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Entropy, why dS = dQ / T

  1. Jul 14, 2007 #1
    While I think I have somewhat an understanding of Entropy. (A measurement of a given state by its tendency proceed to it's most "probable" state), I have somewhat of a problem reconciling it with temperature.

    Specifically why:
    dS = dQ / T

    rearranging the equation a bit...

    1/T = dS/dQ, which appears to me that to indicate that at smaller temperature, it is "easier" to increase the entropy of a system. Easier in this case means requires less energy. Am I correct in this assumption?

    Now, anticipating a bit, suppose this was qualitatively true, why the directly inverse relation 1/T, why not 1/T^2, or e^(-T). Also, supposing this were to be true, is there any quantitative usefulness to this? That is what sort of phenomenas can we predict defining Entropy this way. Is Entropy conserved? What sort of useful properties does it have?
  2. jcsd
  3. Jul 14, 2007 #2


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    That equation does not always hold. In the situations where it does hold you may take the equation [tex]\frac{dS}{dE}=\frac{1}{T}[/tex] as defining the temperature.
  4. Jul 14, 2007 #3


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    The rule of thumb I remember from my grad thermo was:

    1 kJ of heat from a 1000 K reservoir is more organized than 1 kJ of heat from a 300 K reservoir.

    The definition was somewhat explained to me by using a carnot cycle and the efficiency. Here's the Reader's Digest very condensed version:

    [tex]\eta = \left[\frac{W_{net}}{Q_{in}}\right][/tex]

    [tex] = \left[\frac{Q_{in}-Q_{out}}{Q_{in}}\right][/tex]

    [tex] = \left[1 - \frac{Q_{out}}{Q_{in}}\right] = \left[1 - \frac{T_{l}}{T_{h}}\right][/tex]


    [tex]\left[\frac{Q_{out}}{Q_{in}}\right] = \left[\frac{T_l}{T_h}\right][/tex]

    [tex]\left[\frac{Q_{out}}{T_l}\right] = \left[\frac{Q_{in}}{T_h}\right][/tex]
  5. Jul 14, 2007 #4


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    Fred, I like your quote about engineering... of course there is also a plus side to being an engineer as opposed to being a student: they pay you instead of you paying them
  6. Jul 15, 2007 #5


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    Thanks. I wish I could remember where it was from. I grant you that the getting paid part does soften the blow.
  7. Jul 16, 2007 #6
    Reply to question by Laix:

    The essential content of the second law of thermodynamics is
    encapsulated by the statement that all actual processes are
    irreversible in the sense that it is impossible to restore **all**
    systems to their original states. The entropy, S, is a measure of the
    degree of irreversibility of any particular process. Furthermore, it
    is formulated in such a way as to be a function of state, (e.g.
    S(E,V)); i.e the change in entropy in any process can be computed
    from the values of the state variables before and after the process
    without paying attention to the particulars of the process. This is
    Since total entropy increases in all spontaneous processes, one can
    calculate the entropy change for some feasible process to see whether
    it would actually occur. This is also useful. There are axillary
    functions of state that can be derived from the entropy which can
    tell us whether a given process will occur or not: the Gibbs free
    energy could tell whether a particular chemical reaction will occur
    spontaneously under conditions of constant temperature and pressure
    and it tells us also how much useful work can (in principle) be
    extracted from the process. This is useful. Would you try to put out
    a fire of burning charcoal by pouring petrol on it? Obviously not.
    But would you be tempted to pour water on it to put it out? Be
    careful! If the temperature is high enough, there is a
    spontaneous process leading to the production of carbon monoxide and
    hydrogen, so you would be adding fuel to the flames! Wouldn't it be
    useful to know what this temperature is before trying to extinguish a
    fire in a nuclear reactor? Useful!

    To get a quantitative measure of the degree of irreversibility of a
    spontaneous process, consider restoring the system to its
    original state by putting it in contact with a standard system made
    up of weights and pulleys and reservoirs such that the restoration
    process is carried out *reversibly*. The measure of irreversibility
    (in the original process) is then taken to be the heat absorbed by
    the standard system divided by some single-valued function B(T) of the
    temperature (B(T) could itself be used as a measure of temperature).
    Division by B(T) is necessary if we are to be able to distinguish
    between the degrees of irreversibility for the same amount of heat
    absorbed by reservoirs of differing temperatures.

    Now, why take B(T) to be T instead of, say, exp(T)? It must first
    be emphasised that B(T) is a universal function: it is defined with
    respect to the heat absorbed reversibly by the standard system used to
    restore the original system to its prior state. It does not depend on
    the original system under investigation. This means we can take *any*
    system to find out what B(T) is. In particular, one can take the
    perfect gas for which the energy depends only on the temperature T and
    not on the volume V.
    Since PV=nRT, the differential for the heat absorbed in the
    reversible expansion is dQ = Cv dT + P dV. (This differential is not
    exact). The definition of entropy is dS = dQ/B(T). To ensure that dS
    is an exact differential, it is necessary that d(Cv/B)dV = d(P/B)/dT.
    This in turn implies that (nR/V)[d(T/B)/dT] = 0. This can only occur
    if B is proportional to T; hence, one can take B=T. Thus, the
    thermodynamic termperature B can be taken to be the ideal gas

    I hope this is a useful way of looking at it. There are many other
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