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Well insulated tank with valve entropy problem

  1. Apr 6, 2017 #1
    1. The problem statement, all variables and given/known data
    A well-insulated tank of volume 6 m3 is divided into two equal volumes. The left part is initially filled with air at 100 C and 2 bar, and right side cell is initially empty. A valve connecting two cells will be opened so that gas will slowly pass from cell 1 to cell 2. The wall connecting the two cells conducts heat sufficiently well that the temperature of gas in both cells will always be the same. Plot on the same graph a) the pressure in the second tank versus the pressure in the first tank, and b) the change in the total entropy of the system versus pressure in the tank 1. Do these calculations with the increments of 5% of the gas transferred until all gas is transferred to tank 2. At these temperatures and pressures air can be considered as an ideal gas with constant heat capacity. Before plotting the total entropy change what do you think final pressure will be? Also will the temperature change? Since entropy increases for any natural process, what do you think entropy will be at the equilibrium pressure (minimum or maximum)?


    2. Relevant equations
    PV=nRT
    entropy Balance


    3. The attempt at a solution
    I know that for a valve it is isenthalpic so the enthalpy that that goes from partition1 into partition2 will be the same. I also know that the second pressure or the maximum amount of pressure that would be released into partition2 would be 1 bar as it would have to find equilibrium with that of the pressure coming from partition1 at some point. I know the final temperature would be the same again because it is isenthalpic and "well insulated".

    I just am confused on how they want me to approach this problem. It says to do the increments in 5% of the gas transferred. Are they referring to the mols ? And what about the entropy wouldn't it continuously increase seeing that this partition is being filled?


    here is my attached attempt at a solution.

    ThermoHW7_12.jpg ThermoHW7_11.jpg ThermoHW7_13.jpg

    Best Regards,

    D
     
    Last edited: Apr 6, 2017
  2. jcsd
  3. Apr 7, 2017 #2
    I think you meant constant internal energy, rather than isenthalpic (although, in this case, because the internal energy is constant, the enthalpy is also constant).
    Yes.
    Entropy would increase from the initial state to the final state because, for this spontaneous process, entropy is being generated within the system, while no entropy exchange is taking place with the surroundings.

    On the first sheet, you are supposed to assume 5% increments in mass, not increments in time. And you are supposed to plot the pressure in each chamber as a function of the percentage mass transferred.

    On the second sheet, you did not allow for the heat transfer between the two chambers or for the change in internal energy within each chamber. To do part (a) correctly, you know the number of moles in each chamber as the process progresses, and you know that the temperature in both chambers is equal to the initial temperature. So you can use the ideal gas law to determine the pressure in each chamber as a function of the number of moles transferred.

    The third sheet does not quantitatively address part (b). You need to determine the entropy change as a function of the initial pressure in chamber #1. You know the initial state of the gas, and you know the final state (half the initial pressure, double the initial volume, at the same temperature as initially). Do you know how to determine the entropy change of an ideal gas quantitatively from this initial state to this final state?
     
  4. Apr 9, 2017 #3
    First I'd like to thank you for your clarification and detailed response to the problem.

    To answer your question if I know how to determine the entropy from initial state to final state.

    I am pretty sure I would use the equation

    ΔS=∫(Cp/T)dT-Rln(P2/P1)

    I know that since the process doesn't have a change in temperature that my ∫(Cp/T)dT will zero out and I will be left with

    ΔS=-Rln(P2/P1)

    I know that since my moles being transferred is the only thing changing.and that the volumes of both partitions are the same. So I believe V1=V2 since the second partition is just being filled. So I can have

    PV=nRT Solving for the quantities mols then multiply by 5% to find the increments of transfer.

    I also know that V1=V2 so my PV=nRT==>( PV/n)=RT=Constant since temperature is not changing

    becomes P1/n1=P2/n2

    This will give me my increments of pressure.



     
  5. Apr 9, 2017 #4
    Looks good.
     
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