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!

Free expansion of an ideal gas.

  1. Sep 30, 2009 #1
    I know that the work done on the system in any free expansion is 0 since the external pressure is 0. However.. is q necessarily 0? Does the temperature necessarily stay constant for an ideal gas?

    I've been trying to justify the reasons for which q is necessarily 0, but I can't find a reason... Does anybody know why this is so? I know that U varies only with T for an ideal gas, but that would require justifying that the change in temperature is also 0.. does anybody have a mathematical justification for why q is necessarily 0 for a free expansion of an ideal gas?

    Thanks.
     
  2. jcsd
  3. Oct 1, 2009 #2

    Mapes

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The problem is usually described as adiabatic free expansion of an ideal gas to address this issue.
     
  4. Oct 1, 2009 #3
    It's the definition of an adiabatic course
     
  5. Oct 1, 2009 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Apply the first law. Does the gas do work? Does heat flow into or out of the gas from/to the surroundings? What does that tell you about internal energy of the gas?

    AM
     
  6. Oct 1, 2009 #5
    The gas does no work, but heat could, in theory, flow in or out of the gas. I don't see why it can't... The problem doesn't specifically say that the process is adiabatic, but... I guess it is.. Obviously, that would solve all of the problems.

    Thanks everybody!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Free expansion of an ideal gas.
  1. Free Expansion of gas (Replies: 22)

Loading...