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I'm an undergraduate Engineering/Mathematics Student down in Brisbane, Australia. Interested in learning as much as I possibly can about maths, particularly the ideas/constructions/history/motivation behind the classic undergraduate concepts.


Oct
14
awarded  Popular Question
Oct
14
awarded  Nice Question
Oct
14
comment How did Kelvin make this fascinating calculation?
Thanks, that's beautiful in the terribly-obvious-when-you-see-it kind of way :)
Oct
14
accepted How did Kelvin make this fascinating calculation?
Oct
13
asked How did Kelvin make this fascinating calculation?
Oct
9
asked Is every imaginable quasi-static locus (with non-decreasing entropy) physically realisable?
Sep
24
awarded  Autobiographer
Sep
18
comment Inconsistency between Helmholtz and Gibbs Free Energies
This is probably a terribly silly question, but if two phases are at equilibrium at constant $p, T$ and $N$, doesn't that mean that they have equal Gibbs free energy (per mole), so that irrespective of the fraction of ice or water, the Gibbs free energy is the same single value?
Sep
18
accepted Inconsistency between Helmholtz and Gibbs Free Energies
Sep
18
comment Inconsistency between Helmholtz and Gibbs Free Energies
I can imagine that being a problem. Thanks for your help: I've only skimmed through Chapter 8 of Callen, but it's probably time to study it in a bit more depth :)
Sep
18
comment Inconsistency between Helmholtz and Gibbs Free Energies
OK, I think I'm seeing where you're coming from. Concavity of $A(V)$ would certainly solve the problem, as then both $P(V)$ and $V(P)$ would be single-valued and we'd all be happy. I was under the impression though that they need not always both be single-valued - sometimes at a certain pressure the volume could be multiple-valued - say at a phase transition with liquid and gas co-existing. Ahh, but that would just be at one single value of $P$, wouldn't it? That would be the discrete jump of $V(P)$ you mention. At that discrete jump then in some sense $V(P)$ is multi-valued, but only there.
Sep
18
comment Inconsistency between Helmholtz and Gibbs Free Energies
Thanks for your answer :) Just to clarify, are you saying that $G$ need not be a single-valued function of $p, N$ and $T$, as I'd assumed? I was under the impression that the whole point of making the Legendre transformation was to change coordinate systems, so that we find a new 'thermodynamic potential' $G$ with the same minimization properties as the internal energy, but which is a (presumably single-valued) function of the intensive variables. Thanks
Sep
13
asked Inconsistency between Helmholtz and Gibbs Free Energies
Sep
6
comment Generalisation of Reversible Equation to Non-Reversible Situations Because it Only Contains 'Properties of the System'
I asked this question years ago, and your response is exactly what I'd say to my younger self today :) The only thing I dislike is the use of the word 'reversible' - I prefer quasi-static, and reserve reversible for quasi static isentropic processes, in line with callen's thermodynamics text.
Aug
29
awarded  Popular Question
Aug
12
accepted Why doesn't Euler's theorem provide an *absolute* energy scale
Aug
12
comment Why doesn't Euler's theorem provide an *absolute* energy scale
Ahh, OK, I think I understand: classical thermodynamics is consistent without assuming the 3rd law. Thanks for your help, I think this makes sense now :)
Aug
10
comment Why doesn't Euler's theorem provide an *absolute* energy scale
OK, the more I think about this the more it makes sense: if we add even one more atom, how do we quantify the energy added? For instance, do we count relativistic mass-energy or not? It's impossible to define exactly how much energy we've added, without of course having a reference state telling us how much energy is in a mole of some state of matter. Still not sure on the entropy question though :)
Aug
10
comment Why doesn't Euler's theorem provide an *absolute* energy scale
Sorry, one more question. It seems natural to me that the difference in energy between a 1L balloon and a 2L balloon with the same composition is well defined. But according to what you've said this doesn't seem to be the case. If the 1L is said to have 1kJ of energy the 2L will have 2 kJ, and the difference will be 1kJ, but if the 1L is said to have 3kJ of energy the 2L will have 6 kJ, and the difference will be 4kJ. This is unintuitive to me (I feel this internal energy difference should be well defined) but am I correct to reason that this energy difference is in fact not well defined?
Aug
10
comment Why doesn't Euler's theorem provide an *absolute* energy scale
To emphasize this again, I can develop an experimental method for measuring $S$, by starting at 0 Kelvin and adding energy and monitoring the temperature increase and evaluating $S= \int_0^{T_f} dQ/T $. There's no change in $N$ here, and we only need to measure energy flow and temperature, both of which are absolute. How then isn't entropy 'measurable' like pressure or volume?