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B Entropy of free expansion

  1. Oct 30, 2016 #1
    I learnt that
    $$ dS = \frac Q T$$
    In free expansion of Ideal gas, it is obvious that Q = 0. However, the entropy increases. I guess the reason is that it is because the process is not quasistatic. If I am right, why is this process not quasistatic. If I am not, what's wrong with the formula above. Thanks!
  2. jcsd
  3. Oct 31, 2016 #2
    In free expansion the system does not pass through a series of equilibrium states. The P, T and S do not have the same value throughout the extent of the system at every instant. It starts with equilibrium state ends in equilibrium state but in between we are not sure rather we are sure it is not in equilibrium.
  4. Oct 31, 2016 #3
    The equation you wrote is incorrect. If you learned it that way, then you were taught incorrectly. The correct formula is $$\Delta S=\int{\frac{dQ_{rev}}{T}}$$
    What do you think the subscript "rev" stands for?

    Here is a reference to my recent Physics Forums Insights article the provides a cookbook recipe for determining the entropy change in an irreversible process: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
    See in particular Example 3

    Here is another article on entropy and the second law that should help with your understanding: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
  5. Oct 31, 2016 #4
    Thanks all for your comments
    I am not sure if I am correct. Q may not always be full differential, so I guess we can't always write dQ?
  6. Oct 31, 2016 #5
    As I said, your exact problem is solved in Example 3.
  7. Oct 31, 2016 #6
    But gas is pushing nothing, isn't that the work done = 0?
  8. Oct 31, 2016 #7
    Who said that the work done is not zero? Did you read what I wrote?
  9. Oct 31, 2016 #8
    dq is not perfect differential but [dS =(dq/T)] is perfect differential and T is called integrating factor. One may argue that for infinitesimal dq T can be considered as constant but for the process under consideration at any instant during the process if you were to measure T at different points in the mass of gas you will not get the same value which will convince you that the mass of gas does not have definite value of temperature hence the gas is not in equilibrium at any instant during the process. So formula is not applicable.
  10. Oct 31, 2016 #9
    Thanks for your explanation!
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