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Reversible Process

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the final temperature of a sample of argon of mass 12.0 g that is expanded
    reversibly and adiabatically from 1.0 L at 273.15 K to 3.0 L.


    2. Relevant equations



    3. The attempt at a solution

    n = 12[g]/40[g][mol]-1 = 0.3 [mol]

    pVi=nRTi

    p = 0.3[mol]0.082[L][atm][K]-1[mol]-1/1[L] = 6.7 [atm]

    Tf = pVf/nR = 6.7[atm]3[L]/0.3[mol]0.082[L][atm][K]-1[mol]-1 = 817.07 [K]

    What is wrong? What have I overlooked?
    I think it has something to do with the term 'reversibly', but how?
     
  2. jcsd
  3. Oct 11, 2009 #2

    ideasrule

    User Avatar
    Homework Helper

    How can that be correct? You haven't multiplied in the temperature.

    To solve this problem, you need to use an equation that's specifically meant for adiabatic expansion/contraction. Do you know the equation?
     
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