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How Do I Determine the Change in Entropy of the System?

  1. Feb 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine ΔSsys when 3.0 mol of an ideal gas at 25°C and 1 atm is heatedd to 125°C and expanded to 5 atm. Rationalize the sign of ΔSsys.

    2. Relevant equations
    State Function: dS = (dU)/T + (PdV)/T
    State Function for Entropy of Ideal Gas: dS = (CV,mdT)/T + (nRdV)/V

    Ideal gas law: PV = nRT
    Boyle's Law (I believe this assumes constant temperature): P1V1 = P2V2

    3. The attempt at a solution
    My professor actually worked out this problem for us, but I had a few questions about the solution...

    Entropy problems are always solved reversibly because the formula for this is a state function. Thus, I drew a graph that would hold one of the variables constant while I solved for entropy in regard to changing one variable at a time.

    Path 1: Constant temperature, pressure goes from 1 atm to 5 atm.

    dS = (CV,mdT)/T + (nRdV)/V
    Since temperature is constant, the (CV,mdT)/T term equals zero.

    dS = (nRdV)/V
    ΔS = nRln(Vf/Vi)
    P1V1 = P2V2
    (1atm)Vi = (5atm)Vf
    Vf/Vi = 1/5
    ΔS = (3)(8.31)ln(1/5)
    While Boyle's law in general assumes temperature to be constant and during this part of the path from initial state to final state temperature is indeed constant, but temperature throughout this process isn't constant, and PV = nRT can also be used to solve for the initial and final volumes to plug into this problem, but when I do this, my volume term is then off by a factor of the difference in temperature and this is not made up for later by using CV,m in path 2. Additionally, there is another homework problem I have where temperature is held constant for part of the pathway, but I have to use PV = nRT to find the correct volume, and if I use Boyle's law to find it, my state function answer for entropy will be wrong. How do I know when to use Boyle's law and when to use the ideal gas law?

    Path 2: Constant pressure, temperature goes from 298K to 398K.
    dS = (CP,mdT)/T + (nRdV)/V
    ΔS = CP,mln(Tf/Ti)
    ΔS = n(5/2)Rln(Tf/Ti)
    ΔS = (3)(5/2)(8.31)ln(398/298)
    Why does the (nRdV)/V term disappear? Just because pressure is constant doesn't mean volume is constant, or at least that's what I thought.
    Why are we allowed to change the specific heat variable just because pressure is constant for a single part of this pathway? Again, in a third homework problem, pressure is constant for a single part of a pathway, but we can't change from CV to CP then because we "have to remember that this term was derived from dU, which is a state function in terms of CV." Why don't we have to remember that for this problem?

    In the end the entropies from path 1 and path 2 are summed together to get the final change in entropy.
  2. jcsd
  3. Feb 28, 2015 #2
    You are right, If pressure is constant and you change temperature, then volume changes... Look at the ideal gas equation:


    V/n =Vm (molar volume)
    ... now lets just take all of the constants and make them = c
    V is proportional to T if p is constant

    (therefore as T increases, so does V)

    So you should discuss this with your professor because I believe that term should have stayed.
  4. Feb 28, 2015 #3
    In Path 2, you should be using constant volume not constant pressure. That will alleviate your issue. Also, your equation for dS involving Cp is incorrect. If should read: ##dS=nC_pdlnT-nRdlnP##

  5. Mar 1, 2015 #4


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    Both T and the pressure P change during the process, so you should rather consider the entropy as function of T and P instead of T and V. Use the Ideal Gas Law and find dV in terms of dT and dV : PV=nRT --->
    ##PdV+Vdp=nRdT \rightarrow dV= \frac{nRdT-VdP}{P}##. Substituting back into the previous equation for dS

    ##dS=C_v n \frac{dT}{T}+\frac {nR}{V}\left( \frac{nRdT-VdP}{P} \right)=(C_v +R)n \frac{dT}{T}-nR\frac {dP}{P}##
    Cv+R=Cp so

    ##dS=C_p n \frac {dT}{T}-nR\frac {dP}{P}##.

    This is the same equation Chestermiller wrote, but in a more readable form.
    You can integrate the equation along any path between the initial and final values of the temperature and pressure one part with p=const, from Ti to Tf and the other with T=const, from Pi to Ps.
    Last edited: Mar 1, 2015
  6. Mar 1, 2015 #5


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    No, you have the the differential dS= (nCv/T) dT -(nR/V)dV of the function S(T,V) function of the temperature and the volume.

    Integrate it between Vi and Vf , keeping the temperature constant.

    It is right up to ΔS = nRln(Vf/Vi). You have to plug in the initial and final volumes. Not those, along the integration part, but the real ones.
    Vi=nRT/Pi; Vf=nRT/Pf.

    During the integration, you get the form of the entropy change in terms of Vi and Vf. But Vi and Vf come from the Ideal Gas Law PV=nRT.

    NO, that is wrong. S is function of T and V, Path 2 is integral for constant volume, the integral of dS=(nCv/T)dT+(nR/V) dV.

    At the end, the whole change of entropy is ΔS is expressed in terms of Vi and Vf, Ti and Tf.
    It does not matter, what was the integration path.
    Determine Vi and Vf from PV=nRT from (Pi, Ti), and (Pf, Tf) and plug in the values into ΔS.

    You get the same value when you consider the entropy as function of T and P, dS=(nCp/T )dT - (nR /P)dP and integrate along Path1 with respect to P at constant temperature, and along Path2 with respect to T at constant pressure.
    Last edited: Mar 1, 2015
  7. Mar 2, 2015 #6
    I think what Ehild is saying (please correct me if I'm wrong) is that the entropy change between the initial and final states of the system is exactly the same no matter which reversible path you choose to use to evaluate the change. You (Gwazdzilla) chose to use the exact path followed by the system, involving a constant-volume and a constant-pressure segment. But an infinite number of other reversible paths are also possible. All of them give exactly the same value for the entropy change.

  8. Mar 2, 2015 #7


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    You are right, the change of a state function between two states is the same, no matter, how the function has been obtained.

    Godzilla tried to integrate dS along a constant-temperature segment and after a constant pressure segment, but used the wrong equation and got confused.
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