1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ideal gas problem

  1. Jul 26, 2004 #1
    Okay, I've been working on this for a while:

    Three-tenths mole of an ideal gas at 400K is expanded isothermally from a pressure of 5x10^5 N/m^2 to 1.5x10^5N/m^2. If the gas is then heated to 500K at constant volume and then is compressed isothermally back to 5x10^5 N/m^2, and then isobarically back to the original condition, find the net work of the cycle.

    I understand the whole concept of expanding and contracting, but combining the ways that these happen is very confusing to me.

    first i found the volume for the first part using the formula V = nRT/P.
    so for the first part, when its expanded isothermally i used the formula:
    w = nRTln(V2/V1)
    w= (3/10)*(8.314)*(400)*ln(6.65*10^-3/2*10^-3)
    = 1198.68

    same for the second isothermic expansion, except the temperature changed to 500, and therefrore the volume changed and:

    w = -1503.58

    then for the third part, i used the formula for isobaric gas:

    w = P(V2 - V1)
    = 245

    Then for the net work, I added everything together, and got -364.8

    What am I missing here??

  2. jcsd
  3. Jul 26, 2004 #2


    User Avatar
    Science Advisor

    If you have a set temprtature and pressure, the volume is
    also determined for an equal amount of gas. So, if at the
    end you get the original condition - all of the same parameters,
    what kind of energy difference = work, are you looking for ?
  4. Jul 26, 2004 #3
    it just says find the net work. I stated the problem exactly as it is written. Grr...this one is driving me nuts!
  5. Jul 26, 2004 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good. (You are defining work as work done by the gas on the environment. So + work means that energy is leaving the gas. That's OK. Just be aware that some define work as work done on the gas.)

    Take a second look at this one. I suspect you are using the wrong final volume.
  6. Jul 26, 2004 #5
    daisyi - if you add up all your work terms again, you'll see it will not give you -365. The calculation for all the individual work terms are correct, except for some rounding errors. When I did the calcuations I get: 1201.18 - 1501.47 + 249.42 = -50.87 or -50 after rounding.
  7. Jul 27, 2004 #6

    Doc Al

    User Avatar

    Staff: Mentor

    one more thing...

    I never got to this step in my previous post:
    It's an isobaric compression: the work will be negative.

    And yes, as PureEnergy points out, you did add your terms incorrectly. But that's not the real problem. :smile:
  8. Jul 27, 2004 #7
    I double checked and it still looks right to me. This is how I did it.

    V = nRT/P

    so, when the gas is heated to 500k with constant volume, the volume is:

    V1 = (3/10)(8.314)(500)/1.5x10^5
    = 8.314x10^-3

    and then when the pressure is compressed back to 5x10^5:

    V2= (3/10)(8.314)(500)/5*10^5
    = 2.49x10^3

    I then plugged that into the equation w = nRTln(v2/v1) to get the 1198.68
  9. Jul 27, 2004 #8

    Doc Al

    User Avatar

    Staff: Mentor


    No. V1 = (3/10)(8.314)(400)/1.5x10^5 (You know the temp and pressure before it's heated.)

  10. Jul 27, 2004 #9
    Doc's right, the third step should be negative so the net work is around -550.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook