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

In summary: The tricky bit is usually the circularity in the differentiation.In summary, the conversation discusses finding the enthalpy of CO2 at a specific temperature and pressure. It is suggested to find the reference state used by the software and then integrate the equation to determine the enthalpy. However, it is noted that if using package software, the integration is not necessary as the enthalpy value is provided. The conversation also delves into the calculation steps and the elusive part of the integral. A derivation is provided, which uses the equations for entropy and Gibbs free energy to derive the final equation for enthalpy.
  • #1
dingweiyang
2
0
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
 
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  • #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).
 
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  • #3
Chestermiller said:
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).

Thank you, could you please educate me more?
 
  • #4
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##
 
  • #5
BvU said:
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##
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.
 
  • #6
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.
 
  • #7
BvU said:
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.
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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What is gas enthalpy?

Gas enthalpy is the measure of the total energy content of a gas system at a given pressure and temperature. It includes both the internal energy of the gas molecules and the work required to expand or compress the gas.

How do I calculate gas enthalpy?

The gas enthalpy can be calculated using the following formula: H = U + PV, where H is the enthalpy, U is the internal energy, P is the pressure, and V is the volume of the gas system. The values of U and PV can be determined using thermodynamic tables or equations of state.

Why is it important to calculate gas enthalpy?

Calculating gas enthalpy is important in many industrial and scientific applications. It helps to understand the energy content of a gas system, which is crucial in processes such as combustion, refrigeration, and chemical reactions. It also allows for the efficient design and operation of gas systems.

How does pressure and temperature affect gas enthalpy?

The pressure and temperature of a gas system have a direct impact on its enthalpy. As the pressure increases, the gas molecules are compressed and their internal energy increases, leading to a higher enthalpy. Similarly, as the temperature increases, the gas molecules gain more kinetic energy, resulting in a higher enthalpy.

What are some common units of measurement for gas enthalpy?

The most commonly used units for gas enthalpy are joules (J) and kilojoules (kJ). However, in some industries, units such as British thermal units (BTU) and calories (cal) may also be used. It is important to ensure that all units in a calculation are consistent to get accurate results.

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