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!

Monatomic gas, Isochoric Process.

  1. Dec 16, 2012 #1
    1. The problem statement, all variables and given/known data

    n = 1.46 moles of ideal gas are heated isochorically (at constant volume) from tepmerature To = 649 oC to temperature Tf = 1184 oC. Find:

    a) Work done on the system.
    b) Change in Internal Energy of the system.
    c) The total heat, Q, added or removed from the system.

    2. Relevant equations
    Change in I.E. = Q + W
    KE (avg) = T = 3/2kT, where k = 1.38 x 10^-26 kJ/K

    3. The attempt at a solution

    a) Since this is an isochoric system, work done is ZERO! (No issues here.)
    b,c) Change in Internal Energy, therefore is equal to Q, by the first equation, Change in IE = Q + W.
    So, we can calculate change in IE, by change in KE.

    KE = (3/2kT) for a monatomic gas.
    so, change in KE = 3/2(k)(Tfinal - Tinitial), = (3/2)(1.38E-26)(1500 K - 704K) = change in IE = Q = 1.65 x 10^-23.

    Am I reading the problem incorrectly? Many thanks in advance once again!
     
  2. jcsd
  3. Dec 16, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    (3/2)kT is the average translational KE of just one molecule.
     
  4. Dec 16, 2012 #3

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It appears you have calculated the change in KE for a single molecule of the gas.
    Remember, you have 1.46 moles of gas to start with.
     
  5. Dec 16, 2012 #4
    Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Monatomic gas, Isochoric Process.
  1. Monatomic Gas (Replies: 2)

  2. Isochoric process (Replies: 2)

Loading...