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Adiabatic expansion at constant pressure.

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data

    A sample of 4.00mol of oxygen is originally confined in a 20L vessel at 270K and then undergoes adiabatic expansion against a constant pressure of 600torr until the volume has tripled. Find q, W, dT, dU, dH.

    2. Relevant equations

    U=q+w
    H=U+PV

    3. The attempt at a solution

    q=0 (Adiabatic)

    w=-pexΔV

    U=w

    ΔT= ΔU/Cv

    ΔH=0


    The reason I'm posting this is because I believe that all of the above is correct, but my friend claims that ΔH is not zero. First of all, ΔH=q for a constant pressure expansion, and q=0, so ΔH=0, right?

    Also:

    ΔH=ΔU+Δ(PV)=w+ PΔV= -PΔV+PΔV=0

    I believe oxygen is considered a perfect gas in this question, because we are expected to calculate C(v) from C(p)-C(v)=nR

    So, am I right, does ΔH=0?

    EDIT:

    So, I get different answers depending on whether I use ΔH=ΔU+Δ(PV) or ΔH=ΔU+Δ(nRT)

    Shouldn't they be the same? shouldn't Δ(PV)=Δ(nRT) ?

    I've been obsessing over problems like this for weeks now. Every time I figure one thing out another thing pops up that just confuses everything for me.
     
    Last edited: Oct 24, 2011
  2. jcsd
  3. Oct 24, 2011 #2

    Ygggdrasil

    User Avatar
    Science Advisor

    No, ΔH = q only if the pressure of the system stay constant. Although the external pressure is constant in this problem, the pressure of the system is not.
     
  4. Oct 24, 2011 #3
    Are you talking about a reversible vs. irreversible change here?

    So why does Δ(PV) not equal Δ(nRT)? Is it because to use Δ(PV), the P that you use has to be constant for the whole system? I didn't see that anywhere in my textbook.

    So the P in the work equation is p(external) and in the enthalpy equation it is the pressure of the system?

    EDIT: Ok, I see see that my last statement is correct, and Δ(PV) does indeed equal Δ(nRT), but since P is changing Δ(PV)≠PΔV because P is the pressure of the system. So we just use nRΔT because we know all those quantities.

    I'm glad I finally cleared this up.
     
    Last edited: Oct 24, 2011
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