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Thermodynamics: ideal gas undergoing an isothermal process

  1. Oct 11, 2012 #1
    1. The problem statement, all variables and given/known data

    For an ideal gas, undergoing a quasistatic process, the equations below are correct. Evaluate them given that we have an isothermal process

    2. Relevant equations

    [itex]PV^\alpha=K[/itex] where K is a constant and [itex]\alpha=C-C_P/C-C_V[/itex]

    [itex]W = \frac{K}{\alpha -1} (\frac{1}{V_f^{\alpha-1}}-\frac{1}{V_i^{\alpha-1}})[/itex]
    [itex]Q = C(T_f -T_i)[/itex]
    [itex]\Delta S= Cln\frac{T_f}{T_i}[/itex]



    3. The attempt at a solution
    For an isothermal process, ΔT = 0, but what does that mean for the equations given? First I though it would mean that C→∞, but that would mean that Q=0 and W→∞ (because [itex]\alpha[/itex]→1), which clearly is not correct.

    How should I argue?
     
  2. jcsd
  3. Oct 12, 2012 #2
    Or does these equations not apply for isothermal processes?
     
  4. Oct 12, 2012 #3

    rude man

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    This looks wrong. For example, if it's an isothermal process, α = 1 but then (C-Cp)/(C-Cv) = 1 or Cp = Cv which is definitely not true for an ideal gas.
     
  5. Oct 12, 2012 #4
    I though α=1 because C→∞ (C=Q/dT, where dT→0)?

    Anyways, I'm thinking that these equations are not good to use when the process is isothermal, as the equations for the work, heat and entropy becomes of the type: ∞ muliplied with 0. Do you think this is a good conclusion?
     
  6. Oct 12, 2012 #5

    vela

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    The point of the problem might be to figure out how these indeterminate forms can be evaluated and to show the result is what you'd expect for an isothermal process.
     
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