1. Not finding help here? Sign up for a free 30min 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!

Work done by gas

  1. Dec 4, 2006 #1
    Hello! Any ideas how I could solve this exercise:

    5 mol of gas is in temperature of 326K and the same time the pressure is expanded from 1atm to 3atm. How much work is done by the gas?

    Detailed instruction would be good, thanks :)
  2. jcsd
  3. Dec 4, 2006 #2
    W/mole = dPV = Pdv + Vdp.

    it should be pretty obvious from here.
  4. Dec 4, 2006 #3
    Yes, it probably should, but I'm very dumb with physics :P I already tried to find some help from Google but I didn't have any luck :/
  5. Dec 4, 2006 #4
    okay, I assume you are in like a freshman/sophmore level physics course right.

    Let's assume an ideal gas.

    we know the standard molar volume right? (22.414 L/mole). so v=nRT/P.
    We are assuming that the size of our container is constant. (isochoric condition)
  6. Dec 4, 2006 #5
    Oh, yes it's ideal gas. I forget to mention that :)

    Do I need to calculate v1= nRT/1atm and then v2= nRT/3atm?
  7. Dec 4, 2006 #6

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    The question does not have sufficient information. You have to know how the pressure is changed and whether heat is being added or released.

    If the external pressure is increased, the volume must decrease, in which case the work done by the gas is negative. If pressure is increased because heat is added, the volume can be held constant, in which case, no work is done. Or it could increase, in which case the work done by the gas is positive.

    Work = Pdv not [itex]\Delta (PV)[/itex]

    If you assume that the external pressure is gradually changed from 1 atm to 3 atm, and the temperature is kept constant, the work done is:

    [tex]W = \int_{P_i}^{P_f} PdV = \int_{P_i}^{P_f} nRTdV/V = nRT\ln(\frac{V_f}{V_i}) = -nRT\ln(\frac{V_i}{V_f})[/tex]

    If it is adiabatic (temperature will increase) it is more complicated.

  8. Dec 5, 2006 #7
    Thanks Andrew! That works great! :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Work done by gas
  1. Work done by a gas (Replies: 6)