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Homework Help: Thermodynamics: p/V = constant

  1. Jul 12, 2010 #1
    First of all, i want to begin with saying that this is my translation of the problem in english (the original is in romanian) so i don't know if some words are correctly used...

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

    A monoatomic ideal gas expands under the law [itex]p=a*V[/itex] with a = constant. Find the molar heat in this transformation.

    A) 6 R
    B) 2 R
    C) 0.5 R
    D) R
    E) 3 R
    F) 5 R

    2. Relevant equations

    http://img710.imageshack.us/img710/8069/asd3432re.jpg [Broken]

    I forgot this: [itex]C_V=\frac{3}{2}R[/itex]

    3. The attempt at a solution

    http://img534.imageshack.us/img534/6503/32ef5456436tgr.jpg [Broken]

    Where's my mistake :-s ?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 12, 2010 #2
    From line 5 to line 6: 1/V is not p/a. Because p=aV, it should be a/p instead.
  4. Jul 12, 2010 #3
    Oh yea, thanks for pointing this out...

    Now i'm stuck at [itex]\frac{T}{pV}=\frac{1}{a}[/itex] = constant
  5. Jul 12, 2010 #4
    You end up at the first place, right? :biggrin:
    Why don't you try to do something about the 2nd relevant equation you gave above? They ask for C, right? :smile:
  6. Jul 12, 2010 #5
    ye but i don't know which Q should i "use" for a p/V transformation... if it was isobar it would be [itex]c=\frac{\nu C_p \Delta T}{\nu \Delta T}=C_p=\frac{5}{2} R[/itex] if it was isochoric it would be 3/2 R
  7. Jul 12, 2010 #6
    The textbook derives C for you in special cases. This case doesn't fall into any of those, so you must derive it from the start, i.e. from the definition of the molar heat capacity, the ideal gas equation, the 1st law of thermodynamics and most importantly the law of the process :smile:

    You have: [tex]dQ = nCdT[/tex].

    Moreover: [tex]dU = dQ - dA = dQ - pdV[/tex].

    You also have: [tex]dU = nC_vdT[/tex]

    Therefore: [tex]C = \frac{dQ}{ndT} = C_v + \frac{1}{n}\frac{pdV}{dT}[/tex]

    Now try to find pdV/dT from the 2 equations: pV/T = const and p/V = const :smile:
  8. Jul 12, 2010 #7
    I don't want to be rude, but can you solve it without using those "d's" (differentiation) because i wasn't taught at school thermodynamics with derivatives (or integrals) and i don't understand what you did over there :-s
  9. Jul 12, 2010 #8
    Oh I see. Then just change [tex]d[/tex] into [tex]\Delta[/tex] and you should get the same result:
    [tex]C = \frac{\Delta Q}{n\Delta T} = C_v + \frac{1}{n}\frac{p\Delta V}{\Delta T}[/tex]
    There is nothing about differentiation here; just the notation :smile:
  10. Jul 12, 2010 #9
    i found out that [itex]\frac{1}{\nu}\frac{p\Delta V}{\Delta T}[/itex] is R ... but the answer's not good :-s ...

    well isn't [itex]\frac{pV}{T}[/itex] constant? equal to [itex]\nu R[/itex]
  11. Jul 12, 2010 #10
    Yes. But not [tex]\frac{p\Delta V}{\Delta T}[/tex].

    We can change it a bit: [tex]\frac{p\Delta V}{\Delta T}=\frac{aV\Delta V}{\Delta T}[/tex]

    We need to find the relation between [tex]\Delta V[/tex] and [tex]\Delta T[/tex]

    [tex]pV/T = nR[/tex] so [tex]V^2 = (nR/a)T[/tex]. Therefore: [tex]\Delta (V^2) = (nR/a)\Delta T[/tex], agree?

    As the volume changes slightly from [tex]V[/tex] to [tex]V+\Delta V[/tex], we have:
    [tex]\Delta (V^2)=(V+\Delta V)^2-V^2=2V\Delta V + (\Delta V)^2[/tex]

    Because [tex]\Delta V << V[/tex], we can omit the 2nd term, and thus: [tex]\Delta (V^2)=2V\Delta V [/tex]

    The rest should be easy :smile:
  12. Jul 12, 2010 #11


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    You can't assume [itex]\Delta V \ll V[/itex].

    Draw a P-V diagram showing the path the system follows going from V1 to V2. The work done is given by the area under that curve, which you can calculate geometrically from the diagram.
  13. Jul 13, 2010 #12
    Why not?
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