Chemistry

posted by Lauren

Consider an ideal gas encloesd in a 1.00 L container at an internal pressure of 10.0 atm. Calculate the work, w, if the gas expands against a constant external pressure of 1.00 atm to a final volume of 20.0 L.
w = ____ J

now calculate the work done if this process is carried out in two steps.
1] first let, the gas expand against a constant external pressure of 5.00 atm to a volume of 4.00 L.
2] From there, let the gas expand to 20.0 L against a constant external pressure of 1.00 atm.
w = _____ J


i know that the formula to solve this is w=-(p)(Delta V) but when they give you 2 pressures does that mean the formula turns to w=-(delta p)(delta v)?

  1. DrBob222

    1)
    -Pdelta V = -1 atm x (v2-v1) = -1(20-1) = -19 L*atm work.
    2)
    The first one is -5 x (4-1) = -15 L*atm
    Second stage is -1 x (20-4) = -16 L*atm
    Total = -31 L*atm
    Here is a good site you can read; in fact there is an example problem that is ALMOST the same as this one.
    (Broken Link Removed)

  2. Lauren

    Thank you Dr. Bob! That was very helpful

  3. DrBob222

    No prob. I always help sexy ladies ~

  4. Long D

    2=2

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