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February 16th, 2017, 01:39 AM   #1
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Entropy Calculation

Can any one please help me to solve this?

Calculate the entropy change if one mole of water evaporate at 373 K and 1 atm. If you look up the enthalpy of vaporization, for water in a table, you would get a molar enthalpy of 40.7 kJ/mol.
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February 17th, 2017, 07:05 AM   #2
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For the case of constant temperature:

Change in entropy = Change in heat content / T

$\displaystyle \Delta S = \frac{\Delta Q}{T}$

For latent heat transfer (evaporation):

$\displaystyle \Delta Q = \Delta m \times l$
= 1 mole x latent energy per mole
= 40700 J

$\displaystyle \Delta S = \frac{\Delta Q}{T} = \frac{40700}{373} = 109.12 $ J/K
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February 20th, 2017, 10:15 PM   #3
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Entropy Calculation

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Originally Posted by Benit13 View Post
For the case of constant temperature:

Change in entropy = Change in heat content / T

$\displaystyle \Delta S = \frac{\Delta Q}{T}$

For latent heat transfer (evaporation):

$\displaystyle \Delta Q = \Delta m \times l$
= 1 mole x latent energy per mole
= 40700 J

$\displaystyle \Delta S = \frac{\Delta Q}{T} = \frac{40700}{373} = 109.12 $ J/K

Well I have another entropy related question, I was searching for material for more information about entropy conversions. I come to know about MM-PBSA Charmm, I read out some information and have some query related entropy conversion.

My question is:-

What is the conversion that charmm uses for convert the entropic units to kcal/mol? In the scripts I found the MM-PBSA script and I see that to obtain the entropy in kcal/mol only multiply by temperature, is that correct? That may implicated the 1 eu = 1kcal/mol.

Last edited by JamSmith; February 20th, 2017 at 10:24 PM.
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February 21st, 2017, 02:24 AM   #4
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I have no idea what CHARMM is, so I can'#t really answer your question, but the response to (presumably) your post here:

https://www.charmm.org/ubbthreads/ub...at&Number=5712

seems correct. It's just the formula in my post; you're basically calculating heat from entropy.
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