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!

An entropy change question?

  1. Aug 3, 2010 #1
    1. The problem statement, all variables and given/known data

    One mole of a monatomic gas first expands isothermally, then contracts adiabatically. The final pressure and volume of the gas are twice its initial pressure and volume. (Pf=2Pi and Vf=2Vi)

    Find the net change in the entropy of the gas.

    2. Relevant equations

    3. The attempt at a solution

    I found the entropy change for the first step. Let the volume between the two steps be Vm.

    TVγ-1 = constant, so Vm = 22.5 Vi

    ΔS = nR ln(Vm/Vi) = 1.73 nR

    But I have no idea about the entropy change of the second step. What is the entropy change for an adiabatic process? Isn't dQ=0 ?
  2. jcsd
  3. Aug 3, 2010 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    There is no entropy change in a reversible adiabatic process, so you just have to determine the change in entropy of the isothermal process. You appear to have tried doing this by applying the adiabatic condition (after finding the final temperature) to get the volume at the end of the isothermal process, but I can't figure out how you get your answer. If you apply the adiabatic condition:

    [tex]\left(\frac{V_m}{V_f}\right)^{\gamma-1} = \frac{T_f}{T_m} = 4[/tex]

  4. Aug 3, 2010 #3

    Pi Vi Ti ---> Punknown Vm Ti


    Punknown Vm Ti ---> 2Pi 2Vi 2Ti

    So, for the adiabatic process I applied TVγ-1 = constant.

    Ti Vm2/3 = 2Ti (2Vi)2/3

    Isn't this correct?

    Another thing, is the change in entropy ΔS zero for adiabatic processes?
  5. Aug 3, 2010 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    It is not correct. What is the final temperature? (hint: apply the ideal gas law). The final temperature is not twice the initial temperature.

    Yes. See my previous post.

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