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Isobaric/isochoric (?) heating of an ideal gas

  1. Oct 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider an ideal gas with [itex]C_V=6.76 \frac{cal}{mol \cdot K}[/itex]. Calculate [itex]\Delta H[/itex] and [itex]\Delta U[/itex] when ten moles of this gas are heated from 273.15 K to 373.15 K.

    2. Relevant equations
    [tex]\Delta H = \Delta U + P\Delta V[/tex]
    [tex]Q=n C_V \Delta T[/tex]

    3. The attempt at a solution
    As I'm given the heat capacity at constant volume I'm assuming this is an isochoric process. That means [itex]W=0[/itex]. Therefore, [itex]\Delta U=Q[/itex].
    [tex]\Delta U=Q= (10 \ mol) \left(6.76 \frac{cal}{mol \cdot K} \right) (100 \ K) = 6760 \ cal[/tex]
    Now, for the change in enthalpy we have:
    [tex]\Delta H = 6760 \ cal + P\Delta V[/tex]
    This is where I'm having trouble. Should I cancel the second term in the above equation? And so have: [itex]\Delta H=\Delta U[/itex]. But, if the gas is heated at constant volume pressure should increase, that means I should consider the PV term in the last equation. I wasn't provided with initial or final values for pressure and volume, so there's not enough information to use PV=nRT. What should I do?
     
  2. jcsd
  3. Oct 16, 2014 #2
    But the last term isn't V, it's ΔV, so no matter what P is, PΔV for an isochoric process will be 0, because volume doesn't change (ΔV=0). I feel that this is a safe assumption, since you were given CV, which is the specific heat at constant volume.
     
  4. Oct 16, 2014 #3
    Could the last term be Δ(PV)? I just realized I can get (PV)1 and (PV)2 from PV=nRT with the data I was provided.
     
  5. Oct 16, 2014 #4
    Sorry to double post but I think I figured it out. The standard definition of the change of enthalpy: ΔH = ΔU + Δ(PV). The case ΔH = ΔU + PΔV is only for isobaric processes. The process here is isochoric, so we have no work done by the system, but we had a change in pressure. Therefore, ΔH = Q + PΔV + VΔP. I can cancel out the PΔV term, and I can get VΔP with the ideal gas law. So:
    [tex]\Delta H = 6790 \ cal + (10 \ mol) \left(1.987 \ \frac{cal}{mol \cdot K} \right)(100 \ K) = 8777 \ cal[/tex]

    Is it right this time?
     
    Last edited: Oct 16, 2014
  6. Oct 16, 2014 #5

    ehild

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    Nothing is said about the process of heating. The internal energy of n mol of an ideal gas is U=nCvT, so
    ##\Delta U=nCv \Delta T##
    for any process. H is defined as H=U+PV. Substituting P=nRT/V from the ideal gaw law, H=nT(Cv+R) =nTCp.
    ##\Delta H=nCp \Delta T##.
    Your method and result are correct.

    ehild
     
    Last edited: Oct 16, 2014
  7. Oct 16, 2014 #6
    Thank you!
     
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