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Question regarding change in enthalpy

  1. Jan 18, 2014 #1
    This isn't a formal homework question so much as a conceptual question for my own edification.

    I'm reading my textbook's section on enthalpy and energy, and given the expression:

    ΔH=nCpΔT

    It states that, "we can use this expression to represent the change in enthalpy when n moles of an ideal gas are heated, regardless of any conditions on pressure or volume."

    I know that the ideal gas law stats that PV = nRT, and thus T is proportional to PV.

    How can it be, then, that enthalpy change is only affected by temperature change and not affected by changes in pressure and/or volume? :confused:
     
  2. jcsd
  3. Jan 18, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    First, that equation for enthalpy holds for constant pressure. At const. P, the volume will change when T changes of course. But it is taken care of by considering only T because P is fixed.

    What the book mentions are "conditions" on P and V. This means that you do not need to know what P is, or what the initial and final volumes are. So long as you know ΔT, you can calculate ΔH.
     
  4. Jan 18, 2014 #3
    What they are saying it that the enthalpy of an ideal gas is independent of pressure. If we regard the enthalpy (per unit mass) of a pure substance to be a function of pressure and temperature, the we can write:

    H = H(T,P)

    From this it follows that:

    [tex]dH=\frac{\partial H}{\partial T}dT+\frac{\partial H}{\partial P}dP[/tex]
    But, by definition,
    [tex]C_p=\frac{\partial H}{\partial T}[/tex]
    so
    [tex]dH=C_pdT+\frac{\partial H}{\partial P}dP[/tex]
    For real gases in the limit of low pressures, it has been found experimentally that:
    [tex]\frac{\partial H}{\partial P}→0[/tex]
    But real gases approach ideal gas behavior in the limit of low pressures. So, for ideal gases, the enthalpy is independent of pressure.
     
    Last edited: Jan 18, 2014
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