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Thermodynamics atmospheric pressure Question

  1. Oct 19, 2014 #1
    1. The problem statement, all variables and given/known data
    A liter of air, initially at room temperature and atmospheric pressure, is heated at constant pressure until it doubles in volume. Calculate the increase in its entropy during this process.

    so Ti= 300K, Volume which is 2Vi=Vf; Pressure is constant
    2. Relevant equations
    ΔS @constant pressure is = ∫Cp/T dT (where Ti -> Tf

    3. The attempt at a solution
    so since I don't know what T is but I know how the volume changes is there a way I can relate it? is it just PV=nkT so use V as an analog to gauge the proportional change in T? so would that give me something like Cp*Ln[2]?
     
  2. jcsd
  3. Oct 19, 2014 #2

    Simon Bridge

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    But you do know what T is - it's the temperature. You mean you are not told what the final temperature is?
    Then - yes - you would need to draw information in from another source - the equation you are using for entropy change is for an ideal gas, so it seems reasonable to use the ideal gas law to relate temperature and volume. Unless you have some notes about how air behaves?

    http://www.thebigger.com/chemistry/...anges-of-an-ideal-gas-in-different-processes/
     
  4. Oct 19, 2014 #3
    Yeah, sorry; the final T is unknown. So do I write it as Ln[Tf/Ti]? where Tf= 2PVi/Nk ? how do I solve for N?
     
  5. Oct 19, 2014 #4
    OK, I think I figured it out. I got it to be Tf=600K so it becomes 5/2*R*Ln(600K/300K)
     
  6. Oct 19, 2014 #5

    rude man

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    What happened to n? Your answer is for 1 mole, not 1 liter.
     
  7. Oct 19, 2014 #6
    wouldn't n be the same for both sides of the ideal gas law, which would lead them to cancel out?
     
  8. Oct 19, 2014 #7
    I also never had n, so I assumed you hold is constant, which would still give me the same result, since all other variables are held constant. Unless there's something else I'm missing.
     
  9. Oct 20, 2014 #8

    Simon Bridge

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    At constant pressure $$\frac{PV_f=nRT_f}{PV_i=nRT_i}\implies \frac{V_f}{V_i} = \frac{T_f}{T_i}$$ ... it's a law that got named after someone.
    It means you don't actually have to calculate the temperatures here if you know the volumes.


    Of course you can also look up the molar density of air "at room temperature and atmospheric pressure".
     
  10. Oct 20, 2014 #9

    rude man

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    Does it make sense to you that the change in entropy of 1 mole of gas is the same as the change in 1 liter? Your formula gave the change for 1 mole. So figure out how many moles in 1 liter of air at STP.

    Your formula for ΔS = Cp ln(T2/T1) is correct. But Cp = ncp and you only hav cp. BTW air is essentially a diatomic gas for which cp ~ (7/2)R.
     
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