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Entropy change

  1. Apr 14, 2017 #1
    MENTOR NOTE: NO TEMPLATE BECAUSE SUMITTED TO WRONG FORUM.

    3.1) A quantity of 0.10 mol of an ideal gas A initially at 22.2 degrees C is expanded from 0.200 dm3 to 2.42 dm3 . Calculate the values of work (w), heat (q), internal energy change (delta U), entropy change of the system (deltaSsys), entropy change of the surroundings (deltaSsurr), and total entropy change (delta S univ) if the process is carried out isothermally and irreversibly against an external pressure of 1.00 atm.

    The question is how do I calculate entropy? I tried something, and this is what I got:
    delta S system=2,0728 JK-1
    delta S surroundings=0.76 JK-1
    delta S universe=2,8328 JK-1

    Can you please check this out?
     
    Last edited by a moderator: Apr 14, 2017
  2. jcsd
  3. Apr 14, 2017 #2

    Andrew Mason

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    Hi Chemochem. Welcome to PF!

    It is difficult to check what you have done since you have not explained what you did.

    To calculate the change in entropy, you need to calculate ##\Delta S = \int_i^f \frac{dQ_{rev}}{T}## where dQrev is the heat flow during a reversible process from the initial to final states. (Note: For irreversible processes, the reversible path between initial and final states of the system will not be the same path as the reversible path between the initial and final states of the surroundings). In this case, what would be a reversible path between the initial and final states for the system? For the surroundings? Calculate ##\Delta S_{sys} \text{ and } \Delta S_{surr}## and add them together to the ##\Delta S_{univ}##.

    AM
     
  4. Apr 14, 2017 #3
    Well, your delta S for the surroundings is definitely wrong. It should be negative. Show us the details please.
     
  5. Apr 14, 2017 #4
    I used delta S=nRln(Vf/Vi) for delta S system and then I found a reversible path and calculated the value as being 0.76 JK-1 for delta S surroundings (it turned out to be -w/T). I may have forgotten to write the minus sign, I don`t know.
     
  6. Apr 14, 2017 #5
    The change in entropy for the surroundings should be -Q/T for the actual irreversible path.
     
  7. Apr 14, 2017 #6
    Then if it`s isothermal, dU is equal to zero, so the value for Q is just -w, right?
     
  8. Apr 14, 2017 #7
    Yes. Do you understand why, for the surroundings, ##\Delta S## is determined by the heat transferred in the irreversible path?
     
  9. Apr 14, 2017 #8
    Yes, I used w because it had the same value and then forgot the sign, so I got the wrong result. Thank you!
     
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