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How to calculate the gas enthalpy in a particular pressure and temperatur?

  1. Feb 18, 2017 #1
    Hi All,

    Is there anybody who can help me with this question?

    Example:
    What is the enthalpy of CO2 in 200psig and 80deg.F?

    I have a software, but I'd like to see the calculation steps.

    Thanks
    Ding
     
  2. jcsd
  3. Feb 19, 2017 #2
    The first thing you need to do is find out what the temperature and pressure are for the reference state of zero enthalpy (that the software uses).
     
  4. Mar 26, 2017 #3
    Thank you, could you please educate me more?
     
  5. Mar 26, 2017 #4

    BvU

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    Yes, but you could also take a textbook (e.g. Atkins, physical chemistry) and find out how enthalpy is determined. You would find the enthalpy is usually defined zero for elements at a standard T and p (most of the time ##10^5## Pa and ##25\ ^\circ##C). CO2 has a standard enthalpy of formation under those conditions and a next step is to go to your 80 ##^\circ##F and 200 psig by integrating ##dH = c_p \,dT +\displaystyle \left ( \displaystyle \partial H\over \partial p \right )_T \, dp##
     
  6. Mar 26, 2017 #5
    Hi BvU.

    If you are using package software, then there is no need to integrate anything. It just gives you the enthalpy value at the specified temperature and pressure (i.e., it has done the integration for you). But you must know the reference state used by the software.
     
  7. Mar 27, 2017 #6

    BvU

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    Thanks Chet,

    I was catering tot he OP's desire to see the calculation steps. Very commendable to try and understand what the program actually does !

    At the risk of hijacking ding's thread: I can clearly see and imagine the ##\int c_p \, dT## part of the integral, but the ## \left ( \displaystyle \partial H\over \partial p \right )_T \, dp## part is much more elusive.
     
  8. Mar 27, 2017 #7
    Here are the steps:
    $$dH=TdS+VdP$$
    $$dS=\left(\frac{\partial S}{\partial T}\right)_PdT+\left(\frac{\partial S}{\partial P}\right)_TdP$$
    $$dH=T\left(\frac{\partial S}{\partial T}\right)_PdT+\left[V+T\left(\frac{\partial S}{\partial P}\right)_T\right]dP=C_pdT+\left[V+T\left(\frac{\partial S}{\partial P}\right)_T\right]dP$$
    $$dG=-SdT+VdP$$
    So,
    $$S=-\left(\frac{\partial G}{\partial T}\right)_P$$and$$V=\left(\frac{\partial G}{\partial P}\right)_T$$
    So, $$\frac{\partial ^2 G}{\partial T\partial P}=-\left(\frac{\partial S}{\partial P}\right)_T=\left(\frac{\partial V}{\partial T}\right)_P$$
    So, finally, $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$

    This derivation is in most thermo books.
     
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