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Relation between entropy and pressure at constant temperature

  1. Dec 20, 2007 #1
    if there is a decrease in pressure at constant temperature will there be an increase in entropy?
     
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  3. Dec 20, 2007 #2

    Hootenanny

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    Maybe....
     
  4. Dec 20, 2007 #3

    LURCH

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    Very interesting question, Bill. My first knee-jerk reaction is to say yes, because energy is lost from the system. But the fact that it's mechanical potential energy (pressure) rather than electromagnetic energy (heat) is a bit strange, so maybe the question requires more discussion.
     
  5. Dec 20, 2007 #4

    siddharth

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    Increase in the entropy of what? Only the system or the system and the surroundings together?

    I think it depends on how the system is defined and what process the system is subject to (reversible/irreversible).
     
  6. Dec 20, 2007 #5
    I would say yes.
    Starting from
    [tex] T dS = dE + P dV [/tex]

    and dividing by dP at constant T gives
    [tex]T\left( \frac{\partial S}{\partial P} \right)_T = \left( \frac{\partial E}{\partial P} \right)_T + P\left( \frac{\partial V}{\partial P} \right)_T[/tex]

    If you use the idea gas equations [tex] E = \lambda NkT[/tex] and [tex] PV = NkT[/tex] then energy derivative reduces to zero and you're left with the pressure/change in volume term. The final result I get is
    [tex]\left( \frac{\partial S}{\partial P} \right)_T = -\frac{V}{T}[/tex]

    which is negative, indicating that as pressure goes up, entropy goes down, or vice versa.

    For a real system, the energy derivative is likely to be positive and the volume derivative will always be negative (for P > 0) so these two terms will compete.
     
  7. Dec 20, 2007 #6

    Andrew Mason

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    The only way to have a decrease in pressure at constant temperature is to have an increase in volume. Work must be done in order to increase the volume (unless it is a free expansion). From the first law, if temperature remains the same (ie. internal energy remains constant) there must be heat flow into the gas. Since entropy is dQ/T, and dQ is positive (ie. into the gas) there is an increase in entropy of the gas.

    If there is an adiabatic free expansion of a gas, the temperature is unchanged (isothermal) so you might think that there is no increase in entropy (dQ = 0 so dQ/T = dS = 0). However, this is not correct. dQ = TdS is defined only for reversible processes. While the gas is expanding, the speed distribution of the molecules is no longer a Maxwell-Boltzmann speed distribution. So temperature is undefined. In order to determine the entropy (which since it is a state function it is independent of the process by which the state changed) you have to determine an equivalent reversible process. In any reversible process, work must be done during the expansion, so heat must flow into the gas. Therefore, dS>0.

    So in all cases, there is an increase in entropy of the gas where there is a decrease in pressure but no change in temperature.

    There is a very easy conceptual way to see this. Entropy measures the energy concentration. If the same amount of energy disperses into a larger volume, entropy must increase.

    AM
     
  8. Dec 20, 2007 #7

    siddharth

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    That's true for an ideal gas. But for a non-ideal gas (or a general homogeneous substance), the heat absorbed will be,

    [tex]dQ= C_p dT - TV \alpha dp[/tex]

    where [tex]\alpha[/tex] is the coefficient of volume expansion. So, if you look at a quasi-static equilibrium process for a substance with a negative coefficient which expands on cooling in a certain temperature range (like water), the entropy change can be negative depending on the value of the integral,

    [tex] \int TV \alpha dp[/tex]
     
  9. Dec 20, 2007 #8

    Andrew Mason

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    But in this case, temperature is constant. I am not aware of any substance which, if pressure is decreased its volume reduces. So if pressure decreases and temperature is constant, you have expansion which does work. Therefore, heat must flow into the system, regardless of the type of substance.

    AM
     
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