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Work and Volume in Adiabatic Expansion

  1. Nov 19, 2017 #1
    1. The problem statement, all variables and given/known data
    A sample of 1.60 mol H2 (Cv = 20.5 J K-1 mol-1) at 21°C and 1.50 atm undergoes a reversible adiabatic compression until the final pressure is 4.50 atm. Calculate the final volume of the gas sample and the work associated with this process. Assume that the gas behaves ideally.

    2. Relevant equations

    3. The attempt at a solution
    pv=nRT → (1.5 atm)(V0=(1.6 mol)(0.08206 atm mol K)(294 K)
    V0=25.734016 L (tried this initially as I was given Cv, but that was wrong)
    P1V1=P2V2→(1.5 atm)(25.734016 L)=(4.5 atm)(V) →8.57800533 L (which was wrong)

    ΔE=-(4.5)(17.1501067)= -77.202 J (nope)
    ΔE=(1.6 mol)(20.5 K mol)(294 K)= 9643.2 J (Nope)

    I'm not really sure where I went wrong with this, any input would be really helpful
  2. jcsd
  3. Nov 19, 2017 #2
    Is the process isothermal, or is it adiabatic?
  4. Nov 19, 2017 #3
  5. Nov 19, 2017 #4
    For an adiabatic reversible process, is this equation correct: p1v1=p2v2 ?
  6. Nov 19, 2017 #5
    That's what I wrote in my notes at least. I suppose not though?
  7. Nov 19, 2017 #6
    You suppose correctly. Do you know the correct equation? It can be derived by applying the 1st law of thermodynamics to this adiabatic reversible compression. What would the 1st law give you for this?
  8. Nov 19, 2017 #7
    The first law of thermodynamics would be conservation of energy, which would be Internal Energy=Q+W. Since it's an adiabatic system, Energy would just be equal to work. E=p deltaV and nCv delta T.... Right?
  9. Nov 19, 2017 #8
    Yes! Except that, because the process is reversible, those deltas should be d’s. That is, differentials. And there should be a minus sign. Now substitute p=nRT/V into the equation. What does that give you?
  10. Nov 19, 2017 #9
    Do you mean the volume value I calculated before or the actual equation itself?
    E= -(4.5 atm) (d/dv 25.734016 L) =....
    E= nRT/v x d/dv V -> E= nRT d/dv
    (I'm sorry I lost momentum here I'm unsure what to do.)
  11. Nov 19, 2017 #10
    What I'm saying is $$nC_vdT=-\frac{nRT}{V}dV$$ Does this equation make sense to you?
  12. Nov 19, 2017 #11
    I thought I did but based on what I came up with it seems I've shamed my calculus teacher

    ∫nCvdt=-∫nRT/v dV → nCvT1-nCvT2=-(nRT)ln(V1)-(nRTln(V2)
    So if I plug what I know
    (1.6 mol)(20.5 J K-1 mol-1)(294K) - (1.6 mol)(20.5 J K mol)(T2)=-[(1.6)(8.315)(294)ln(25.734016 L)-(1.6)(8.315)(294)(lnV2)]

    -3060.220545=(1.6 mol)(20.5 J K mol)(T) - (319.296)(ln (25.734016 L/ V2)

    So I'm assuming this is super wrong?
  13. Nov 19, 2017 #12
    This is not integrated correctly. And, please stop substituting numbers. Do it all algebraically.
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