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Calculate heat supplied

  1. May 2, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    An adiabatic cylinder of length 2l and cross-sectional area A is closed at both ends. A freely moving non-conducting this piston divides the cylinder in two parts. The piston is connected with right end by a spring having force constant K and natural length l. Left part of the cylinder contains one mole of helium and right part contains 0.5 mole of each of helium and oxygen. If initial pressure of gas in each part is P0, calculate heat supplied by the heating coil, connected to left part, to compress the spring through half of its natural length.


    3. The attempt at a solution

    Let initial temperature in both cylinders be T0 and initial volume of both parts be V.
    γleft = 5/3 and γright = 3/2=γ'
    For left part
    [itex]T_0 = T_2 \left( \dfrac{3}{2} \right) ^{\gamma-1} \\ [/itex]

    For right part
    [itex]T_0 = \dfrac{T_1}{2^{\gamma' -1}} [/itex]

    [itex]ΔU = ΔU_{left} + ΔU_{right} \\
    =\left\{ \left( \dfrac{2}{3} \right) ^{-1/3} + 2^{3/2} - \frac{7}{2} \right\} T_0[/itex]

    ΔW = kl^2/4

    According to First Law
    ΔQ=ΔU+ΔW

    But this does not give the correct answer. I also want to know why this process is considered adiabatic even when heat is being supplied to the system.
     
  2. jcsd
  3. May 3, 2014 #2
    The cylinder is insulated, but there is an electrical coil inside of it. So for the left side, you can assume that the only heat that enters is via the heating coil. In this case, you are not including the electrical coil as part of your system.

    My advice is to focus on the right compartment first. You know everything you need to solve this part. Don't forget that oxygen is diatomic, so its molar Cv is (5/2)R, and the molar Cv of the helium is (3/2)R, so you can get the molar Cv of the mixture in the right compartment.

    Chet
     
  4. May 3, 2014 #3

    utkarshakash

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    The Cv for the mixture will be 2R. Work done = kl2/4
    ΔQ=ΔU+ΔW
    =CvΔT + kl2/4

    I found out ΔT using adiabatic formula and then the final expression becomes

    [itex]ΔQ = 2P_0V(\sqrt{2} -1)+ \dfrac{kl^2}{4} [/itex]

    Is this correct?
     
  5. May 3, 2014 #4
    I don't think so. I confirm your result for the work done by the piston on the right chamber contents. But the internal energy of the gas in the left chamber changes because you added heat. From the force balance on the piston in the final equilibrium state of the system, what is the final pressure in the left chamber in terms of the final pressure in the right chamber 2P0√2 and the spring force kl/2? From the ideal gas law on the left chamber, what is the final temperature of the gas in the left chamber? What is the change in internal energy of the gas in the left chamber? Now, you're ready to apply the first law.

    Chet
     
  6. May 3, 2014 #5

    utkarshakash

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    I found out the change in internal energy to be
    [itex]ΔU_{left} = \dfrac{3}{2} Al \left\{ P_0(1-3\sqrt{2}) - \dfrac{3kl}{4A} \right\} [/itex]

    I hope I'm correct this time.
     
  7. May 3, 2014 #6
    I get the exact same result, but with the opposite sign. The final temperature has to be higher than the initial temperature.

    Chet
     
  8. May 3, 2014 #7

    utkarshakash

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    OK. Finally applying first law

    [itex]Q=\dfrac{11kl^2}{8} -\dfrac{3AlP_0 (1-3\sqrt{2}) } {2} [/itex]

    But it's still incorrect!
     
  9. May 3, 2014 #8

    TSny

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    In getting your answer for Q, it looks like you took the work done by the gas on the left to be ##\small kl^2/4 ##. But I don't think that's right.
     
    Last edited: May 3, 2014
  10. May 3, 2014 #9
    Yes. You left out the pressure term from the work.

    Chet
     
  11. May 3, 2014 #10

    TSny

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    Also, the work done to compress the spring is not ##\small kl^2/4##.
     
  12. May 3, 2014 #11
    Yes. You're right. Good eye. We missed that one.

    Chet
     
  13. May 3, 2014 #12

    utkarshakash

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    But the pressure is changing. How do I evaluate the work done?

    [EDIT] Here's what I tried.
    Since the process is quasistatic, the piston will be in equilibrium throughout the process.
    [itex]P_1A = kx + 2\sqrt{2} P_2 A [/itex]

    If I multiply dx to both sides and then integrate, from 0 to L/2,

    [itex]W_{left} = \dfrac{kl^2}{8}+ \int_0^{L/2} 2\sqrt{2} P_2 A dx [/itex]

    But P2 is not constant.
     
    Last edited: May 3, 2014
  14. May 3, 2014 #13
    You already calculated total the work done by the gas in the left chamber on the gas and spring in the right chamber as [itex]W=2P_0V(\sqrt{2} -1)+ \dfrac{kl^2}{8}[/itex], but, when you calculated the heat Q (from W + ΔUleft), you omitted the first term of W. (I have also corrected the second term to take into account the error we made in determining the spring energy, as pointed out by TSny.).

    You also calculated ΔUleft and corrected the sign. So, to get the heat added to the left chamber, all you need to do now is:

    Q = ΔU+W

    Chet
     
  15. May 3, 2014 #14

    utkarshakash

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    I still don't understand why this term denotes the work done by the gas. Please help.
     
  16. May 3, 2014 #15
    When we analyzed the right chamber, we determined the final pressure and final temperature using the formulas for adiabatic reversible compression (PVγ=C) and the ideal gas law. The equation for adiabatic compression inherently included the work done on the gas by the piston. Once we knew the temperature, we calculated the change in internal energy of the gas in the right compartment, which was equal to the adiabatic work done by the piston on the gas in the right compartment. But this is also equal to the work done by the gas in the left compartment on the piston. So this is how we got the work done by the gas in the left compartment. (We could have gotten the same results by saying that the sum of the internal energy changes in the two compartments plus the energy stored in the spring has to equal the heat added, since, for the combination of the two compartments, no work is done on the surroundings).

    Chet
     
    Last edited: May 3, 2014
  17. May 3, 2014 #16

    utkarshakash

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    Thanks for your help. :smile:

    P.S.- Is there any good book which you'd like to recommend on Thermodynamics? You see, I'm very weak in it.
     
  18. May 3, 2014 #17
    I don't think you are weak in it at all. You seem to have a better sense than most people learning the subject about the fundamentals, and you seem to have good intuition (and mathematical ability). I haven't really ever found a thermo book that I'm happy with. My daughter used Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness, and I thought that was pretty OK. The best thing you can do is practice doing a lot of problems. I think the problems I've seen you working on are very instructive.

    Please also note that I added a few words about the spring to my previous response.

    Chet
     
  19. May 4, 2014 #18

    utkarshakash

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    Thanks for your valuable suggestions.
     
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