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Old 07-08-2012, 12:47 PM   #28
Imiweevierm

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Oct 2005
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When a physicist or a chemist is looking at the phase diagram of water then for him or her water, water vapour and ice are synonymous. Of course a common man will think that you have gone round the bend.

Anyway I would like to get a technical thing out of our way.

According to Wikipedia an HO-H bond of a water molecule (H-O-H) has 493.4 kJ/mol of bond-dissociation energy, and 424.4 kJ/mol is needed to cleave the remaining O-H bond.

Now if even if one O-H bond is broken then it is no more water.
So if we supply 424.4 kJ of energy then one mole of water will be dissociated - it will not be water.

Thus to dissociate 6.022×10^23 bonds we need 424.4 kJ of energy.

Or to dissociate one bond you need (424.4 divided by 6.022×10^23) kJ of energy.
That is roughly 7×10^(-19) J of energy.

Now Boltzmann's constant tells us that 1.38×10^(−23)J of energy equals one degree K.

Thus 7×10^(-19) J of energy will equal (7/1.38)×10^(23-19)K in terms of temperature.

This is 51068 K. Assuming I have not committed obvious mistakes.

This is ten times the solar surface temperature. So at the solar surface water vapour can exist.
Wow Maripat, are you a physics-guy?

Anyway, I'll go with your explanation, even though I dont understand a lot of it
Imiweevierm is offline


 

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