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Work done by an Ideal Gas

  1. Jan 18, 2010 #1
    See my reply
    Last edited: Jan 18, 2010
  2. jcsd
  3. Jan 18, 2010 #2
    I'm trying to calculate the work done when 50.0 g of tin dissolves in excess acid at 1.00 atm and 298.15 K


    Here is what I've done:

    [tex]\Delta V=\frac{\Delta n RT}{P}[/tex]

    [tex]\Delta n=n_{H_2(g)(after)}-n_{H_2(g)(before)}=2 mol H_2[/tex]

    It seems that the moles of hydrogen gas will apply pressure to the air, so the work will be

    [tex]w=-P\Delta V[/tex]

    [tex]\frac{-P(2 mol H_2)RT}{P}[/tex]

    [tex]=-(2 mol H_2)RT[/tex]

    This makes sense in terms of dimensions, but I don't have the answer in my book, so I can't tell if I did this right.
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