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What is enthalpy for an adiabatic process

  1. Feb 14, 2017 #1
    I have a question I hope someone may be able to answer.

    I am currently working out an adiabatic compression process that gives me an initial pressure,volume, and temperature. Along with a final pressure and given that the cylinder/piston assembly does not conduct any heat. Another given is that the Cp=3.5R

    I have done the work to solve for my gamma and the final volume and temperature for the system. With that I was able to calculate work done by the system which is also the change in internal energy.

    My question is how do I find ΔH?

    I know that ΔH=ΔU+ΔPV, but I just don't think my way of thinking is correct as I would assume that I would substitute for ΔU=Q-W where I know Q will equal zero.

    leading me to an equation of ===> ΔH=dQ-dW+ΔPV leading me to an answer of 0.

    Is this a correct assumption? I look forward to any and all productive feed back.

    Best Regards,

    Dylan
     
  2. jcsd
  3. Feb 15, 2017 #2

    Andrew Mason

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    Be careful: ##\Delta (PV) = \int PdV + \int VdP##

    Since ##\Delta Q = \Delta U + \int PdV## (first law) and ##\Delta H = \Delta U + \int PdV + \int VdP##, for adiabatic processes where ##\Delta Q = 0##, this means ##\Delta H = \int VdP##.

    AM
     
  4. Feb 15, 2017 #3
    Thank you Andrew for your prompt response.

    My only question is with what volume will I be using for my calculations and why? Will I be using the final Volume or the Initial?
     
  5. Feb 15, 2017 #4

    Andrew Mason

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    You have to integrate VdP over the process. If you know the relationship between V and P during the process you can work this out.

    If the gas is ideal, you can try using the adiabatic condition: ##PV^\gamma = P_0V_0^\gamma##. The adiabatic condition assumes a reversible process but it is often a good approximation for a even quick non-reversible adiabatic expansions or compressions involving small volumes (where the average molecular speed is such that the distance covered by an average molecule during the process is several times the distance between walls).

    AM
     
  6. Feb 15, 2017 #5
    since it is an ideal gas would I be able to just fit the expression so that V=RT/P that way I get the equation into the form of

    RT∫(1/P)dP

    And calculate from there.


    Also I appreciate your answers Andrew and was wondering if you had any recommended sources to practice single region partial derivatives.

    I am still trying to fine tune my working with the subject.

    Best Regards,

    D
     
  7. Feb 15, 2017 #6

    Andrew Mason

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    T is not going to be constant. Work is done by or on the gas, so U must change, which means T must change. If T were constant, heat flow would have to occur, which means it is not adiabatic.

    AM
     
  8. Feb 15, 2017 #7

    DrDu

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    You just have to calculate delta pV, which onli depends on the initial and final state.
     
  9. Feb 16, 2017 #8
    If you are using ##\gamma##, you must be treating it as an ideal gas. The enthalpy of an ideal gas is a function only of temperature. It is given by:
    $$\Delta H=nC_p\Delta T$$
    You know the number of moles, the molar heat capacity at constant pressure, and the temperature change. So, in addition to DrDu's method, this method will also give the same result.
     
  10. Feb 18, 2017 #9

    Andrew Mason

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    That's right. ##\Delta H = \Delta U + \Delta(PV) = \Delta U + \int PdV + \int VdP##. For an adiabatic process, ##\Delta U = -\int PdV## so ##\int VdP = \Delta(PV) - \int PdV = \Delta (PV) + \Delta U = nR\Delta T + nC_v\Delta T = nC_p\Delta T = \Delta H##

    AM
     
    Last edited: Feb 19, 2017
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