My Math Forum Entropy Calculation

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 February 16th, 2017, 12:39 AM #1 Newbie   Joined: Feb 2017 From: Lebanon Posts: 18 Thanks: 1 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.
 February 17th, 2017, 06:05 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,090 Thanks: 701 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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 Thanks from JamSmith
February 20th, 2017, 09:15 PM   #3
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Entropy Calculation

Quote:
 Originally Posted by Benit13 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 09:24 PM.

 February 21st, 2017, 01:24 AM #4 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,090 Thanks: 701 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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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