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Adiabatic process, calculating final T, enthalpy etc

  1. Apr 7, 2017 #1
    1. The problem statement, all variables and given/known data

    Find the final temperature, Q, ΔU, ΔH given the following

    Initial state of gas
    Ti = 353K
    Pi = 250000Pa
    2.5mols of gas
    Cv = 12.47Jmol-1

    Final pressure = 125000Pa

    2. Relevant equations
    PV = nRT
    W = -PΔV
    ΔH = ΔU + Δ(PV)
    PVγ = constant

    3. The attempt at a solution
    Cv / R ≈3/2 so monatomic => γ = 5/2
    PV=nRT => V1 = (2.5)(8.314)(325)/(250000) = 0.027 m3
    using PVγ = constant => V2 = (P1 / P2 *Vγ )1/γ = (2.5/1.25)2/5 * (0.027)= 0.0356 m3

    so Tfinal = P2 *V2 /(n*R) = [(125000*0.0356)/(2.5*8.314)] = 214K
    Q = 0 adiabatic
    W = ? nCv ΔT = 2.5*12.47*1(-139) = -4333J?
    I get a different answer if I use W = PVγ * (V21-γ - V11-γ)/(1-γ) = 1527
    ΔU = W = -4333J
    H = H = ΔU + Δ(PV), should this be = ΔU + nRΔT or = ΔU + PΔV + VΔP? They dont give me the same answer

    nRΔT = 2785J, PΔV + VΔP = 5525J but I used P = 125000Pa and V = 0.0356, i dont know if this is the right formula let alone the right parameters to enter.

    I didnt even use the fact that the external pressure was 1bar either.

    All round confused with this part of my thermal course and just applied random formulas so I think I botched it.
    PVγ = constant is something from another course I did and wasn't in my notes so is it the correct route or maybe an there's an easier method? Any confirmation/corrections would be much appreciated.

    Edit: made numerical errors that I corrected
    Last edited: Apr 7, 2017
  2. jcsd
  3. Apr 7, 2017 #2
    Would it be possible for you to provide the exact problem statement?
  4. Apr 7, 2017 #3

    this is an image of it, it was a past exam question. http://i.imgur.com/KT8XOOx.png
    sorry I definitely wasn't clear enough in my initial post, I see that now
  5. Apr 7, 2017 #4
    edit: sorry posted same comment twice..
  6. Apr 7, 2017 #5
    OK. Thanks. That helps.

    There is a little bit of trickiness to this problem. Here's what happens:

    To start with, the (massless, frictionless) piston is sitting on top of the gas. The pressure outside the cylinder is 1 bar, and the pressure inside the cylinder is 2.5 bars. So the piston has to be held in place. At time zero, the piston is released, and the gas expands against a constant external pressure of ##P_{ext}=1## bar. But, the gas is not allowed to fully expand to match the external pressure of 1 bar. Instead, at some final volume ##V_f## the piston is again constrained. Once this happens, the gas re-equilibrates uniformly to a new final pressure of 1.5 bars.

    So the volume change occurs at a constant applied external pressure of 1 bar (which matches the local gas pressure on the gas side of the piston face). So the work done by the gas on its surroundings during this process is ##W=P_{ext}(V_f-V_i)##

    ##V_i## is known from the ideal gas law, but, since the final temperature ##T_f## is unknown, the final pressure ##V_f## is also unknown. So there are two unknowns in this problem. These two unknowns must be determined by combining the equation for the first law of thermodynamics with the ideal gas law. Please (algegraically) write down the equation for the first law of thermodynamics that would apply to this system, in terms of n, ##C_v##, ##T_i##, ##T_f##, ##V_i##, ##V_f##, and ##P_{ext}##.
  7. Apr 7, 2017 #6
    Thanks for explaining whats actually physically going on, I had no idea.

    OK so,
    ΔU = Q - W, adiabatic => Q = 0
    ΔU = -W

    ΔU = nCVΔT = -W = -Pext(Vf - Vi)

    so nCv(Tf - Ti) = -Pext(Vf - Vi)

    Would that be correct?
  8. Apr 7, 2017 #7
    Yes. Good job. Now for applying the ideal gas law.

    Algebraically, from the ideal gas law, what is the initial volume ##V_i## in terms of ##T_i##, ##P_i##, and n?
    Algebraically, from the ideal gas law, what is the final volume ##V_f## in terms of ##T_f##, ##P_f##, and n (where ##P_f## is the 1.25 bars)?
    If you algebraically substitute these volume equations into your result from the ideal gas law, algebraically, what do you get?
  9. Apr 8, 2017 #8
    Thanks so much for the help! Sorry, awkwardness of timezones means this late reply.

    OK so,

    Vi = nRTi/Pi = (2.5*8.314*325)/(250000) ≈ 0.027 m3
    Vf = nRTf/Pf = (2.5*8.314*Tf)/(125000) ≈ 1.66*10-4*Tf

    and so using nCv(Tf - Ti) = -Pext(Vf - Vi)

    2.5*12.47*(Tf - 353) = -100000(1.66*10-4*Tf - 0.027)

    31.18 Tf + 16.6 Tf = 11005 + 2700

    ≈47.8 Tf = 13705 => Tf ≈ 289K

    so barring mistakes
    a) Tf = 289K

    b) Q = 0

    c) W, work = PextΔV

    Vf = nRTf/Pf = 2.5*8.314*289/125000 ≈ 0.048 m3
    so W = PextΔV = 100000*(0.048 - 0.027) = 2100 J

    d) ΔU = just minus the work

    e) ΔH = ΔU + Δ(PV) I'm unsure how to use this formula in terms of the Δ(PV), should I use product rule or just nRΔT
  10. Apr 8, 2017 #9
    You can use ##(P_fV_f-P_iV_i)## of ##nR\Delta T## or ##\Delta H=nC_p\Delta T##. They should all give the same result.
  11. Apr 8, 2017 #10
    Ah yes they do, thanks a lot much appretiated.

    Out of interest, is TVγ-1 = constant applicable to this question? When I tried just there for T1, T2 and V1 to find V2 it didnt work, I got V2 = 0.0365 m3
  12. Apr 8, 2017 #11
    That equation is only valid for an adiabatic reversible process. This uncontrolled expansion against a constant pressure is a spontaneous irreversible process.
  13. Apr 8, 2017 #12
    Thanks for helping my understanding, it's all a lot clearer now. Cheers for your time.
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