Additivity of thermodynamic potentials?

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The discussion centers on the additivity of thermodynamic potentials, specifically the Helmholtz free energy (F). It is noted that F is not additive, meaning F does not equal the sum of F1 and F2 for two separate systems. However, if the two systems are at the same temperature and have negligible interaction energy, F can be considered additive. The conversation emphasizes the conditions under which thermodynamic potentials like F, enthalpy (H), and Gibbs free energy (G) may or may not be additive. Understanding these conditions is crucial for analyzing thermodynamic systems effectively.
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Homework Statement
Two ideal gases, e.g. helium and argon with NA and NB atoms, are mixed, NA + NB = 1 mol. Determine the helmholtz energy of the entire system.
Relevant Equations
F = E - TS
My professor said that F is not additive, meaning F ≠ F1 + F2, where F1 is the helmholtz energy of system 1 and F2 is the helmholtz energy of system 2. So my question is, how can I decide wether a thermodynamic potential (F, H, G) is additive or not?
 
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If the two systems have the same temperature and have a negligible interaction energy, then F additive.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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