1

Physics, Astronomy & Cosmology / Re: Is there a substance where entropy decreases at higher temperature?

« on: 17/12/2022 16:42:12 »

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2

Physics, Astronomy & Cosmology / Re: Is there a substance where entropy decreases at higher temperature?

« on: 16/12/2022 05:51:33 »

Hi.

    I've had a bit of time to think.   As you ( @chiralSPO ) implied:  It's very much a case of needing to keep our definitions of temperature straight (and it does become a little dull then).

    The situation with a simple two-energy level system does illustrate a situation where temperature can increase (in this case it is increasing by becoming less negative) while the entropy decreases.   As you ( @chiralSPO ) implied it's clear that energy is always being added to the system as more particles are driven to the higher energy state,  however that wouldn't guarantee that the temperature was increasing.   
    For a typical substance like an ideal gas held at constant volume, it's apparent that increasing the internal energy will increase the temperature.   For a more arbitrary substance and where statistical temperature is used, you really do have to directly check the gradient  ∂S/∂E.   A diagram will save 1000 words here:

Suppose the entropy, S, of a system of N particles (each particle having only two energy levels) varied with E, the total energy of the system, as follows:


Untitled.jpg (49.3 kB . 1152x648 - viewed 2373 times)
   An increase in the the total energy of the system would not produce an increase in temperature once you've entered the region with most of the particles in the higher energy state.  The temperature is negative in that region but perfectly constant.

  Fortunately the entropy of the simple 2-energy level system doesn't look like that,  it has this sort of shape:


entropy2.png (24.19 kB . 882x680 - viewed 2453 times)
   Where the gradient =  1/T   is  monotonically decreasing with E.

  With that established:   The simple 2-energy level system does have the usual relationship where increasing its energy remains synonymous with increasing its temperature.   The note of caution in your ( @chiralSPO ) first sentence is then substantially laid to rest:

I think you need to be careful about specifying whether you mean temperature or thermal energy.

   More generally, the simple 2-energy level system ticks all the boxes that I was asking for in the original post.  So, well done and thank you.   You can have the best answer award.
-------
     I don't think I need to ask anything else.   If you restrict attention to just non-negative temperatures, then I'm now fairly sure that the answer rests entirely upon finding a substance with specific heat capacity, C_V < 0.
     While I don't know of any substance with that property, in any situation (e.g. in any range of temperature, pressure or other conditions),  if someone else does -  Please let me know.

Best Wishes.

3

Physics, Astronomy & Cosmology / Re: Is there a substance where entropy decreases at higher temperature?

« on: 15/12/2022 02:39:37 »

Hi and thanks.

   Lasers and population inversion was exactly what I had in the back of my mind.

This sentence

One reasonable definition of energy is: T = ∂U/∂S (constant volume).

   might be a typing error.   That's a perfectly good thermodynamic definition of temperature.

Obviously I'm going to have to take a bit longer to consider everything.  Thank you very much for your time and it's exactly the sort of thing I was looking for.

Best Wishes.

[Spoiler:]

4

Chemistry / Re: How can I find the optimum ΔH and ΔS for passive T control?

« on: 01/06/2022 02:39:40 »

Chunk 2

Spoiler: show

So the question becomes. For a given high and low temperatures (T_high and T_low), what is the optimal choice of ΔH and ΔS, such that the system is most stable between those to temperatures?

    Stability is not something you've decleared very well.   I'm going to interpret the notion of "stability" as implying that the reaction would absorb (or release) as much heat as possible to counter a temperature change in the room.
   Let's just spend a moment to see that this is a sensible approach:

Diagram 2:   How the room loses / gains thermal energy

   Thermal Energy change
   in the Room,  cΔT               ----> heat lost to the outside world, ΔE   (due to... poor insulation etc.)
       /\
        |
        | 
  Heat released by chemical equillibrium system, ΔU


    Under the situation shown in the above diagram we have     
Change in thermal energy of the room  =   c.ΔT  =    ΔU - ΔE                 

[eqn 10]

  with c = heat capacity of the room;   ΔE = energy transferred out of the room;   ΔU = heat released by the chemical system into the room.
   Note the sign conventions we have established:    Room cooling: ΔT<0   ;  ΔU >0 the chemical system will release heat (thermal energy) into the room  ;  ΔE > 0  heat is lost to the outside world.    Let's not spend too long discussing this, it should be apparent that ΔE can be considered as the sum of natural heat exchange to the world outside the room + energy exchanged due to heating or cooling systems like radiators or air-conditioning units that a person has inside the room.  A negative value indicates a net heat input to the room (e.g. from the house central heating system).
    Note also that the chemical system is considered as part of the contents of the room.   In the most degenerate situation the room is only the chemical system but usually the room is the chemical system + the rest of the room (like walls and furniture) such that the chemical system is just a minor contribution to the mass and heat capacity for the room.   (LATE EDITING:  Stuff removed - the above is just conservation of energy ideas and doesn't need further comment).

   The following results follow from [Eqn 10]
(i) ΔU ≠ ΔE  iff the room changes temperature.
(ii)  .. result removed in late editing... serious error here, sorry.   The following result still holds but for different reasons...
   Therefore,  |ΔU| ≤ |ΔE| always and equality holds if and only if ΔT =0.   

[Result 11]

   We can now see that no chemcal equillibrium system A↔Z   exists which would perfectly maintain a fixed temperature in the room.    Assume some energy ΔE≠0 is removed from (or added to) the room.  For ΔT = 0, we must have ΔU = ΔE  but  the chemical equillibrium would not change at all if the temperature didn't change, so we would have ΔU = 0.    Therefore, when some energy enters or leaves the room we are always in the situation with |ΔU| < |ΔE| and the best we can hope for is that the chemical equillibrium opposes  ΔE as best it can (so that the temperature change will be less severe than if the chemical equillibrium system wasn't there).

    I would suggest that the best interpretation of "maximum stability" for our temperature buffering system will be trying to keep |ΔT| as low as possible for a given amount of ΔE (energy transferred to/from something outside the room).  For our purposes, the quantity ΔE is beyond our control or influence, that requires someone to insulate the room and turning on/off the central heating system of the house etc.

    From [Eqn 10] ; [Results 11] and recognising that under our sign conventions ΔU opposes the sign of ΔT we have,
   |ΔT| = 

[Equation 12]

So, that minimising the size of the temperature change for the room due to a given Energy transfer into / out of the room, requires the size of |ΔU| to be MAXIMISED.

     CLAIM  that ΔU = heat released from the chemical system = function of the current Temperature T, the parameters ΔH° and ΔS° of the reaction A→B   and, of course, of the temperature change ΔT that occurrs.  PROOF: We have the following,
   ΔU =  ΔN_A . ΔH° / (Avogadro)   

[Equation 13]

 
   Where, ΔN_A = the change in N_A (change in the number of molecules of A);     ΔH° is the specific enthalpy of the reaction change (heat change for 1 mol of A being converted to Z);     (Avogadro) = Avogadros number.   The signs work (I hope),  recall ΔN_A is negative, N_A decreases, when molecules change from A to Z   and  ΔH° = negative when  A→Z is exothermic,  making  ΔU = heat released to the room = positive.
    This is taking too long.... let's just speed this up.... we can obtain an expression for ΔN_A   from  [Equation 4] in terms of the change in the equillibrium constant K.   We have a thermodynamic expression for K from [equation 5].   That establishes the "CLAIM" that ΔU is a function of the things mentioned above.  The dependance on ΔT (the change in temperature) came from ΔN_A but that will be clearer in a moment...

    During the actual operation of the chemical equillibrium device,  you cannot change the parameters ΔH° or ΔS°  of the reaction or add extra molecules of A,  or do very much to change the way the device can abosrb or release heat.   The ony changes that occurr are temperature changes in the room.  So in calculating the quantity ΔN_A during the actual operation of the device,   we can treat all other parameters as fixed constants.   This saves a mess writing derivatives as partial derivatives,  for example  K = K(T) = a function only of Temperature during actual operation,   even though  K has dependance on ΔH and ΔS  looking at [eqn 5].   This should also help to explain why finding partial derivatives w.r.t.  variables like  ΔH°   is not something we will do.   Just to be clear on this:  It might seem that since we are trying to maximise ΔU  (amount of heat released by the chemical system) with a carefull selection of parameters like ΔH° for the reaction, then we would want to find derivatives w.r.t. ΔH° and treat the problem as a typical maximisation problem.   However, that is not what we will do.   Instead, we treat the reaction parameters as a fixed constant (don't specify the actual value - just recognise that it is a constant) and see how the system will naturally behave in operation.   After that, we are free to see how those results change or what they will mean when the reaction parameters like ΔH° and ΔS° are given a particular numerical value.   This method seems to work better than just directly trying to find maxima from derivatives w.r.t.  reaction parameters.   (For example, there doesn't seem to be any local maximum.   This might be discussed later -  the larger ΔH° can be the better, the progression is endless and no local maxima exists  BUT also  you just don't get much useful information about the system anyway).

   Let's see [Eqn 13] again:
δU =  δN_A . ΔH° / (Avogadro)   

[Equation 13' ]

     We'll drop the Avogadros number (as if ΔH° was stated per molecule not per mole), it only scales the value of ΔU,  the maximum won't be affected without it anyway.   Note that we are now considering only small changes, as marked with δ symbols -   the quantity  ΔH°  remains with a big delta because it is just a reaction parameter, a constant and not something that changes during the operation of the device.  We'll also just write ΔH instead of ΔH°  because it's shorter and should be clear enough that this is a fixed constant, a reaction parameter not something that changes dynamically during operation.
   From [Eqn 4] we can differentiate, noting that temperature is the only variable that changes during operation of the device:
   δN_A   =   
K and the derivative dK/dT can be found from [Eqn 5] and [Eqn 7]
    = 

   

[Check the signs work here:  If ΔH >0,  then A→Z is endothermic.  K = [Z]/[A]  and we expect K to increase when T increases.  dK/dT >0  is what we have.  Next up δN_A has a negative sign out at the front, so it's negative and yes we have A→Z so the number of molecules of A would decrease.   This all looks OK to me. ]

  Substitute this into [Eqn 13' ] and we have:

δU =      

[Eqn 14]

   where  ζ =  e^ΔS/R  =  a convenient constant to hold all the entropy information.   Note that a lot of the quantities appearing in {Eqn 14] are constants,    T is the only variable.

Recall that we wish to maximise δU  given a fixed transfer of energy δE from the room.  Straight away from [Eqn 14]  we can see that  if N (the total number of molecules of A and Z) is increased then δU is increased.   There is also a term on the numerator  ~  (ΔH)²   and although ΔH does appear elsewhere its effect is most noticeable here. *LATE EDITING:  Reasonably true for small to medium ΔH.   However, when ΔH becomes very large, then the exponential function e^-ΔH will dominate and bring the numerator towards 0.   This effect is more important than I first thought because ΔH usually is large - hundreds of thousands in standard units of Joules/mol. * 
   We can tidy up our expression a little.   We are trying to maximise δU  for a fixed value δE of energy transferred out of (or into) the room.   However, we can imagine the maximisation requirement differently.   Assume that a small change in temperature (δT) of the room has happened.   We would like to have this cause a release (or absorption) of energy by the chemical system  (δU)  that is as large as possible because that would have forced δE to be as large as possible.   [Eqn 12]  can be used to formalise this condition and establish that it is equivalent.  The post is too long, I'm skipping that.  We will just re-state the maximisation problem as required:   To buffer the temperature of the room as much as possible,  we wish to maximise δU  for a given fixed change in temperature δT.    Anyway, with that modification we can divide by δT  and obtain the following result (where δT→0 to form the derivative on the R.H.S.)

   

[Eqn 15]

     Our objective is to maximise this quantity.

.....  I stopped writing here.... there are some derivatives and some graphical plots to show what is happening that i would have inlcuded.... but by now   no-one is reading this   and the whole set of expressions have become sufficiently messy,   no-one wants to see them anymore.   They belong on a computer system like SymbolLab or Mathematica... or whatever your preferred mathematical manipulation software might be.   They do not belong on a text document for people to read or differentiate by hand.

    Minor Notes:    We were really maximising  |δU|  ,  the size of a change.  As it happens  δU  will oppose  δT in sign,   which means that we are often actually trying to minimise  δU  because  its negative and a lower negative has a greater size.   I started modifying everything to |δU|  in the early part of the post... but time limits cut in... I never carried it through to the end of the post.
    Here are some plots of what we get from [Eqn 15].

Parameters   N=1  (arbitrary scalar anyway).         ΔH = 10 000   J/mol       ΔS = 33.33  J/mol/K    (deliberately chosen so that  ΔH/ΔS ≈ 300 kelvin ≈ approx. room temperature  and ΔG would become 0 at that temp).

Plot-300.JPG (66.42 kB . 1363x418 - viewed 2405 times)

   Note that although I would have expected the minima   (maxima of |dU/dT| ) to be  around  T = 300 k,  it was actually a bit below that.


Paremeters as before,   ΔH raised to 20 000 J/mol.    ΔS = 33.33.     Expected  T = 600 kelvin but the maxima is still slightly off.

plot-600.JPG (82.3 kB . 1401x535 - viewed 2419 times)


Anyway, the general  behaviour of the chemical system seems to be reasonable.....   it always has maximum temperature buffering ability at a particular value of temperature.   Changing the values of ΔH  and  ΔS   move that  maxima   up or down.     You can probably also see why you can't get clear results by taking something like [Eqn 15] and directly differentiating w.r.t.  reaction parameters like ΔH.....      You can have a system that is actually tuned to buffer around a temperature of  100 kelvin *   instead of  300 kelvin   BUT   if  the value of N (the number of molecules in the system) is large enough and also  ΔH   is large eneough then it would still be a numerically better buffer at room temperature than a smaller and more sensibly ΔH valued system.    * LATE EDITING in the previous sentence:  It's more noticeable where the system is tuned to buffer below room temperature not above.  See the actual plots with given parameters.  It is, in my opinion, only by analysing the situation with ΔH and ΔS as fixed parameters and recognising that T is the only thing that can chnage in operation of the device where the natural behaviour of the device is observed.   For a realistic situation, where you can't have an infintely sized system or infinitely high values of ΔH,  you would want to work with the natural behaviour of the system and set it up with ΔH and ΔS  parameters that put it's maximum buffering ability right at the room temperature you wish to try and maintain.

Best Wishes.

[Spoiler:]

5

Chemistry / Re: How can I find the optimum ΔH and ΔS for passive T control?

« on: 01/06/2022 02:37:00 »

Hi again.

   @chiralSPO  - this is what I had started writing but stopped.   It's rough, not spell-checked and just not finished.   However, if it helps you,  then you can see it.     You'll realise why I stopped.... it's too long and I started by explaining things too simply, as if other users might also be reading it.

    Actually it just is too long....  the forum won't allow me to post anything with too many characters.....   I'll split it into two chunks.   I'll also put it under a "spoiler" so it won't clog up all the space for people who are reading this thread.
- - -  - - - - - - - - -

Chunk 1

Spoiler: show

But what I am most interested in at the moment, is just the math involved in figuring out how to maximize the ability of the solution to absorb/release energy over a desired temperature range.

    Let's minimise peoples disinterest and just do something for the simple situation.
   
You have the simple chemical reaction forming an equilibrium   A Z

   Since the reaction is 1 molecule  to/from  1 other molecule,  we have

N_A   + N_Z  =  N   = a constant at all times 

[Eqn 1]

   with  N_A = number of A molecules;   N_Z = no. of Z molecules;   N = total number of molecules.

We're going to tackle this situation with the equilibrium constant.  I've considered alternative strategies but this one is reasonably easy, makes good progress and produces answers that seem to agree with what you might have expected.

K  =   [Z] / [A]   =  ratio of concentrations at equilibrium.   

[Eqn 2]

      The way you've described it we can assume the reactants and products are in the same container,  possibly in some solvent....  Whatever details apply - they slosh around in or contribute to some fluid of volume V.   The volume might change slightly as products and reactants come and go, or temperature causes expansion and contraction, that won't matter.  All that matters is that at any instant, the products and reactants are in the same volume.   So we get some cancellation and can replace concentrations with numbers of molecules in the equilibrium constant.

     Hence,    K  =  N_Z  / N_A.     

[Eqn 3]

Combine [Eqn 1]  and [Eqn 3]  to obtain,

    N_A  =     

[Eqn 4]

That's a decent start,  we know the number of A molecules that should be present at any time.   Note that the equilibrium constant K  is a not a constant, it changes depending on various parameters.

   The balance of the reaction equilibrium will be entirely determined by K.   We take this expression for K
   

[Eqn 5]

A reference for this result was given in a previous post.   Here's a slightly different one:  https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Fundamentals_of_Thermodynamics/Free_Energy_and_Equilibrium
     This is where we will make contact with your (@chiralSPO) ideas of varying the enthalpy and entropy.  We haven't forgotten about it.  All the information and contact we shall need will come through the equilibrium constant.  That constant tells us how the equilibrium shifts. 
     We're going to need a moment to explain this formula [Eqn 5].  As already mentioned in an earlier post, some texts would have used ΔG° in place of ΔG in that formula and everyone has a slightly different view on what ΔG° is supposed to be.   In the worst case, some texts insist that ΔG° has been determined at a standard reference temperature, in that situation ΔG° is nothing more than a fixed value you can find in a book, a constant regardless of whatever else is happening.
    Now we MUST HAVE a quantity,  ΔG,   which depends on temperature.   Consider for a moment that it doesn't but instead we are using a quantity which I'll call ΔG" and which is just a constant regardless of whatever else is happening.  An equilibrium with K described by [Eqn 3] achieves a perfect balance with an equal amount of product and reactant   if and only if     K = 1.     Next use [Eqn 5] with ΔG = ΔG" = a fixed constant:
    K =  e ^(-ΔG"/RT).   

[Eqn 6]

So   K = 1   if and only if    one of the following holds:   (1)  Temperature T →∞  ,  or else  (2) ΔG" = 0.   We're not interested in infinite temperatures, so we would want ΔG" = 0.   However ΔG" was assumed to be a constant, so using  [Eqn 6] again,  we have   K = e⁰ =  1   regardless of temperature.    The equilibrium never changes, it doesn't respond to temperature at all.   We've discussed this before - it's no use.
   The approximation where a fixed standard value ΔG" was used in [Eqn 5] is acceptable for many situations especially where you are working around standard temperatures AND  ΔG" is not approximately 0.   However, that is not our situation.  We are very likely to want an equilibrium operating around the perfect balance point with K ≈ 1 and hence ΔG" ≈ 0.    We must have a better approximation for the equilibrium constant.
 
   Here's [Eqn 5] again:

[Eqn 5]

That's a perfectly good expression where ΔG  is allowed to be determined by temperature.   Many texts will still call the value that will be inserted into that formula ΔG° but it is fully assumed to be dependant on temperature.  It is just the free energy change that occurs for the forward reaction  A → Z  at a given temperature.   (Further discussion of this removed, this post is already too long).
Anyway, this ΔG is calculated as follows:
     ΔG =  ΔH - TΔS

[Eqn 7]

   As @chiralSPO  mentioned,   ΔH   and  ΔS  do have some temperature dependence but over the range of temperatures we are considering it's not important.   ΔH ≈ ΔH° = constant   and  ΔS ≈ ΔS° = constant.     We obtain,
    ΔG°  =   ΔG°(T) =     ΔH° - TΔS°

[Eqn 8]

    Note that this ΔG°  does retain some temperature dependence and it is sufficient for our purposes,   the quantity T is a variable and appears as the multiple of ΔS°.   (It's only the more extreme definitions of ΔG° where the temperature would have been locked at one value and become what I previously described as ΔG" ).

   I appreciate we might lose some people here.  It would help to get a diagram to show what we're doing:

Sketch to show  ΔG°   and  K  as functions of T  given by Eqn 8 and Eqn 5

graph2.png (19.6 kB . 1495x935 - viewed 2402 times)

Features of the sketch:    K = 0 at T=0 kelvin.      K  → finite value = e^ΔS°/R  as T →∞.   It has a point of inflection around  T=ΔH° / 2R.     
   ΔG°  is just a straight line,  with  ΔG° = 0  at  temp. T₀ = ΔH°/ΔS°   (easily obtained from Eqn 8 ).
   The point of inflection for K(T) is not too important but is easily obtained from the expression   K = e^-ΔG°/RT  =  (e^ΔS°/R) .  e^-ΔH°/RT    =    (constant) . e^-(ΔH°/RT)   and differentiating twice with respect to  T.   
     The most important characteristic is that K passes through 1  at T = T₀  this is where the equilibrium would have an equal amount of product and reactant.

---End of Chunk 1 ---

6

Chemistry / Re: How can I find the optimum ΔH and ΔS for passive T control?

« on: 30/05/2022 15:25:22 »

Hi again.

I'll try and get it done but I'm off to the big city tomorrow and I know I won't be writing much here for at least a day.

    I'm really sorry but I'm not going to get this done any time soon.

   I'm already on pages of work.   Therefore no-one will want to read it, not even @chiralSPO .
I do not have the time to complete.  I'm way behind on several other important tasks.  The whole thing is now begining to cause me some stress and that's not how a forum should be. 
   I'm very sorry @chiralSPO  but you should not wait for a detailed response from me, it would safer to assume it won't happen.

Apologies,  Eternal Student.

7

Chemistry / Re: How can I find the optimum ΔH and ΔS for passive T control?

« on: 27/05/2022 09:00:21 »

While it's true that ΔS  ≈  ΔH / 300 is pretty much the criterion for optimal regulation at 300K, it is just restating the requirement that the melting point is near 300.

And tables of melting points are easier to find than tables of ΔS

It's also worth thinking about the "units".
As Alan didn't notice,  ΔH is typically given per mole or per unit mass.
On a practical basis you might want to look at the value per cubic metre.
And then you need to look at   ΔH per £.

 

8

Chemistry / Re: How can I find the optimum ΔH and ΔS for passive T control?

« on: 27/05/2022 04:40:39 »

Hi again.

   I've had another look at the mathematics and tried a few approaches.  I've got something that looks reasonable but it's taking too long to write it down and mark up the equations with LaTex.    I'll try and get it done but I'm off to the big city tomorrow and I know I won't be writing much here for at least a day.

    The final results I can write down.   They are dissapointing and probably only tell you what you should have already been able to guess.   On the other hand... they make sense and that suggests I haven't gone too far wrong.   Glass half-full or half-empty.

1.   You want to have as large an amount of Substance A (or equivalently of substance Z) as possible.    i.e.  Whatever you do, do lots of it -  make the biggest system you can.

2.    You want a reaction with as large a ΔH  as you can get.   Infinity would be great but otherwise just as large as you can get.

3.    What about ΔS?    If you want maximum response from the system at room temperature   (say about 300 Kelvin).   Then you will need ΔS  ≈  ΔH / 300.       More generally the system will act most effectively as a buffer for the room temperature when  ΔS  =   (whatever your value of ΔH is)   divided by (whatever value of T you're trying to hold the room at).   Changing the value of ΔS from that value has a rapid and detrimental effect:   The system becomes almost non-changing (and hence unable to buffer the temperature) at your desired room temperature.   Instead it works well as a buffer at a different temperature   (which will be precisely T = ΔH / ΔS ).

    You could have guessed these couldn't you?   The mathematics will show it but it's going to be a day or more.

Best Wishes.

9

Chemistry / Re: How can I find the optimum ΔH and ΔS for passive T control?

« on: 25/05/2022 23:55:43 »

Hi.

This question is inspired

   Well, it is quite a good idea.

ΔG = –RTln([Z]/[A])

    Could you clarify this please?   I'm not sure what your ΔG is,  is it actually ΔG° ?   Are  [Z] and [A] concentrations at equillibirum only?   i.d.k.

This is the conventional equation:
    ΔG =   RT   Ln (Q/K) 
Where ΔG = Gibbs free energy change for the system, (in the forward direction and at the specified concentrations).
Q = quotient of concentrations of products / reactants =  [Z] / [A]
K = chemical equilibrium constant =  Quotient as above but AT EQUILIBRIUM.
   
    Just to clarify this,  this ΔG is a function of 3 variables:   The temperature, T,  and the concentrations [Z] and [A].

[reference:  https://chem.libretexts.org/Courses/Grand_Rapids_Community_College/CHM_120_-_Survey_of_General_Chemistry/7%3A_Equilibrium_and_Thermodynamics/7.11%3A_Gibbs_Free_Energy_and_Equilibrium ]

   There seems to be a K missing in your expression,    much as if you were assuming K always = 1.   
This could be enough to stop your idea working completely.    If  K = 1 always, then the net reaction never shifts forward or backward -  the equillibrium point remains with equal concentrations of products and reactants [Z] = [A]  regardless of what happens.   In this way it won't respond to changes in temperature at all.   
    To re-phrase this  K ≠1.  It is essential that K = K(T) = some function of temperature.

     Using conventional theory,   it seems that we can approximate  K(T) = equillibrium constant at temperature T  as
K(T)     ≈   e ^{-(ΔG°/ RT )}

    This quantity, ΔG°  is not a function of the concentrations of the products and reactants.   At most it is a function of the temperature, T, but more usually the temperature and pressure are also assumed to be standard temp. and pressure.   Since you're interested in changes occurring around room temp. and pressure, it shouldn't be a problem to assume  ΔG°  is just a constant  which you can find in a book for the reaction A → Z.

   Anyway, re-arranging that equation we obtain:   ΔG°  = -RT Ln (K)  =   -RT Ln ([Z]/[A])     where  [Z] and [A] are now only to be taken as the concentrations at equillibrium.     That might have been the equation you were suggesting in your original post.  It matters a lot because, if that was what you were doing,  then when you re-arranged it to find ΔH  I don't think it was the ΔH that you were actually hoping or thinking you'd find.

   Summary:  Sorry that was confusing.  I'm confused and just trying to match up your notation with that used in some other texts on the subject.  I need you to check or explain what it was you were hoping to suggest with your formula  ΔG = -RT Ln ([Z] / [A])  . 

Best Wishes.

10

Physiology & Medicine / Re: why is my skin so sensitive when I have a fever?

« on: 19/05/2022 16:14:22 »

Is this a known effect?

Very much so, especially for flu. It is similar to heightened sensitivity to sound and light, especially when feverish.

Is there a known (or likely) mechanism?
Is there anything I can do to limit it while recovering?

Apparently staying hydrated is a good way to limit it. Ibuprofen helps reduce inflamatory related symptoms, including the skin sensitivity. I found that acetaminophen does a nice job on headaches and fever, but not so helpful with the inflamation.

Benefit of covid: Our altered social practices have seemingly prevented about two years of all the common stuff I/we usually contract each year. Sorry this hasn't been entirely true for you.

11

General Science / Re: Is 2 really prime? If so, why isn't 1?

« on: 01/05/2022 04:11:50 »

There are also prime polynomials (if you ignore imaginary zeroes).
- These are important in telecommunications and encryption schemes

[Spoiler:]

12

General Science / Re: Is 2 really prime? If so, why isn't 1?

« on: 30/04/2022 00:19:02 »

Hi.

I came across another wrinkle: what about –1?

    Your general arguments after this are reasonable.   However, I think it is again just a matter of simplicity and having a set of numbers that are useful for something.   It is possible and useful to confine your attention to what people might call the counting numbers or the Natural Numbers, so we do.   That doesn't mean that mathematicians have never considered generalising the idea of prime numbers and investigating properties like prime factorisation in a structure bigger or more abstract than just the positive counting numbers - they certainly have.

   There is already some terminology you could use to describe a set of things that behave like prime numbers but apply to a much more generalised set of objects than just the Natural numbers.   These things are called "prime elements" and the parent algebraic structure is known as a "Ring".   You seem to be interested in the Ring which is the Integers (positive and negative Naturals with 0,  under  conventional binary operations of + and x).

   See   Wikipedia entry:   Prime elements, if you're interested.   https://en.wikipedia.org/wiki/Prime_element
However, you should note that they exclude "units" which would  include -1  in the ring of Integers,   i.e.   they would directly exclude both  +1  or  -1   from the prime elements in the ring of Integers.  (For what reason?   Similar to excluding 1 from the primes,  it makes it much easier to state an equivalent unique factorisation theorem for the ring of Integers).

    Here's a quick question or puzzle, just for fun.   It relates to the idea you mentioned earlier of eliminating the number 2 from the prime numbers.   You also seemed keen to extend beyond the positive numbers and consider negative numbers but you really don't have to stop there - you can consider Complex integers.
    The Complex Integers or "Gaussian Integers"   are the  Complex number equivalents of integers.   Specifically, the Gaussian integers are the set of all complex numbers of the form   a+bi   where  a and b are integers.
    Just like in ordinary arithmetic with Natural numbers, a prime  (or prime element) of the Gaussian integers is a Gaussian integer,  p,  that is irreducible or cannot be factorised.   Specifically,   if we have  p =  q × r   (where × is just ordinary multiplication of the complex integers q and r)  then  at least one of  q or r must be a unit element.     A  "unit"  is any complex number that lies on a unit circle around 0,  so the only  units in the  Gaussian integers are    +1, -1,  +i, -i.
   The number 2   is a prime in the ordinary integers.   Is it still a prime in the Gaussian integers?   To say that another way, can you factorise the number 2 in the Gaussian integers?

Spoiler: show

  2  =   (1+i) (1-i)   = the product of two Gaussian integers, neither of which are unit elements.   So 2 is not prime in the Gaussian integers.
  As it happens, the Gaussian integers do form a Unique Factorisation Domain.   This means there is a set of prime elements often called Gaussian Primes, all the Gaussian integers can be written as a product of those prime elements and, as always, that factorisation is unique.   However, its prime elements are quite different to the prime numbers of the ordinary Integers.

Best Wishes.

[Spoiler:]

13

General Science / Re: Is 2 really prime? If so, why isn't 1?

« on: 28/04/2022 18:22:32 »

Hi.    Fantastic diagrams.   Great that you're making an effort to engage the audience with some Mathematics etc.
I wish you well.

I'll hide everything else under a spoiler because it's a bit dull and might prevent others from making their comments.

Spoiler: show

   The main reason for not counting 1 as a prime number is that most of the results we have about prime numbers, or more generally about whole numbers, won't work if you tried to state them as they are now and continued to use the term "prime number" in that statement of the result.   The decision not to include 1 as a prime number wasn't really done because of some elaborate definition or way of identifying what the primes are supposed to be.   I don't think following a pattern that emerges from dots had a lot to do with it.    Instead, it was done because it's not all that useful to have 1 included in the set.

    Another way to say this is that there's no reason you couldn't include 1 as a prime number if you want to.   You go right ahead and do that.   You don't even need to make up a good reason like drawing an arrangement of dots.   For whatever reason, you can put the number 1 into the primes if you like.   The only change that will result is that mathematicians will stop quoting their results by referring to "prime numbers" .  Instead they will identify a slightly different set of numbers, let's call them "Q-rimes" and their results will be stated with respect to that.   The Q-rimes will naturally be your Primes excluding the number 1.  So, the only thing that will have happend is that you will have changed the name we apply to describe what is currently called the prime numbers.

    I suppose to finish this I should give at least one example of a result that is useful and easily stated with reference to prime numbers (with 1 excluded but not if 1 is included).

     The fundamental theorem of arithmetic
Every counting number can be written as a product of prime numbers each raised to an appropriate (Natural number) power.   Furthermore, that respresentation is unique up to changing the order in which you perform the multiplication.

Example:     40  =  2³ x  5
   If you try to write 40 as some other product of primes, let's say you allow yourself to use three prime numbers   p, q, r    such that   40  = p^a x q^b x r^c     for some exponents  a,b,c   then you find that you can't,  there's no solution for that.     The fundamental theorem of arithmetic holds.
    However, if you allowed 1 to be a prime number then you can.....   One solution is to set p = 2,   q = 5, r = 1    and a= 3, b = 1, c= 2 .  That will be another representation of the number 40 as a product of primes:    40  =  2³ x  5  x  1²     and so the fundamental theorem of arithmetic doesn't hold.

Is 2 really a prime number?

    Actually 2 is another number that has very unusual properties even though it is prime.   It is often very useful and desirable for mathematicians to consider a subset of primes that doesn't include 2.   They call this set the "odd primes" and several theorems are stated with reference to  "odd primes" instead of just "the primes".  Alternative terms exist for this set and it's quite common not to bother naming the subset and just write a result as holding  "for all primes, p > 2".
   So 2 is a prime number but its certainly not typical of primes and there is a similar set, the odd primes, where you do just exclude it.

Best Wishes.

14

Physics, Astronomy & Cosmology / Re: Can sand/salt permanently molecules absorb resonate frequency?

« on: 22/02/2022 20:28:31 »

The patterns formed on Chladni plates are a result of the properties of the plates themselves, and has nothing to do with the sand/salt/sugar/dust/etc. placed on top for visualization purposes. ..... The pattern that the nodes make is a function of the frequency of the vibration and the size/shape of the plate.

Fully agree, acid test is that the pattern changes with frequency, so there is no memory effect.
PS I use tea leaves, but don’t read anything into that 

15

General Science / Re: co2 bomber extinguisher

« on: 16/01/2022 16:01:51 »

Hi. I come up with extinguisher that can kill forest fires. I call it CO2 bomb.

Looks like American-football ball, but bigger. inside is lots of liquid CO2.
Drop one in the forest fire. Valves will open and release huge amount of liquid CO2, which will become gas. 

Fires need heat fuel and oxygen, you need to remove one to extinguish the fire. You cannot reliably remove the oxygen from a fire for a sustained period, nor the fuel in any reasonable scenario using fire suppressants etc, so you need to remove the heat. Extinguishing the fire may remove the heat from a small fire long enough to cool the fire but a large fire needs cooling down a great deal, this is why the fire brigade spend long periods dampening fires and raking through material piles.

Co2 from solid to gas I would bet has a far smaller  energy requirement than liquid water to gas.

16

General Science / Re: Is this a feasible system for recycling CO2?

« on: 07/01/2022 16:55:47 »

Alan, don't you work with MRIs? I assume you know what happens when a magnet quenches...

Indeed. But suffocation incidents are usually associated with the loading and cooling process, not a subsequent quench. We have exhaust stacks to vent quench gas safely once the magnet is assembled.
My own MRI units  used room-temperature resistive magnets or high-temperature supercons cooled with gaseous helium, but now I'm working with other people's kit, fraught with the dangers of liquid refrigerants.
Fortunately modern MRIs don't use nitrogen - one less problem - and capture helium boiloff, saving a lot of money. Time was that liquid helium was cheaper than beer when vast quantities were used for North Sea divers maintaining oil and gas rigs, but so much was exhausted to the cosmos that it is now more expensive than champagne.

17

Physics, Astronomy & Cosmology / Re: Energy loss in electrolysis

« on: 11/11/2021 08:41:49 »

If you want light energy...

...Heat a piece of quicklime with your oxyhydrogen torch.

18

Technology / Re: What are some low-tech ways to address climate change?

« on: 08/11/2021 07:34:39 »

A great way to reduce co2 or methane if you believe both to be the cause is to shoot animals, people will even pay you for the privilege.

We fish and shoot already.
Animals aren’t a problem because they release only “neutral” carbon into the atmosphere. They release large amounts of CO2 and methane but their carbon source is limited to plants or animals that eat plants and plants get their carbon from the atmosphere. So, their only source of carbon is indirectly from the atmosphere and they only cycle carbon back to the atmosphere from whence it came. They do not cause a net gain in the amount of carbon. They recycle what is already there.

If it were not for animals, decomposition, insects, and fire would continue to recycle CO2 and methane back into the atmosphere in their place. If animals ate non-neutral carbon sources like coal or oil, that would be a problem because they would be releasing carbon from 300 million years ago back into the atmosphere and contribute to the net amount of atmospheric carbon. Fossil fuels are the problem. 

19

General Science / Re: What's 0^0 ?

« on: 03/11/2021 00:53:20 »

Hi.

You're doing well there @hamdani yusuf .
   You've already got enough to see that real numbers raised to some exponenet aren't always equal to a real number.

Sometimes, you can't find any real number that would be suitable.   Sometimes you can find many suitable Real numbers.  It's quite natural to extend the scope of the problem to consider complex numbers but then you can sometimes find an infinite set of numbers as a solution.

You already have enough evidence to recognise that 0⁰  was never required to be a unique Real number.  It's not even required to be a unique Complex number.  You've also presented enough examples to see that it cannot be defined as a real number in any consistent way.   If it has to be anything, it could be a bicycle.  It's just a collection of mathematical symbols we can write down but it's not representative of any numerical value.   ∀6� 12  is another set of symbols that doesn't equal or represent any numerical value.

   Technically,  x^1/n  is defined to be the positive root wherever there was a choice.  So that  9^1/2 is +3 and nothing else.   This is done because it maintains exponentiation as a well defined function for as long as possible.   Anyway, using that it would mean that your answer to q. 4 is wrong.
   If you were given the expression    (x²)^1/2  = x      and then set x= -3    then you would have to deduce that  +3 = -3.   
   Obviously we don't really want anything that silly, so the only possibile resolutions are that we give up on considering a^b  as a well defined function for all a,b ∈ Z   OR ELSE   accept that the given equation   (x²)^1/2  = x  was not valid for all x∈ℜ.
   Mathematics has taken the second option,   the rules of manipulating indicies that we were taught in school do not hold for all real numbers as a base for the exponentiation.  A good teacher might have brought that to the attention of their students but it wouldn't matter much anyway:  As human beings we want to follow patterns and we want to extend these patterns wherever we can, so we would have ignored any warning.
   Thus (x^m)^1/n  = x^m/n  is only a true statement for some values of x.  The original question 4 that I presented was a little misleading.   The most appropriate response should be "Exponentiation cannot follow this rule even though it seems like it should (because otherwise +3 = -3)". 
    I needed this to be considered because there are so many misconceptions and false proofs based on using "rules of indicies" even though these rules do not and cannot hold in the system of mathematics we commonly use.

Best Wishes.

20

General Science / Re: What's 0^0 ?

« on: 02/11/2021 17:49:40 »

Hi.

   The original question was:

What's 0 to the power of 0?

    This is actually a very good question and it's something that isn't easily resolved. 
    It also turns out that whenever something like 0⁰ is encountered in science, it has nearly always arisen as a Limt of x^x  as  x→0+    (approaching 0 from the right).   This limit is defined and does equal 1.   As a consequence of this it has become an un-official convention that  0⁰ = 1   and  regrettably many calculators, like the one you were using, will show you that result.

   Too often we start from a false assumption.  It is easy to imagine that just because we can write some mathematical term down it must have some numerical value.  For example, I have never calculated the value of 10067 + 765409 but I might assume it is some Real number.  We also frequently assume that when there are patterns to follow we must be able to extend those patterns.  For example, whatever my answer to that sum might be it should be bigger than the first number, 10067.
  For exponentiation it's actually much safer if we start by assuming nothing at all.  Do not assume a^b defines any function from (a,b) → ℜ   and  don't even assume that the Real numbers exist.  Instead start from more basic assumptions (axioms).   If we do build up the Real Numbers and develop enough real Analysis to construct the exponential series then we will see that 0⁰ was never defined and indeed it cannot be defined in any consistent way as a Real number.

    I could just spit out some chapters from a textbook on Real Analysis or Complex Analysis that talk about the exponential series but I can't do that any better than the textbooks.  Instead let's put out some minor problems to consider, which might help to identify just how complicated it is to raise numbers to an exponent:

   1.    What  is   4^1/2 ?     Why?
   2.    What is    (-4)^1/2 ?   Is there no real solution?
   4.    By the rules of indicies we have (x²)^1/2  = x   for all x.  This seems reasonable but what happens to the LHS and  RHS  when you let x = -3 ?    Don't we obtain  +3 = -3  ?
   5.    3^π  cannot be written as the  integer root of any integer power.  Specifically  3^π ≠ (^a√3)^b  for any inetgers a,b.   So what is the value of  3^π?   Could it be a negative number?    If that's too easy  consider   (-3)^π .

Best Wishes.

PS,  yes I know question 3 was missing.