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Is there any process adiabatic and isothermal?

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Is it possible to have a process that is both adiabatic and isothermal.


    2. Relevant equations



    3. The attempt at a solution

    I think the answer is no. Because when a system is doing an adiabatic process, the total energy of this system must be changed. For an adiabatic process, there is no heat transfer to or from the system. So the temperature of the system must be changerd.

    But, an adiabatic process only means that there is no heat transfer to or from out side. So when the out side environment is doing work on the system and the system transfer whole work out of itself. the temperature of the system will be same. but i am not sure is this ture.

    could somebody explane this for me?

    Thanks.
     
  2. jcsd
  3. Jan 24, 2010 #2

    kuruman

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    If a process is both adiabatic and isothermal, what form does the first law of thermodynamics take?
     
  4. Jan 24, 2010 #3

    Mapes

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    Why?
     
  5. Jan 25, 2010 #4
    thanks. I am not sure that I understand completely what u said. But I think u mean that there is impossible a process is both adiabatic and isothermal. That is because when outside environment doing work to the system, the internal energy must be changed. And work can be change to heat 100% but heat can not change to work 100%. Right?

    but, if there is a cylinder (cacuum, use a spring to replace gas ), when the work is done by outside. The energy of spring must be changed. and, in a ideal environment, this spring can do 100% work to outside when release. In this process, there is no heat, only work. So is this an adiabatic and isothermal process?

    I am confused. pls give me some advice. thanks.
     
  6. Jan 25, 2010 #5
    thanks.
    when a system is doing an adiabatic process, there is no heat can trasfer from or to outside. so the total energy of this system must be changed.
     
  7. Jan 25, 2010 #6

    Mapes

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    What about letting two gases mix, or a gas expand into an empty chamber? These processes can be made adiabatic and have no associated energy change.
     
  8. Jan 25, 2010 #7

    D H

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    That is not what kuruman asked. He asked a simple question. Try answering that simple question.

    Wrong. What is the first law of thermodynamics?
     
  9. Jan 25, 2010 #8
     
  10. Jan 25, 2010 #9
    so? still confuesd...
     
  11. Jan 25, 2010 #10

    kuruman

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    Remember that the change in internal energy ΔU is proportional to the change in temperature ΔT. What is the change in temperature, what is the change in internal energy?
     
  12. Jan 25, 2010 #11

    D H

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    Correct.

    Correct.

    Whoa! How do you draw that conclusion?

    Hint: delta T=0 => delta U=0.

    What does this say about work?
     
  13. Jan 26, 2010 #12
    adiabatic process

    delta U=Q+W, and Q=0 => delta U=W

    isothermal process

    delta U=Q+W, and delta U=0, => Q=-W

    combine these two factors, so conclusion is impossible a process adiabatic and isothermal. Right?
     
  14. Jan 26, 2010 #13

    Mapes

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    [itex]\Delta U=0[/itex] only implies [itex]\Delta T=0[/itex] for an ideal gas. Also, processes exist where the work and heat transfer are both zero, as I described above. The processes are both adiabatic and isothermal.
     
  15. Jan 26, 2010 #14

    D H

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    Wrong!

    There is one number that is equal to its additive inverse. What is it? (i.e., solve [itex]W=-W[/itex] for [itex]W[/itex].)
     
    Last edited: Jan 26, 2010
  16. Jan 26, 2010 #15
    sorry, i dont know what u meant

    adiabatic process
    delta U=Q+W, and Q=0 => delta U=W
    isothermal process
    delta U=0
    so if a process is adiabatic and isothermal, when the work done by system or at system, it is impossible to satisfy U=W and U=0
     
  17. Jan 26, 2010 #16

    D H

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    Who said work was being done on the system? Your original question was
    The answer to that question is definitely "yes".
     
  18. Jan 26, 2010 #17
    Mapes was right.

    In a free expansion problem, no heat is exchanged nor is any work done. Thus, dU = 0 for the system ...and so is dT... but it is a real system, and it does create new entropy. So, there is at least one example of such a process.
     
  19. Jan 27, 2010 #18
    yes! got it

    thanks very much to all
     
  20. Jan 27, 2010 #19
    check this...
    http://www.goiit.com/posts/list/physical-chemistry-name-one-process-where-both-adiabatic-isothermal-916569.htm [Broken]
     
    Last edited by a moderator: May 4, 2017
  21. Jan 27, 2010 #20
    check http://www.goiit.com/posts/list/physical-chemistry-name-one-process-where-both-adiabatic-isothermal-916569.htm [Broken]
     
    Last edited by a moderator: May 4, 2017
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