are referring to (##TV^{\gamma-1} =C##) and plugging in the values for both end points does not result in equality.

    Edit: I think I've made a mistake about the temperature change.
     
  12. May 4, 2015 #11
    You're just checking to see if it is consistent.
     
  13. May 5, 2015 #12
    Ok thank you. I can sort of see how this would work. It seems that it would be possible to reach the same end point without adiabatic processes, but ##TV^{\gamma -1} = C## is necessarily true for adiabatic processes.

    Does the fact that the process only involved reversible pressurisation imply that it is adiabatic? Reversible pressurisation involves no heat loss and since no heat was added I'm sure that it means no heat exchange.

    Also the temperature was 4T at the end.
     
  14. May 5, 2015 #13
    No. It is true only for adiabatic reversible processes.
    No. You've heard of reversible isothermal compression, correct?
    No. Reversible pressurization does not necessarily involve no heat being removed or added. Again, consider isothermal reversible compression.
    If the temperature increased by a factor of 4, and the volume decreased by a factor of 8, is that consistent with ##TV^{\gamma -1} = C## , given that, for a monatomic ideal gas, γ = 5/3?

    Chet
     
  15. May 5, 2015 #14
    Then aren't we merely confirming that it the ends points are consistent with reversible adiabatic processes, not showing the process must be adiabatic? What I meant to say with the second point is that the reversible pressurisation was not coupled with exchanging heat.

    Edit: also I was wrong about the temperature change previously, which is why the the two values didn't coincide.
     
    Last edited: May 5, 2015
  16. May 5, 2015 #15
    Hmmm. This is a very interesting point. The question is whether it is possible to have a reversible path between these same two end points without the process being adiabatic. To do this, one would have to conceive of at least one path between the same two end points that is not adiabatic. I think I have been able to do that. Can you? The person who formulated the problem probably did not think in those terms.

    Chet
     
  17. May 5, 2015 #16
    I'm not sure. It must be a reversible pressurisation, what sort exist apart from isothermal and adiabatic?
     
  18. May 5, 2015 #17
    Suppose, at the initial or the final end point of an adiabatic reversible path, you add a reversible Carnot cycle. (This can also be done at any point along the adiabatic reversible path). Then the two end points will still be the same, because the Carnot cycle is a cycle, but you have done some reversible work and transferred a corresponding net amount of heat reversibly. These are just examples of how, for the same two end points, you can have a non-adiabatic reversible path. But they suggest that many others may exist.

    Chet
     
  19. May 6, 2015 #18
    The next question is what further assumptions must be made so the statement does become true?
     
  20. May 6, 2015 #19
    I'm not sure it is worth spending your valuable time thinking about the answer to this question. I recommend focusing on problems closer to real world relevance.

    Chet
     
  21. May 6, 2015 #20
    Especially if you have exams coming up....
     
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