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Homework Help: Work done on compressed gas

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Pt. A) A gas is compressed at a constant pressure of 0.632 atm from 7.11 L to 5 L. In the process,
    440 J of energy leaves the gas by heat.
    What is the work done on the gas?
    Answer in units of J

    Pt. B)

    What is the change in its internal energy?
    Answer in units of J

    2. Relevant equations

    PV = nRT
    U = Q + W
    1 atm = 101325 pa
    1L = .001m^3

    3. The attempt at a solution

    .632atm = 64037.4 pa
    7.22L = .00711 m^3
    5L = .005 m^3

    W = P(Vf -Vi)

    U = Q+W

    U = -135.189 -440

    OR (done other ways)
    : 304.881
    : -304.881
    : 575.118914

    All of these were wrong. Please help?

    B) ...
    Last edited: Oct 2, 2012
  2. jcsd
  3. Oct 2, 2012 #2

    Andrew Mason

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    Is this the work done ON the gas or the work done BY the gas? What does the question ask for?
    What is W in U = Q + W? (ie is it work done ON or BY the gas?).

  4. Oct 2, 2012 #3
    It asks for the work done ON the gas. The reasons its W = P(Vf-Vi) and not W = -P(Vf-Vi) is because its work ON gas not BY the gas. In U = Q + W, W is work on gas presumably, though I'm not 100% sure.
  5. Oct 3, 2012 #4

    Andrew Mason

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    This is important. Positive work is done BY the gas when it expands. The work done BY the gas is PΔV where ΔV>0. Positive work is done ON the gas when it is compressed. So the work done ON the gas is -PΔV.

    So, in the expression U = Q+W, the W is work done ___ the gas.(hint: if W is positive is there an expansion or compression?).

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