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Change in enthalpy

  1. Aug 26, 2014 #1
    I was asked to find the change in specific enthalpy of constant volume process.

    Mass of 10 kg of a gas is heated to a pressure of about 2 times the initial pressure. The initial temperature was 20°C.

    The process is a reversible, non-flow, constant volume process.

    I found out the final temperature which is 40°C

    Change in enthalpy Δh=Cp*Δt

    My question is, is it correct to use Cp which is defined for a constant pressure process in a constant volume process?

    If it's correct then how it is correct?
     
    Last edited: Aug 26, 2014
  2. jcsd
  3. Aug 26, 2014 #2
    How can the gas be compressed if its volume is constant?

    Chet
     
  4. Aug 26, 2014 #3
    Sorry it's is heated not compressed.
     
  5. Aug 26, 2014 #4

    jack action

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    You need to look at the formal definition of enthalpy on Wikipedia to understand.

    At some point you will find the following equation:

    c25908f7c6c63ac4b7422cd7d91b2c16.png

    The left part is the definition of enthalpy. The first two terms of the right part are the internal energy dU and the third term is the work done.

    The second term of the right part is the internal work done (which affects the internal energy and the enthalpy) and the third term could be seen as the total work done (I'm not sure it's the proper terminology).

    The thing is that the pdV term is also included in the d(pV) term (= pdV + Vdp). This is why we end up with:

    8e15be705c669a27bdd04c2882bf26bc.png

    In an isochoric process, TdS is the internal energy because there is no internal work done, but dp is not equal to zero so the enthalpy is still higher than the internal energy.

    The way to think of it is, even though there is no volume change, that change in exterior pressure to keep the volume constant requires some energy from an exterior source.
     
    Last edited by a moderator: Apr 19, 2017
  6. Aug 26, 2014 #5
    There is an important point that, in my opinion, is always omitted, or at least, under-emphasized in thermodynamics texts. Enthalpy represents a physical property of the particular gas, and is not intrinsically related to the characteristics of any particular process. It is a function only of the temperature and pressure of the gas. So, for example, if you tell me the temperature and pressure of a particular gas in two different equilibrium states, I can tell you the change in enthalpy (per unit mass) between these states. All I need to do this is a table of enthalpy (per unit mass) as a function of temperature and pressure. For an ideal gas, things are even simpler. At low pressures, where ideal gas behavior prevails, the enthalpy is a function only of temperature.

    So where does this constant pressure business come into play? Well, this is how we experimentally measure the enthalpy values that go into the table (once and for all). By doing an experiment at constant pressure, the change in enthalpy between equilibrium state A and equilibrium state B is equal to the amount of heat we have to add in the process to get from state A to state B. But, once these measurements have been performed, they never have to be performed again.

    What we call the heat capacity Cp is simply equal to the partial derivative of the enthalpy with respect to temperature.

    Hope this helps.

    Chet
     
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