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Gas is isothermally and reversibly condensed

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data
    gas is isothermally and reversibly condensed
    find w, q, U, H

    2. Relevant equations

    3. The attempt at a solution
    w=-n(g)RT n(g)=-2 w=5.6kJ
    So U=H-w U Should be -70.6-5.6=-76.2, but the book gives -65, Where am I wrong?
    And why is H=q in this process?
  2. jcsd
  3. Nov 29, 2008 #2


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    Science Advisor
    2015 Award

    Re: Thermodynamics

    In the condensation from a gas to a liquid, what is the sign of ΔV? What then, should be the sign on w? Is work being done on the gas by the surroundings or is work being done on the surroundings by the gas?

    Also, to answer your second question, ΔH = q for any reversible process done at constant pressure.
    proof: From the definition of enthalpy, we know that
    ΔH = ΔU + Δ(PV)
    Next, we combine this expression with the first law of thermodynamics, ΔU = q + w, to get:
    ΔH = q + w + Δ(PV)
    At constant pressure, w = -integral(PdV) = -PΔV, and Δ(PV) = PΔV. Plugging these two expressions into the expression above gives the desired result:
    ΔH = q
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