Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Reversible adiabatic expansion of ideal gas, entropy change?

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data
    One mole of an ideal gas at 0 celcius is subjected to changes below, calculate the change in entropy of the gas:

    i) Gas is expanded reversibly and isothermally to twice its initial volume. DONE - 5.76 J/K
    ii)A similar expansion to i. is performed reversibly and adiabatically [hint, for an ideeal gas undergoing reversible adiabatic expansion TV^(gamma - 1) = constant]


    2. Relevant equations

    I know delta Q = -nRT Ln (Vf/Vi) for a reversible isothermal expansion

    delta S = delta Q / absolute T

    dE = dW + dQ

    3. The attempt at a solution

    for a isolated system i know the entropy change would be zero, but as it doesnt state this i dont feel that i can just state "assuming system is isolated entropy change is zero"
    so im stuck how to go about this..

    Khris
     
  2. jcsd
  3. Jan 31, 2010 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    The entropy change of the gas is not zero.

    Work it out (Hint: you don't actually have to do any integration).

    [tex]\Delta S = \int_1^2 dQ/T + \int_2^3 dQ/T[/tex]

    AM
     
  4. Jan 31, 2010 #3
    i know
    [tex]
    \Delta S = \int_1^2 dQ/T + \int_2^3 dQ/T
    [/tex]
    but dQ is zero for an adiabatic process isnt it?

    Khris
     
  5. Nov 15, 2010 #4

    bon

    User Avatar

    Sorry to bring this up again, but doing a similar question...

    Surely here dQ = 0 since the gas is expanded adiabatically?
     
  6. Nov 16, 2010 #5

    bon

    User Avatar

    heloooo? *bump*
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook