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I Enthalpy and ideal gas law

  1. Feb 29, 2016 #1
    Hi! I'm new to the forums and currently reading about Chemical Thermodynamics. So here's what I know:

    ΔE = q - w

    So for constant volume reactions, no work is done hence:

    ΔE = q

    But for constant pressure reactions, heat be may released (for exothermic reactions) and work is done hence:

    ΔE = q - pΔV

    so ΔE ≠ q because some of the heat does work.

    So we define enthalpy H = E + PV, so, ΔH = ΔE + PΔV = q - pΔV + pΔV = q

    and hence for constant pressure reactions ΔH = q.

    My confusion is that the book says that the work term pΔV = (Δn)RT for these processes, but how can we hold the temperature constant since it is obviously changing as a result of the heat of the reaction?

    Any help would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Mar 1, 2016 #2
    The formula pΔV = (Δn)RT must be wrong, I would say. Could you tell me which book you have and the equation number? Normally for a constant pressure process the number of moles is constant. That should mean the temperature changes (for an ideal gas). Unless they are thinking of a chemical reaction, of course, but with a constant T??
     
    Last edited: Mar 1, 2016
  4. Mar 1, 2016 #3
    Everything you said is correct. With regard to your question, the temperature is being held constant at the initial temperature by either adding heat or removing heat from the reaction mixture as necessary (say, by contact with a constant temperature bath held at the original temperature). When they talk about THE "Heat of Reaction," they are referring specifically to holding both the temperature and the pressure constant. The term you are referring to has to be included if the number of moles of products differs from the number of moles of reactants. In such a case, the change in internal energy will differ from the change in enthalpy (even if the temperature and pressure are constant).
     
  5. Mar 1, 2016 #4
    Thank you very much for your help. I think the ideal gas term term is just included as a convenient conversion between change in internal energy and enthalpy change when the heat is measured in a bomb calorimeter (constant volume). What do you think?
     
  6. Mar 1, 2016 #5
    No. Although applicable to a bomb calorimeter, the relationship is much more general than just this one application. If you know the ΔH for the reaction, you can always get the ΔU, and vice versa.
     
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