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[Thermodynamics] Calculate change in entropy of closed reversible system |
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| May19-12, 01:38 PM | #1 |
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[Thermodynamics] Calculate change in entropy of closed reversible system
1. The problem statement, all variables and given/known data
Mercury is a silvery liquid at room temperature. The freezing point is -38.9 degrees celcius at atmospheric pressure and the enthalpy change when the mercury metls is 2.29 kJ/mol. Wat is the entropy change of the mercury if 50.0 g of mercury freezes at these conditions? The molarmass of mercury is 200.59 g/mol. Assume the process is reversible 2. Relevant equations Q=m(h_2-h_1) (enthalpy equation) Q=mT(s_2-h_1) (entropy equation) 3. The attempt at a solution first I calculated what the energy change is per kg with what the enthalpy change is. h2-h1 is 2.29 kJ/mol and since there are 4.0118 mols I got 9.187 J/kg. So then I used Q/m = T(delta S) I rewrote Q/m to 459.35 J (since there are 50 grams of the substance) and divided that by T which is 234.1. The answer I get is 1.96 J/K whereas the answer is -2.44 J/K. Not sure what I'm doing wrong. |
| May19-12, 06:27 PM | #2 |
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Recognitions:
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Hi Ortix!
![]() Perhaps you calculated the number of moles the wrong way around? If 1 mol is 200.59 g, how many moles is 50.0 g? |
| May19-12, 06:32 PM | #3 |
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Hey Serena,
Very stupid mistake indeed. It would be 50/200.59=0.249264669 but this would result in a much smaller solution and still not negative (however, it would be negative entropy since the mercury is being frozed, in other words, energy is being taken out) I quickly worked it out on my laptop calculator (in bed on my laptop) and my answer is: 0.0488 J/K which is obviously wrong. |
| May19-12, 06:41 PM | #4 |
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Recognitions:
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[Thermodynamics] Calculate change in entropy of closed reversible system
Yup, that's why it's negative.
Did you take the 2.29 kJ/mol into account? Or else what did you calculate? |
| May20-12, 10:24 AM | #5 |
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I got the answer! Pretty simple calculation, but could you perhaps tell me what "m" exactly is in the equations and what the units are?
Q=m(h_2-h_1) (enthalpy equation) Q=mT(s_2-h_1) (entropy equation) Because I ended up with the right answer being in the form Q/mT when it should be Q/T. I think I'm doing something wrong in the conversion of mol to gram. EDIT: I think I got it. I equated both equations and let the m's drop out. Since enthalpy is given as specific enthalpy per mol, i just multiplied it by the amount of moles present in the substance and got the total enthalpy change. Divide that by T and that is the answer! Can anyone tell me if my reasoning is correct? :) |
| May20-12, 10:40 AM | #6 |
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Recognitions:
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Yep. That's it. :)
m is the mass in kilograms. And for the record, your 2nd equation should read: Q=mT(s_2-s_1) (entropy equation) |
| May20-12, 10:46 AM | #7 |
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Oh yeah, that was a typo :) Thanks for putting me on the right track! :D
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