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Thermodynamics Troubles

  1. Nov 12, 2015 #1
    [Mentor's note: moved from another forum, so homework template missing.]

    An upright and ideally heat-insulated cylinder with a diameter of 30 mm is tightly sealed at the bottom by a piston of the mass mk = 2 kg. The cylinder contains 15 g of air. The ambient pressure is pu = 1 bar, the initial temperature in the cylinder is 300 K.

    Calculate internal pressure and the work necessary to reach ambient pressure in the cylinder.

    I'm pretty sure that I have the correct internal pressure. I just used p=F/A. But, I am really unsure of where to begin on calculating the work since I only know -p*dV=W. That equation doesn't seem like it would work here. Please don't solve it for me (I'll never learn that way :wideeyed:) just tell the basics of what I can do to get to that answer. And thank you in advance for your help! :)
    Last edited by a moderator: Nov 12, 2015
  2. jcsd
  3. Nov 12, 2015 #2

    first of all could you please provide your calculations for finding the initial pressure in the cylinder? Secondly, to apply the right formula for calculating the work, it is necessary to know how the change of state is achieved (isentropically, isothermally, ...). Do you have any information about that?
  4. Nov 12, 2015 #3
    It would be isentropic!
    Also, I have recently found out that my original answer to the calculation of the internal pressure was incorrect. Is there any chance you could also help me with that part? Originally, I put p=F/A= (5kg*9.81ms^2)/(pi*0.015^2)= 27.8kPa. But, now I am not confident in the answer. Thank you again for your time!
  5. Nov 12, 2015 #4
    If I don't misread the statement, the piston is kept in the cylinder due to the different pressures of the ambiance and in the cylinder. All forces are acting in vertical direction. Which forces are pointing down and which forces are pointing up?
  6. Nov 12, 2015 #5
    gravity down (naturally), piston acts upwards but is stationed at the bottom, the internal pressure would pull the piston upwards, we are assuming the system is at rest and no mechanical forces are acting on the piston
  7. Nov 12, 2015 #6
    That's not properly expressed. In fact the (higher) ambiant pressure exerts a higher force on the piston compared to the internal pressure, therefore the net force generated by the two pressures points upwards. How could you calculate this net force?
  8. Nov 12, 2015 #7
    What I mean is the force from the net pressure is in the upwards direction (acting against gravity).

    I have F= -19.62N (downwards being negative, force from the weight of the piston).
    Total P = F/A = -19.62/(π*(0.015^2)) = -0.2776 bar
    External pressure is 1 bar and thus I now believe that internal pressure is 0.7224 bar.
  9. Nov 12, 2015 #8
    That seems to be correct, very good. As you already mentioned, your change of state will be isentropic - I suppose to consider the air as an ideal gas. Do you know how to calculated the work of an isentropic process for ideal gases?
  10. Nov 12, 2015 #9
    I'm not sure I do.
    I have the equation that W= -p*dV, but I am not sure if that's correct
  11. Nov 12, 2015 #10
    Well, yes it is, but it is quite difficult to apply. For ideal gases there are simple formulas for typical changes of state, dependent on the initial and the final state of the gas and its properties. For an isentropic change of state the work can be calculated:

    W = (m⋅R⋅ΔT) / (κ-1) = (p1⋅V1-p2⋅V2) / (κ-1)

    EDIT: κ ... isentropic exponent (sorry wrong expression)

    Any ideas how to proceed?
  12. Nov 12, 2015 #11
    Is k=PV?
    As for finding volume values, I can use pV=mRT where (0.7224)(V1)=(15)(0.285)(300)
    That gives us V1= 0.01775
    Then I can apply p1(V1^n)=p2(V2^n) where n=1.41 to solve for V2.
    V2= 0.01409
    Then I guess the rest is plug and chug? (Assuming I'm on the right path here...)
  13. Nov 12, 2015 #12
    κ = cp / cv (= n considering your nomenclature, I'm sorry I'm still not used to the english one)

    That's correct!

    And that's correct too! Well done!
  14. Nov 12, 2015 #13
    Oh, so k=n. Now I understand.
    W = -0.02154J
  15. Nov 12, 2015 #14
    Sorry, I didn't recognise that in the first place. Watch out for your units: The pressure must not be used in bar in this equation
  16. Nov 12, 2015 #15
    Ah okay, so W= -309.98kJ (?)
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