1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Entropy for an irreversible adiabatic process?

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Calculate the entropy for an irreversible adiabatic expansion of an ideal gas which goes from (Vi, Pi) = (9.01L, 3.00 bar) to (Vf, Pf) = (15.44L, 1.50 bar)

    2. Relevant equations

    [tex]\Delta[/tex]S = [tex]\int dq_{reversible}/T[/tex]

    3. The attempt at a solution
    I understand that entropy is a state function, and thus independent of path. So the above equation can be used to calculate the change in entropy for an irreversible adiabatic expansion. However, what I don't understand is how the entropy can possibly be non-zero. I mean, in a reversible adiabatic expansion which has the same (Vi,Pi) and (Vf,Pf) as the irreversible case, q = 0 by definition. So if [tex]dq_{reversible}[/tex] = 0 then how can [tex]\Delta[/tex]S be non-zero?

    Any help as to what I'm missing here will be highly appreciated!

    Edit: having written this question out, I think I might be on to something. I don't think it's possible for the reversible expansion to be adiabatic because U is a path function, and must be the same for reversible and irreversible case. In irreversible case, delta U = w. In the reversible case, w is different, so the remainder of the energy must be necessarily in form of heat, which is q(reversible). Am I on the right track?
    Last edited: Mar 8, 2010
  2. jcsd
  3. Mar 21, 2010 #2
    Your problem doesn't state how the volume was increased, whether it was done by the gas, or by opening a divider of some sort. . The first law for non-diffusive systems is $d U = d Q -d W$. In this problem, $d U = 0$ so that $d Q = d W = T d S = P dV$. From the ideal gas law, $PV = nRT$. Using this to substitute for $P$, we have $dS = \frac{nRdV}{V}$. Integrating yields $\Delta S = nR\;ln \left( \frac{V_f}{V_i} \right)$. Thus, $Q$ and $S$ are not zero.

    The reason I asked about how your volume was increased is because there is an irreversible process called free expansion where a divider is removed in order to increase the volume. The gas does no work in increasing the volume. The kinetic energy of the molecules stays the same. This is a question in the GRE Physcis practice bulletin, question 47. I have three different thermo texts that do not agree. It seems to me that the molecules hit the inner surface of the container with the same amount of force as before, but now the force is applied to a larger inner surface area, so the pressure, which is force per unit area, would go down. Since the kinetic energy of the molecules stays the same, it would seem that the temperature must stay the same because of the kinetic theory of gases. What stumps me is that the texts say that $dQ$ and $dW$ are zero, so the above calculation would not be correct, but they give this result as the correct answer to the question. Can anyone explain this?
  4. Mar 23, 2010 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    The reversible path between these two states involves doing work (ie expansion occurs against an external pressure infinitessimally lower than the internal gas pressure). Since the temperature, hence internal energy does not change ([itex]\Delta U = 0[/itex]), [itex]\Delta Q = W[/itex]. So heat must be added to reach the final state. If heat is added, entropy increases: dS = dQ/T. Simply put, the reversible path between an adiabatic irreversible expansion is not adiabatic.

    Last edited: Mar 23, 2010
  5. Mar 23, 2010 #4
    http://en.wikipedia.org/wiki/Talk:Work_(thermodynamics [Broken])

    Last edited by a moderator: May 4, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook