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I thought enthalpy meant total internal energy?

  1. Feb 2, 2014 #1
    Then I see this line in a Wiki article:

    For an ideal gas, the heat capacity is constant with temperature. Accordingly we can express the enthalpy as H = n C_P T and the internal energy as U = n C_V T, where n is the amount of substance in moles. Thus, it can also be said that the heat capacity ratio is the ratio between the enthalpy to the internal energy:
    \gamma = \frac{H}{U}

    Is this statement false? If not, some clarification would be greatly appreciated.

    Thanks,
    Chris Maness
     
  2. jcsd
  3. Feb 2, 2014 #2

    Andrew Mason

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    I am not sure where you got the impression that H and U were the same. Enthalpy, H, is a state variable, like U, but it is defined as: H = U + PV. If P is constant, then ΔH = ΔU + PΔV, in which case ΔH = Q i.e. the total heat flow during the process (first law: Q = ΔU + W where W is the work done by the gas).

    AM
     
  4. Feb 3, 2014 #3
    This would be true:

    [itex]\gamma= \frac{H_s}{U}[/itex],

    where [itex]H_s[/itex] is the static enthalpy of the gas. If the gas is in motion with velocity v, then the total enthalpy of the gas is

    [itex]H=H_s+\frac{v^2}{2}=c_pT[/itex], where T is the total temperature of the gas; while the static enthalpy is
    [itex]H_s=c_pT_s[/itex]. Since [itex]U=c_vT_s[/itex], one can write [itex]\gamma= \frac{H_s}{U}[/itex].

    But [itex]\gamma= \frac{H}{U}[/itex], is wrong , if H denotes the total enthalpy. Total enthalpy contains both static enthalpy and the kinetic energy of the gas.
     
  5. Feb 3, 2014 #4
    Actually, for an ideal gas, the heat capacity is not constant with temperature. I think you are thinking of what physicists call a perfect gas, rather than an ideal gas. At low pressures where real gases approach ideal gas behavior, the heat capacity does vary with temperature.

    For ideal gases, the enthalpy H and the internal energy U are independent of pressure, but are functions of temperature, with

    dH = Cp(T)dT

    dU = Cv(T)dT

    So, when you integrate to get H and U, you need to take into account the temperature dependence of the heat capacities. However, you can write:
    [tex]\frac{dH}{dU}=γ(T)[/tex]
    Hope this helps.
     
  6. Feb 3, 2014 #5
    I was deriving P/V and T/V relationships for an adiabatic process for a refresher. I was successful, but I assumed Cp=Cv+R. I would like to derive that too, but there was some holes in my understanding (obviously). When I get home, I will take a closer look so that I can glean more insight. Then take a shot at Cp=Cv+R. I would imagine these "constants" are approximately constant over a range. At least for understanding a non real gas.

    Thanks,
    Chris Maness
     
  7. Feb 3, 2014 #6
    Try this:

    dH = CpdT=dU+d(PV)=CvdT+d(RT)

    This only applies to an ideal gas.

    Chet
     
  8. Feb 4, 2014 #7
    I guess the whole time I thought U contained all forms of potential and kinetic energy. I didn't know it excluded this term PdeltaV. I guess it is useful in chemistry where a reaction can/cannot outgas and expand into free space. The change in energy from the gas leaving the system would no longer be internal. I am thinking you would no longer consider the gas part of the system, but part of the environment.

    Did I get the touchy feely explanation bit down? Or did I bodge it?

    Thanks,
    Chris Maness
     
  9. Feb 4, 2014 #8
    I can't really follow what you are saying. The form of the changes in internal energy and enthalpy I wrote down for an ideal gas apply to a closed system (no exchange of mass with the surroundings).
     
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