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Work done by monoatomic ideal gas

  1. Mar 11, 2015 #1
    1. The problem statement, all variables and given/known data
    The volume of 30.0 moles of a monoatomic ideal gas is reduced at a uniform rate from 0.616m3 to 0.308m3 in 2h. Its temperature is increased at a uniform rate from 27.0◦C to 450◦C. The gas passes through thermodynamic equilibrium states throughout.
    (a) Write down explicitly how the temperature and the volume of the gas depend on time (in hours). Then find the dependence of T on V explicitly.
    (b) Calculate the cumulative work done by the gas.
    (c) Calculate the cumulative energy absorbed by the gas as heat.

    2. Relevant equations


    3. The attempt at a solution
    For part a), pretty sure dV/dt = 0.154 and dT/dt = 211.5. If that's the case then making dt the subject and equation both gives
    0.154 dT = 211.5 dV
    b) I can't add together the work done in the volume change, then add work to change temperature, because of the path dependence of work. This is definitely an integral! Which I am really struggling to form...
     
  2. jcsd
  3. Mar 11, 2015 #2

    TSny

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    For part (a) your expressions look OK except for a couple of sign errors. Note that the volume is decreasing while T and t increase.
    They want explicit expression for V as a function of t, T as a function of t, and T as a function of V. So, you will need to go further than writing expressions for dV/dt etc.

    For part (b) you are right that you will need to set up an integral. Your result for part (a) should be helpful.
     
  4. Mar 11, 2015 #3
    So I need to integrate my expressions for dV/dt and dT/dt? Which should be dV/dt = -0.154 and dT/dt = 211.5. An indefinite integral, would it be?
     
  5. Mar 11, 2015 #4

    TSny

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    You can use indefinite integrals. You will then need to determine the arbitrary constants of integration.
     
  6. Mar 11, 2015 #5
    Right. So dV=-0.154 dt, integrate to get V=-0.154t+c, and v=0.616 when t=0.
    So V= -0.154t+0.616
    dT = 211.5dt , so T=211.5t+c. T=300.15K when t=0 (I assume I have to convert to kelvin), so
    T = 211.5t + 300.15
    Then rearrange these expressions to get the dependence of T on V.
     
  7. Mar 11, 2015 #6

    TSny

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    Yep. Good.
     
  8. Mar 11, 2015 #7
    The answer to part a, then, is
    (T-300.15) / 211.5 = (0.616 - V) / 0.154

    300.15 + 211.5*(0.616 - V) / 0.154 = T
    Still not sure what to actually integrate though. Pressure is changing too, since it doesn't state otherwise. I was expecting to integrate pressure with respect to volume.
     
  9. Mar 11, 2015 #8
    Use pV=nRT to find an expression for pressure in terms of volume maybe?
     
  10. Mar 11, 2015 #9

    TSny

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    Yes. That's it. (Your expression for T as a function of V can be simplified.)
     
  11. Mar 11, 2015 #10
    Simplifies to 1146.15 - 211.15V/0.154 = T. Sub in:
    p = nR(1146.15 / V) - 211.5nR/0.154
    And n=30, R=8.31, but I think that just makes it more complicated. So leaving them as n and R and integrating dV gives

    W = 1146.15nR ln(V) - 211.5nRV/0.154 + c, but I can do a definite integral between V2 and V1
     
  12. Mar 11, 2015 #11

    TSny

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    Yes.
     
  13. Mar 11, 2015 #12
    Which means work is -32926.85...
    And I probably have to use that to work out the next part too. Is this one the first law of thermodynamics? I don't have internal energy so maybe not. And heat is Q=mc dT, but I don't know what c is because the question justs says 'an ideal gas'.
    Some sort of conservation of energy equation... Heat is energy in transit, isn't it?
     
  14. Mar 11, 2015 #13
    No, not energy conservation, not a closed system, heat is added.
     
  15. Mar 11, 2015 #14

    TSny

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    I don't agree with your numerical answer for the work.

    You should know how the internal energy of a monatomic ideal gas depends on T.
     
  16. Mar 11, 2015 #15
    I'm a little confused. I know the equation
    change in internal energy = nCv dT, but Cv= 3R/2 is the heat capacity at constant volume... Except my textbook did a u-turn and says it applies whether volume is constant or not. So I can use that equation after all.
     
  17. Mar 11, 2015 #16

    TSny

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    Right. Internal energy is a state function: U = nCvT.
    ΔU is independent of path.
     
  18. Mar 11, 2015 #17
    Oh! Ok, I remember that. And I don't agree with my previous numerical answer either... Now I have -92602.6448...
    Since I can calculate the internal energy, Q = dU +W, so sub in T1 and T2 to work out dU. etc.
     
  19. Mar 11, 2015 #18

    TSny

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    Yes. I agree with your new value for W. Sounds good!
     
  20. Mar 11, 2015 #19
    dU = 158180.85 and Q = 65578.2052. Provided I typed it in right!

    Thank you for your time and help :) I was so confused!
     
  21. Mar 11, 2015 #20

    TSny

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    OK. Of course, think about significant figures and units. Good Work!
     
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