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Relatively simple question regarding Thermo (entropy and heating)

  1. Feb 18, 2008 #1
    Problem 1:

    1. The problem statement, all variables and given/known data
    An ideal gas is heated reversibly at constant volume from T1 to T2. Calculate the entropy change
    of the system, the surroundings, and the universe.


    2. Relevant equations
    dS = dq/T (reversible change)
    and dq = CvdT (constant V)

    3. The attempt at a solution
    I know how to do the system:
    dS = dq/T, so you can substitute the heat capacity for dq to get
    dS = CvdT/T and integrate to get dS of the system to be Cv * ln(T2/T1)

    Here is where I'm not sure...
    Wouldn't the entropy of the surroundings just be the negative of that? I figure because it's a reversible process the entropy change of the universe is 0...correct? Please correct me on that. Also, is there a better expression or more intuitive solution for the entropy of the surroundings?

    Problem 2:

    1. The problem statement, all variables and given/known data
    For a particular change of state, why is qrev>qirr? That is, why is heat flow greater for a reversible
    change of state rather than when the same change of state is brought about irreversibly?


    2. Relevant equations
    None that I know of.


    3. The attempt at a solution
    Is it because in an irreversible process, heat is often dissipated away in other forms such as work, noise, etc...? Because the system is not in equilibrium for an irr process the heat flow is not always conserved. Is this correct and is there a better explanation for this problem?

    Thanks!
     
  2. jcsd
  3. Feb 18, 2008 #2
    I think you are right in both problems.
    For problem 2, you might be able to add some more quantitative arguments in terms of the second law.
     
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