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Monatomic Ideal Gas volume expansion

  1. May 18, 2008 #1

    TFM

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    [SOLVED] Monatomic Ideal Gas volume expansion

    1. The problem statement, all variables and given/known data

    A monatomic ideal gas expands slowly to twice its original volume, doing 260 Joules of work in the process.

    (a)

    Find the heat added to the gas if the process is isobaric.

    (b)

    Find the change in internal energy of the gas if the process is isobaric.


    2. Relevant equations

    [tex] U = \frac{1}{2}nRT [/tex] per degree of freedom

    pV = nRT

    pV = const.
    p/T = Constant
    V/T = const.

    3. The attempt at a solution

    I am not quite sure what to do for this equation. I now that the pressure is constant, and the the volue has doubled. and it has done 260 Joul;es of work in the process.

    Any ideas where to start?

    TFM
     
  2. jcsd
  3. May 18, 2008 #2

    Doc Al

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    Staff: Mentor

    Try to figure out the change in internal energy. You'll have to play around a bit. Assemble all the equations that apply to this problem and see what you can do.
     
  4. May 18, 2008 #3

    TFM

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    Have I started this the right way.

    The gas is monatomic, so

    [tex] U = \frac{3}{2} nRT [/tex]

    R is a constant, 8.31

    [tex] U = \frac{3}{2} 8.31*nT [/tex]

    However, we don't know thenumber of moles.

    [tex] \frac{v_1}{T_1} = \frac{v_2}{T_2} [/tex]

    We know the volume of v2 is twice that of V1

    [tex] \frac{1}{T_1} = \frac{2}{T_2} [/tex]

    This gives the ratio of temperatures as:

    [tex] \frac{T_1}{T_2} = \frac{1}{2} [/tex]

    Does this look right so far? If so, what could be a possible next step?

    TFM
     
  5. May 18, 2008 #4

    Doc Al

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    Staff: Mentor

    Good.

    Combine this with the ideal gas law and an expression for the work done.

    Don't plug in numbers until the last step.
     
  6. May 18, 2008 #5

    TFM

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    Following on then,

    [tex] pV = nRT [/tex]

    [tex] n = \frac{pV}{RT} [/tex]

    Insert into U ewquation,

    [tex] U = \frac{3}{2} \frac{pV}{RT} RT [/tex]

    Cancels to:


    [tex] U = \frac{3}{2} pV [/tex]

    The Work Equation:

    [tex] W = p \Delta V [/tex]

    thus,

    [tex] p = \frac{W}{\Delta V} [/tex]

    putting into the equation gives:

    [tex] U = \frac{3}{2} \frac{W}{\Delta V}V [/tex]

    Sincve the V is changing anyway:

    tex] U = \frac{3}{2} W [/tex]

    and we have the work.

    Does this look right?

    TFM
     
  7. May 18, 2008 #6

    TFM

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    Sorry, that final equation hould have been:

    [tex] U = \frac{3}{2} W [/tex][/B]

    Is this correct?

    TFM
     
  8. May 18, 2008 #7

    Doc Al

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    Staff: Mentor

    I assume you mean:
    Good! Now use it to find the heat added.
     
  9. May 18, 2008 #8

    TFM

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    Would you use:

    [tex] \Delta U = Q - W [/tex]

    [tex] \Q = U + W [/tex]

    ?

    TFM
     
  10. May 18, 2008 #9

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
  11. May 18, 2008 #10

    TFM

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    I put the values ion and the right answer did come out!!!

    Thanks for your assistance, Doc Al, :smile:

    TFM
     
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