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Volume ratio in an adiabatic gas expansion

  1. Jun 9, 2018 #1
    1. The problem statement, all variables and given/known data
    Consider a perfect monoatomic gas at pressure $p_i$ 1.2 atm and temperature $T_i$ 300K, that is in equilibrium inside a cylinder having a volume $V_i=1L$ and which piston has a mass of 1kg and is at an height of 50 cm. Admit that a mass M=3.13kg is over the piston. When that mass is removed, the gas suffers an adiabatic expansion until a final pressure pf=1.04837, a final temperature Tf and a final volume Vf.

    What is the value of Vi/Vf?

    2. Relevant equations
    3. The attempt at a solution

    So my immediate thought was to use the equation of real gases
    We will have then

    But I also don't know Tf so I'm stuck. Can someone give me a hint?
  2. jcsd
  3. Jun 9, 2018 #2


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    The expansion is adiabatic.
  4. Jun 9, 2018 #3
    Yes I know that means the variation of internal energy is just the work done but how does that help...
  5. Jun 9, 2018 #4
    @TSny if you want to be vague and leave without further explanations, please leave.
  6. Jun 10, 2018 #5
    Forum rules prevent giving very explicit answers. In fact I sometimes think I'm too explicit. :smile::smile:

    There is another important fact in the problem: gas is monatomic.
    Now with this information, what I would do would be to search in Google how the adiabatic processes behave.
  7. Jun 10, 2018 #6


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    You should know an equation relating to adiabatic volume changes.
  8. Jun 10, 2018 #7
    @Granger ,

    You need to have some patience and little more decency if you wish to interact on this forum .

    You are wrong on several accounts .

    1) You said , TSny is vague . I don't think so . You asked for a hint . He gave you a hint . You might not have found that hint helpful . In that case you could have requested him to elaborate his hint . He would have surely given you a more useful hint .

    2) You presumed that he left you without further explanation .Do you realise your 2nd comment after TSny's reply came within 11 minutes of your 1st comment ? He might not have seen your 1st comment at all . Or even if had seen your reply he might have been in the process of replying you .It takes time to think and type in the response . Why be so impatient ?

    Members here are not paid . It's their choice to help . You can't force somebody .

    Even if you required an additional hint , there is a polite way to ask .

    3) Asking TSny to leave , now that is so rude . Have some decency .This is not the way to react on a forum .

    Presuming you have not interacted with TSny before, let me tell you he is a top class contributer on PF , surely one of the best I have seen .

    So next time you write something on this forum , please show patience and decency .

    Hope you will take this message in the right spirit .
    Last edited: Jun 10, 2018
  9. Jun 14, 2018 #8
    @TSny I want to apologize for my reaction on the other day. I was very stressed because I've been working for hours not figuring out how to solve the problem. Not that that can justify my rudeness. So my sincere apologies. I eventually figured out what yo meant with the hint that it was an adiabatic process (Q=0). I was not figuring out the expression for U in the case of an ideal monoatomic gas so I wasn't able to relate it with the work expression.
    Last edited: Jun 14, 2018
  10. Jun 14, 2018 #9


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  11. Jun 14, 2018 #10


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    No problem. Apology accepted. After posting my first reply, I was distracted by some personal matters. So I was not able to keep up my end of the conversation.Sorry for that. Glad you were able to solve the exercise.
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