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Homework Help: Adiabatic expansion, entropy change

  1. Jul 5, 2013 #1
    1. The problem statement, all variables and given/known data

    1 mol of monoatomic ideal gas (temperature T1) is inside a cylinder with a moving piston (all are isolated). The initial external pressure on the piston is P1. at some point the external pressure is changed to (2/3)P1, the gas undergoes (irreversible) adiabatic expansion and reaches equilbrium state at temperature T2.

    write an expression of the entropy change of the system using R,T1,T2.

    2. Relevant equations


    3. The attempt at a solution

    Well, I know that in order to calculate entropy change from state A to state B, I need to construct a reversible path from A to B, and find the heat in this path. A reversible adiabatic expansion from A (P1,T1,V1) to B( (2/3)P1,T2,V2) would mean that there is no heat transfer. why then the entropy change isn't 0?
  2. jcsd
  3. Jul 5, 2013 #2


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    It is not the right relevant equation (it is only part of the full equation). The clue is that the true equation should involve pressure and volume.
  4. Jul 5, 2013 #3
    It's the only equation we learned to calculate the entropy change directly. The solution also uses this form.
  5. Jul 5, 2013 #4
    It is not possible to get from state A to state B in this problem by following a reversible adiabatic path (even though the actual process path was adiabatic). But it is possible to get from state A to state B by following a reversible non-adiabatic path. This is all that is required to get the change in entropy.

    Your first goal should be to express the temperature and volume for state B in terms of the initial parameters for state A and the pressure in state B. If the irreversible process is adiabatic, what is Q? Under these circumstances, what is the relationship between the change in internal energy and the expansion work done on the surroundings. What is the equation for the expansion work done on the surroundings? What is the equation for the change in internal energy in terms of the initial (state A) temperature and the final (state B) temperature? Once you've developed these relationships, you can combine them with the ideal gas law to determine the parameters for state B. Then you need to identify a reversible path between states A and B, or, equivalently, use the formula for the change in entropy for an ideal gas between two equilibrium states.
  6. Jul 6, 2013 #5
    Thanks for the detailed explanation. How can I tell if such a path is possible or not? It really seems weird that the irreversible path is adiabatic but the reversible path is not. What is the explanation for this?
  7. Jul 6, 2013 #6

    rude man

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    If your process in getting from (p1,T1) to (2p1/3,T2) is adiabatic and reversible you will wind up with a different v2 than if the process is adiabatic and irreversible. In other words, you wind up in a different state 2. Since entropy is a state function (depends only on the thermodynamic coordinates p, V and T) for each state, the two entropy changes must be different.

    On the other hand, because entropy is a state function, carrying out the integral ∫dQ/T reversibly will give the same entropy difference no matter what path you take. That can be non-adiabatic, non-isothermal, anything at all as long as it can be done reversibly. Of course, you want to choose a path that makes the integration easy.

    The entropy change of an ideal gas in terms of cp, T2/T1 and p2/p1 is derived the way Chestermiller describes.

    Chet - how'd I do? :smile:
    Last edited: Jul 6, 2013
  8. Jul 6, 2013 #7
    How do I know this?
  9. Jul 6, 2013 #8
    It really doesn't require an explanation. Reversible and adiabatic are two separate concepts, and you need to internalize this idea. To get the change in entropy between two equilibrium states, you need to identify any reversible path between the two states, and then you need to calculate the integral of dQ/T for that path. As Rude man pointed out, there may be multiple paths between the two equilibrium states. However, none of these paths are required to be adiabatic. However, they will all have the same change in entropy.

    Rude man: I really liked your explanation too.

  10. Jul 6, 2013 #9
    I'm familiar with those concepts, my question is: why there is no reversible adiabatic path between A and B? Why it isn't possible for A: p1,t1,v1 and B: 2p1/3,t2,v2
  11. Jul 6, 2013 #10

    rude man

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    Thanks Chet. But I should have said "a different v2 and T2", not just a different v2.
  12. Jul 6, 2013 #11

    rude man

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    OK, show us how you would calulate ∫dQ/T along a reversible adiabatic path?

    If you can't find one, I suggest starting with the 1st law, making the appropriate substitutions and winding up with ds = function of cp, dp, p, dT and T. (Hint: start with du = cvdT and pv = nT so p dv + v dp = R dT. Also, can cp/SUB] be rewritten as a function of R?)

    Then integrate ds from T1 to T2 and from p1 to (2/3)p1 to get ΔS.
  13. Jul 6, 2013 #12
    If you could find a adiabatic reversible path between the two states, the change in entropy would be zero. But since you already can identify a reversible path between the two states for which the integral of dQ/T is greater than zero (i.e., the entropy change is greater than zero), and, since, as rude man pointed out, the entropy is a point function, there can't be any path for which dQ/T is zero (i.e., there can't be an adibatic reversible path).
  14. Jul 6, 2013 #13
    I think it finally settled in my mind. Thank you guys for your patience.
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