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Thermodynamic process

  1. Jan 4, 2007 #1
    is it possible to have a process that is both adiabatic and isothermal?
    i would appreciate if someone could explained that to me.
    if possible, could you please give me an example in real life.

    Thanks
     
  2. jcsd
  3. Jan 4, 2007 #2

    Andrew Mason

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    If the "process" involves change, there can be no process that is both adiabatic and isothermal. You can show this mathematically from the first law and the ideal gas law:

    dQ = dU + PdV (first law)

    PV = nRT so d(PV) = nRdT (ideal gas law)

    d(PV) = PdV + VdP = nRdT

    so the first law becomes:

    dQ = dU + nRdT - VdP

    Now if dQ = 0 and dT = 0 then dU = VdP

    But since dU = nCvdT, and dT = 0 then dP must be 0.

    Also, if dT = 0 and dQ = dU + PdV = nCvdT + PdV = 0, then dV = 0

    So the only adiabatic and isothermal process would be one in which dT, dV and dP are all 0. There is no change at all.

    AM
     
  4. Jan 5, 2007 #3

    Q_Goest

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    I'd agree with most of what Andrew wrote, that's a good primer for understanding the application of the first law here. Note that we're interested here in processes in which pressure will change. However, the caveat I'd add is that a process CAN be isothermal and adiabatic if the fluid is incompressible.

    All fluids, be they liquids or gasses, are compressible to some minor degree. So all fluids will heat up according to the equations provided by Andrew above. However, the more incompressible they are, the less they will increase in temperature.

    Consider the isentropic compression of water. Starting at atmospheric pressure and 70 F and compressing to 50 psig, will result in the pressure increasing by a factor of roughly 500%. The temperature increase on the other hand, is on the order of 0.0053 degrees F, an increase in absolute temperature of only 0.001%.

    So to answer your question, an incompressible fluid can go through a process which is adiabatic and isothermal.
     
  5. Jan 5, 2007 #4

    siddharth

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    You need to be more specific. Is the system open or closed? Does it involve an ideal gas? If it's an ideal gas, what AM posted holds.

    In a more general case, for a single component, single phase closed system, the internal energy will be a function of 2 variables. (say volume and temperature).

    From the first law,
    [tex]\delta U = \delta Q + \delta W[/tex]

    The infinitesimal change in the Internal Energy dU for a general process will be,
    [tex] dU = \left(\frac{\partial U}{\partial V\right)_T dV + \left(\frac{\partial U}{\partial T}\right)_V dT[/tex]

    For an adiabatic process, by definition, [tex]\delta Q = 0[/tex].
    For an isothermal process, dT=0.

    So it's possible to have a process which is both isothermal and adiabatic, and where the internal energy change is non-zero.
     
    Last edited: Jan 5, 2007
  6. Jan 5, 2007 #5

    Andrew Mason

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    I was assuming a process that maintained thermodynamic equilibrium.

    In a dynamic system this may not be the case. For example in the free expansion of a gas into a vacuum, the energy of the dynamically expanding gas does not do work on its surroundings. So its internal energy does not change although its volume and pressure are changing. The problem is that the temperature of the dynamically expanding gas is not uniform (or is undefined).

    If there is no loss of heat (adiabatic), when the dynamic energy of the gas ceases (ie. the gas is confined to a larger volume and its dynamic energy is converted back into heat of the gas) the temperature of the gas will be the same as the original (no change in internal energy because no work has been done).

    AM
     
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