Function for Cp/Cv for H2O(g) accurate to 3000 K

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SUMMARY

The function for Cp/Cv for H2O(g) at 3000 K is estimated to be approximately 1.25, based on treating H2O as an ideal polyatomic gas. At this temperature, it is confirmed that vibrational modes are excited, leading to 8 degrees of freedom, resulting in Cv = 4R and Cp = 5R. If additional modes such as the flapping mode are considered, the ratio can adjust to 1.22. Accurate thermodynamic property tables are essential for precise calculations beyond 1800 K.

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Does anyone know the function for Cp/Cv for H2O(g) accurate to 3000 K ? I would greatly appreciate any replies.
 
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You'll most likely need to find an advanced thermodynamics property table to find that. The one I have is only good up to 1800k

(if I had to make a rough estimate I'd say it should be somewhere around 1.2)
 
ZA,

This has partly been answered to an accuracy of about 1% in the other thread. People have suggested a value of about 1.25

From a theoretic angle, treating H2O as an ideal polyatomic (nonlinear) gas at 3000K (a very good approximation, if you ask me), Cp and Cv values would depend on whether or not you are exciting the vibrational moles in the molecule.

If this temperature is too low (which I doubt, since you typically start to see signs of exciting vibrational modes at about 1000K) to excite vibrational modes, then the molecule has 6 degrees of freedom, and thus Cv = 3R and Cp = 4R, making Y = Cp/Cv = 1.33

If this temperature is high enough (I think this is almost certainly true) to excite both of the dominant vibrational modes of H2O (one where the H atoms have opposite velocities relative to the O atom and the other, where they have the same velocity) then the molecule has 8 degrees of freedom, making Cv = 4R and Cp = 5R, and hence Y = Cp/Cv = 5/4 = 1.25

If you also include a flapping mode (I'm not sure what the activation temperature for this mode would be - just higher than the other modes), then you have Y = 5.5/4.5 = 1.22
 
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