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Adiabatic process-how is internal energy mcv(t2-t1)

  1. Nov 14, 2009 #1
    In polytropic process internal energy change is=mCv(T2-T1)? why is it not mCp(T2-T1)? Why are we not taking specific heat at constant pressure?
     
    Last edited: Nov 15, 2009
  2. jcsd
  3. Nov 15, 2009 #2

    Mapes

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    Are you dealing with an ideal gas? If so, the internal energy is [itex]U=mC_VT+U_0=m(C_P-R)T+U_0[/itex] (the internal energy depends only on temperature), so [itex]\Delta U=mC_V\Delta T=m(C_P-R)\Delta T[/itex]. [itex]C_V[/itex] (or [itex]C_P-R[/itex]) are just constants here. The equations apply to any process.
     
  4. Nov 15, 2009 #3
    mCv(t2-t1) is the heat required to raise the temperature of mass 'm' from t1 to t2 at constant volume. But adiabatic process is not a constant volume process. My question is how does this equation give the value of change in internal energy?
     
  5. Nov 15, 2009 #4

    Mapes

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    Are you dealing with an ideal gas?
     
  6. Nov 15, 2009 #5
    yes. I am dealing with ideal gas
     
  7. Nov 15, 2009 #6

    Mapes

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    It's a unique property of ideal gases that [itex]\Delta U=mC_V\Delta T[/itex] holds for any process. The constant-volume constraint is not required. That's why I also wrote the equation as [itex]\Delta U=m(C_P-R)\Delta T[/itex], which similarly doesn't mean that constant pressure is required just because the constant-pressure specific heat appears. Again, [itex]C_V[/itex] and [itex]C_P[/itex] are just acting as constants, and do not represent constraints on the equation. It's hard for every thermo student to get used to this, but that's how it is.
     
  8. Nov 15, 2009 #7
    Thanks for the answer. But how did they found this relation?
     
  9. Nov 15, 2009 #8

    Mapes

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    It comes from one of the assumptions we make for an ideal gas, that there's no interaction between the atoms or molecules. If this is true, then changing the pressure (while keeping temperature constant) shouldn't have any effect on the total energy (i.e., [itex](\partial U/\partial P)_T=0[/itex]). We're left with a energy dependence on temperature alone.
     
  10. Nov 15, 2009 #9
    Starting from
    [tex]
    \mathrm{d}U=T\mathrm{d}S-p\mathrm{d}V
    [/tex]
    one can in fact show that for the special case
    [tex]
    p=Tf(V)
    [/tex]
    i.e. pressure proportional to temperature at constant volume, the energy is a function of temperature only [itex]U=A(T)[/itex].

    The proportionality to temperature doesn't follow from it yet, but if you also assume that the density of states (statmech treatment) is a power law, the you can show that [itex]U=CT[/itex].
     
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