1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why a cotton wool will not ignite using thermodynamics?

  1. Apr 16, 2015 #1
    So there is a devide for adiabetic expansions where a cotton wool is placed at the end of a plastic tube, and a plunger is pushed quickly down the tube compressing the air and igniting the wool when it reaches the end.

    ''Use the first law of thermodynamics to explain why the cotton wool will not ignite if the plunger is pushed down on the tube very slowly''

    First law is U = Q + W.

    So I know that if it is pushed slowly there will be some heat transfer so the compression wont be adiabatic anymore. I know that temperature is internal energy so in the case of slow expansion there is not enough of internal energy to ignite cotton.

    But I don't know how to explain it in terms of the first law, since I don't know how to relate W and Q into my explanation... Could anyone help please?
     
  2. jcsd
  3. Apr 16, 2015 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Can you relate the words in bold to the equation?
     
  4. Apr 16, 2015 #3
    Suppose that the cylinder were perfectly insulated so that no heat can be transferred from the gas to the cylinder, and so that the compression is adiabatic. Would you still say that if it is pushed slowly, the temperature will not rise enough to ignite the cotton, while, if it is pushed fast, the cotton will ignite. You should say this, because it is true.

    Suppose the initial temperature, pressure, and volume of the gas are T1, P1, and V1. Suppose you apply a constant force per unit area of the piston of Pext > P1 to compress the gas rapidly and irreversibly, and waited until the system equilibrated to a new volume V2 and pressure P2 = Pext. What would the final temperature T2 be?

    Now suppose you start over and compress the gas gradually and reversibly to the same final volume V2. What would the final temperature be?

    How would these temperatures compare? Is more work done on the gas in the irreversible case or in the reversible case?

    Chet
     
  5. Apr 16, 2015 #4
    In reversible case the final temperature for one mole of gas,
    T2 = P2V2/R by ideal gas equation,
    I'm in a bit confusion but for
    Irreversible case it should also be same as temperature is a state variable.

    So internal energies are same in both cases and work done is also same.
    But then pext > p , so work done should be more?
    Chet sir, it is indeed an interesting problem for me.
     
  6. Apr 16, 2015 #5
    Hi Raghav,

    This is not exactly what I had in mind for your involvement. At Physics Forums, we try to help the OP, not by doing the whole problem for him, but by giving hints and asking leading questions. This is what I envisioned your role to be here. So, we need to wait for the OP to respond before we can start to help him.

    Regarding the problem statement that I proposed in post #3, before solving the problem for the reversible case, one needs to solve it for the irreversible case, so that the final volume can be determined for the two cases. This same kind of problem was worked out in the attachment that was submitted in the previous thread that you and I were involved with. Maybe you can help the OP a little at least by writing out the first law equation for the irreversible case that I described. You were correct in reckoning that the work done in the irreversible case will be more (for the reason that you stated). So, what does the first law equation for the irreversible change look like?

    Chet
     
  7. Apr 16, 2015 #6
    Agree with you, that OP must try on his own to solve the problem.
    So the first law equation for irreversible process is
    ΔU = Q + Pext(V2-V1)

    @Tangeton the first law is ΔU = Q + W
    not U= Q + W as stated by you in post number 1.
     
    Last edited: Apr 16, 2015
  8. Apr 17, 2015 #7
    This is not quite correct. There should be a minus sign in front of the second term on the right: ΔU = Q - Pext(V2-V1)

    Of course, we also said that the compression is taking place adiabatically, so that Q = 0. Also, for our ideal gas case, ΔU = nCv(T2-T1)

    Chet
     
  9. Apr 17, 2015 #8
    I think it should be a +ve sign, as the pressure applied is by us. That term would automatically become negative when we substitute values for V2 and V1 , as V2 is less then V1 in compression.

    Post Script: I think OP would be back here again after a long time ( checking some of his statistics ).
     
  10. Apr 17, 2015 #9
    The work done to compress the gas is positive. So the sign should be negative. Otherwise, the final temperature will be lower than the initial temperature.

    Chet
     
  11. Apr 17, 2015 #10
    So that is what I am saying in post number 8.
     
  12. Apr 17, 2015 #11
    If V2 is less than V1, then Pext (V2 - V1) is negative, but you have a + sign in front of it in your 1st law equation. So, with the + sign there, ΔU comes out negative. What I'm saying is that this is incorrect.

    Chet
     
  13. Apr 17, 2015 #12
    Hmm.. That's true.
    But I have learnt a sign convention that when work is done by gas,
    First law is
    ΔU = Q - W
    When work is done by us on gas
    convention is,
    ΔU = Q + W

    So here we are applying external pressure on gas to compress it.
    Why then a -ve sign?
    Although it's true that otherwise ΔU will come negative and final temp. would be less, which is not true.
     
  14. Apr 17, 2015 #13
    In this thread, they're using the +W sign convention, which you seemed to accept in post #6.

    Chet
     
  15. Apr 17, 2015 #14
    Yeah, got it. A small confusion came but it is cleared now.
     
  16. Apr 19, 2015 #15
    Okay...

    The first law is in fact Q = ΔU + W where work is done BY system and Q is heat transfer INTO the system.

    So, in the case I'm given, work is negative as it is done ON the system, and Q is negative as well, since energy is transferred OUT of the system.
    So -Q = ΔU - W so ΔU +Q = W... The heat transfer, Q, take places due to work down so very little temperature increase, which is the increase in internal energy ΔU, will take place since most of the work done will cause heat but it is going to transfer out of the system, so the temperature increase will be 0 or nearly 0 due to all work done W being lost to heat transfer Q.

    Sounds right? Did I correctly say that ΔU + Q = W in this case? Because this is how it looks like...
     
  17. Apr 19, 2015 #16
    No. What does the word adiabatic in the problem statement mean to you?
    (a) Q ≠ 0
    (b) Q = 0

    You are comparing two adiabatic compression processes on an ideal gas, one carried out very rapidly and the other carried out very slowly. Which of these two processes do you think is reversible and which of these two processes do you think is irreversible? Does that mean that there is any significant difference between the final temperatures when the final compressed volumes of the gas are the same?

    In post #3, I laid out a road map for attacking this problem correctly. Apparently, you chose to disregard it. As a result, your analysis was incorrect from the get-go. I hope you decide to reconsider.

    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted