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Entropy change in isochoric process

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi everyone

    In an isochoric process ( a solid heat body with T1 temp. connected with heat bath with T2 temp)
    we could calculate its entropy change(heat body) from dS = Cv/T dT
    and entropy change of heat bath from dS = dQ / T2, using dQ = Cv dT in 1st equation
    ( i read these from notes actually but i am confused with the reversibility in it)

    2. Relevant equations
    but I remember that we have proved before

    dS >= dQ/T for all process, and equality hold when process is reversible(like isotherm and adiabats)


    3. The attempt at a solution

    Then i compute the entropy change of heat body in isochoric system = nR ln(T2/T1)
    and entropy change of heat bath = - Cv(T2-T1)/T2
    so total entropy is not zero when T2 and T1 has a finite change.
    There comes my question, we find out that this isochoric process is irreversible
    but when we use the equation dS = dQ/T, we could only use it in reversible process.

    I try to solve this problem by constructing 1 isotherm and 1 adiabat to replace original isochoric path and find out that dS actually = dQ / T = Cv/T dT ,
    although i know where this come from ( dS = dQ /T) from maths way
    but I think i misunderstand some physical concept
    can someone explain why dS = dQ/T , even when this process is not reversible in some physical way?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 6, 2013 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    The process is not reversible because you cannot change the direction of the process (ie. the direction of heat flow) by making an infinitesimal change in conditions.

    But to find the change in entropy of each body, you only look at the initial and final states - you do not care about the process that occurred in going from the initial to final states. You just have to find a reversible path for each component (ie the solid body and the bath). Since the actual process was NOT reversible, the reversible paths will not be the same path for each component.

    Here you are assuming that the bath has an arbitrarily large heat capacity so its temperature before and after the process is the same. The reversible path would simply be a reversible isothermal heat flow out of the heat bath of Q, where Q is the amount of heat flow into the solid body. So for the bath, ΔSbath=Q/T

    The solid body changes temperature from T1 to T2. A reversible process for this would be to successively put it in contact in series with an arbitrarily large number of heat baths - each of which are at an infinitesimally higher temperature than the body - until it reaches T2. Each step would result in a heat flow dQ = mCdT. You just have to work out the expression for dS and integrate from T1 to T2 to get ΔSbody.

    AM
     
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