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Thermodynamics-a small doubt

  1. Jan 9, 2006 #1
    In thermodynamics it is said that S = dq/T. Then how can we say that S is directly proportional to T.
     
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  3. Jan 9, 2006 #2

    LeonhardEuler

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    That isn't exactly what is said. What is said is that
    [tex]dS=\frac{dQ_{rev}}{T}[/tex]
    The subscript "rev" means that the quantity dq represents the heat that would be transferred if the process were carried out reversibly.

    Also, S is not proportional to T. Suppose you heat a system reversibly at constant pressure. Then we have that
    [tex]dQ_{rev}=C_pdT[/tex]
    The entropy change can be found by integrating the equation for dS. (You may notice that the as T approaches 0, dS seems to approach infinity. This is not actually the case since in reality the heat capacity is a function of temperature and the Debye extrapolation tells us that at very low temperatures the heat capacity varies like T^3) Suppose the heat capacity is constant over a temperature range of interest. Then
    [tex]\Delta S=\int_{T_1}^{T_2}\frac{C_p}{T}dT= C_p \int_{T_1}^{T_2}\frac{1}{T}dT=C_p\ln{\frac{T_2}{_T_1}[/tex]
    So in this case - as in most cases - S is not proportional to T, but does increase with incresing T.
     
  4. Jan 11, 2006 #3
    by what ratio does S increase with increase in T-Sorry Im confused.
     
  5. Jan 12, 2006 #4

    LeonhardEuler

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    There is no definite ratio-it depends on the situation.
     
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