The h Bar

heather

12:03 AM

@DanielSank, death by family, it's terrible =P I figured out the problem, nevermind =)

DanielSank

12:22 AM

@heather good

Bernard Meurer

@alarge Intel, ARM, and I believe AMD all use two's complement in their ALUs, so although it's undefined I preferred to just show the guy what was happening. With that said I've amended the last paragraph with some fixes, take a look :)

@dmckee You too go check it out, you're my low-level reference :P

alarge

12:38 AM

@BernardMeurer We're talking about C here and any UB can and often is optimized by the compiler. e.g. kristerw.blogspot.com/2016/02/…

Bernard Meurer

@alarge Indeed, let me read that article, get back to you in a second.

dmckee

@BernardMeurer As far as I know all major architectures for general purpose CPUs use 2s-complement these days. Don't know about GPUs, the smaller micro-controllers and special purposes chips, however.

Bernard Meurer

@alarge That was a nice read, I've amended that to my answer :)

@dmckee Yeah, all ALUs I know use two's complement for it's integer ops, which is why I hadn't even considered 1's complement or things of the kind

user228700

12:58 AM

Hi, everyone :-)
Can anybody please help me to figure out why, for the process of freezing, the free energy equation is written as follows: $$\Delta G = (H_{liquid} - H_{solid}) -T(S_{liquid}-S_{solid})$$
Freezing is the phase transition of liquid to solid. Therefore, shouldn't $\Delta H = (H_{solid} - H_{liquid})$ (same for $\Delta S$) instead of the other way around?

dmckee

Oh, man. Thermodynamic sign conventions are just messed up and I'm never entirely sure what to expect. I blame those 19th century engineers.

Kenshin

Bonjour all

user228700

:-/

user228700

Hey @Kenshin.

Kenshin

hiya Kaumu

@Kaumudi.H the equation you have written appears to be free energy of melting

user228700

1:03 AM

Yup :-| But no, I found it here:

user228700

8

Q: Justification for Freezing Point Depression & Boiling Point Elevation in Solutions?

I was wondering if the following justification for freezing point depression and boiling point elevation are conceptually correct. The reason why I ask this question is because I have been self studying chemistry for a course I will be taking this fall, and I don't have a human reference to check...

user228700

And the O.P most definitely meant to indicate write it for freezing, not melting.

Kenshin

I have read the post

and the user is wrong

user228700

In any case, I'm only trying to understand why the freezing point drops, in terms of entropy.

Kenshin

notice he says, in order to freeze deltaG = Gl - Gs >=0

so clearly his definition of deltaG is not the conventional one

because conventionally deltaG will be negative, not positive for a spontaneous process

So yes, he has switched around his l's and g's but that's ok because he also switched around his definition of a spontanous process from being -ve deltaG to +ve deltaG

user228700

1:07 AM

Yeah, but if $\Delta G = G_{liquid} - G_{solid} ≥0$, then that means $G_{solid} - G_{liquid} ≤ 0$, which is what we want for freezing.

Kenshin

let me double check

@Kaumudi.H if you listen carefully I can explain why he has siwtched the l's and g's

user228700

Okay...

Kenshin

HE is defining Delta G to be Gl - Gs

By such a definition, of course Delta G = (Hl - Hs) - T(Sl - Ss)

Bernard Meurer

I has new name

Kenshin

but if you are using such a defintion for freezing, thjen you are interested in the case where Delta G is +ve

So therefore the user is fine to define delta G in the way he has done

because he is considering a +ve Change in Delta G to be spontanous

in Physics we approach the problem differently to what he has done (but still arive at the same answer)

in Physics we will define Delta G to be Gs - Gl (for freezing)

and by such a defintion Delta G = (Hs - Hl) - T(Ss - Sl)

but using such a defintion ,then we're interested in the case Delta G is -ve

Notice both approaches yield the same result

user228700

1:11 AM

> "In order to freeze the Gibbs free energy of the frozen (solid) phase must be lower than or equal to the Gibbs free energy of the liquid phase"

user228700

^ This would imply a negative free energy change.

Kenshin

wrong

"change" can be defined as Gl - Gs or Gs - Gl

he has defined "change" in a wierd way as Gl - Gs

user228700

OK, hang on.

Kenshin

where as in physics we define "change" as Gs - Gl

the physics approach is more logical

but both yield consisten results so ultimately it doesn't matter

user228700

@Kenshin No, not the result that I'm looking for.

Kenshin

1:13 AM

Yes both yeidl the same results i can 100% guarantteee it

user228700

Okay. Then answer me this: why does the freezing point drop on adding a non-volatile solute to a pure solvent?

Kenshin

Because approach (1) deinfes "change" as Gl - Gs and stipulates change must be +ve, approach (2) defines "change" as Gs - Gl and stipulates change must be -ve

@Kaumudi.H easy

when you add a non-volatile solute to pure solvent the entropy increases

why?

because entropy is really a measure of the number of possible configurations of a system

when you add salt to water, there are now more possible arrangments the salt-water system can take on

and entropy increases

user228700

The arm-wavy argument would be that the number of possible microstates of the system increase.

Kenshin

that's not arm-wavy

user228700

@Kenshin Right.

Kenshin

1:16 AM

so because the microstates increase, entropy increases

user228700

Right.

Kenshin

now I would define deltaG as Gs - Gl

user228700

Okay.

Kenshin

= (Hs -Hl) - T(Ss - Sl)

now consider a cup with the solute and wihtout

user228700

Dood, -T(S_s - S_l)

Kenshin

1:18 AM

Dood fixed be patient lol

user228700

:-P Okay.

Kenshin

Now Sl is more in the cup with the solute

user228700

Yeah.

user228700

So $\Delta S$ decreases.

Kenshin

and we require delta G to be -ve (by my definition of Delta G)

so with the solute (Ss - Sl) is more negative than without the solute

user228700

1:19 AM

(Whilst $\Delta H$ remains the same, more or less, so let's not worry about it)

Bernard Meurer

Come on people

Kenshin

Therefore - T(Ss-Sl) is more positive with the solute

Bernard Meurer

I'm so close to 1000 rep on SO

gimme love

user228700

@Kenshin I don't think this is a great way to approach it, because $\Delta H$ doesn't change.

Kenshin

@Kaumudi.H we're seeing if we need more or less heat to the system to make the process spontanous now that we have added salt

user228700

1:21 AM

No, not heat.

Kenshin

yes heat

user228700

Let's focus on the temperature.

user228700

$\Delta H$ is enthalpy, not heat.

Kenshin

@Kaumudi.H what does enthalpy mean to you

if not heat?

according to wikipedia, enthalpy is "the thermodynamic quantity equivalent to the total heat content of a system."

thus delta H is the change in heat

user228700

Enthalpy is the total heat content of a given system: the energy in its bonds and such. Heat is energy in transit

Kenshin

1:25 AM

correct, and DeltaH is the "change in heat" which is indeed the energy lost in transit

so indeed we can consider Hs - Hl as being the "heat" gained by the system

user228700

Gimme a minute.

user228700

Alright, I am still an amateur, because of which I dunno how to explain properly, why $\Delta H$ doesn't change despite the fact that heat is supplied to the system, since temperature most definitely changes.

Kenshin

@Kaumudi.H I can explain that

@Kaumudi.H delta H is the heat gained by the system in the process

user228700

$\Delta H$ is roughly a measure of the difference in energies of the bonds b/w the liquid water molecules in the pure solvent and the bonds b/w the liquid water molecules in the solution.

Kenshin

wrong

Delta H is between the SOLID and the LIQUID

Delta H = Hs - Hl

Delta G = Delta H - T Delta S

Now we assume Delta H doesn't depend on whether there is disolved solute or not

that is what we mean when we say Delta H does't change

H does change

user228700

1:35 AM

Sure, yeah, okay, that was wrong (sheesh :-/) but we're comparing $\Delta G_{pure}$ and $\Delta G_{impure}$. I'm saying that in both cases, $\Delta H$ remains the same. Pure ans impure.

Kenshin

yes

but Delta H isn't H pure - H impure

Delta H = Hs - Hl

and you have a Delta H for the pure and you have a Delta H for the impure

we are just assuming the Delta H for the pure is the same as the Delta H for the impure

so @Kaumudi.H let us go back to my explanation of freezing point depression then

user228700

@Kenshin Yeah.

Kenshin

So we know that Delta G = Delta H - T Delta S, where Delta S = Ss - Sl

user228700

So we don't need to consider $\Delta H$ at all, since it's the same in both cases

Kenshin

not quite just wait

user228700

1:38 AM

No, why are u taking it into account when it's the same in both cases?

Kenshin

Now we know Sl in the impure is greater than Sl in the pure

@Kaumudi.H don't worry we'll get to it

user228700

Okay.

Kenshin

So we therefore know that Delta S is more negative in the impure than the pure

So therefore we know that -TDeltaS is more positive in the impure than the pure

user228700

U mean "So therefore we know that -TDeltaS is more positive in the impure than the pure" ?

Kenshin

yep just checking ur paying attention ;p

user228700

1:40 AM

-__- Sure.

Kenshin

ok now for the process to occur, we require Delta G to be -ve

user228700

Yeah.

Kenshin

This means for the process to occur, Delta H must be more negative in the impure than the pure

(since -TDeltaS is more positive in the impure than the pure)

user228700

I don't agree but go on.

Kenshin

No this is a crucial step

Delta G = Delta H - TDeltaS

and if Delta G is negative

and -TDeltaS is more positve in the impure

if the process is to occur (that is if Delta G is to be negative), then Delta H MUST be more negative in the impure

Bernardo Meurer

1:43 AM

@dmckee Help me out here

I got a simple question

user228700

I just asked:

Bernardo Meurer

stackoverflow.com/review/suggested-edits/14627789

Was that a bad edit?

dmckee

Stupid answers and dumb looks are free.

user228700

> "No, why are u taking it into account when it's the same in both cases?

Kenshin

^@Kaumudi.H and this is why

Bernardo Meurer

1:45 AM

@dmckee Is that an answer to my question? :P

dmckee

@BernardoMeurer Mostly a good one, though you changed some stylistic choices that I would have left alone on the basis that deferring to the author is the default.

user228700

Dood, that's not the same as "taking into account when both are the same", that's arguing that both are different!

Bernardo Meurer

@dmckee I did it mostly because he seemed to not have the best ability with English. As someone who once didn't speak the language as well as I do now I remember really appreciating people fixing those things, I'll keep that in mind though :)

Kenshin

@Kaumudi.H I can explain this point later once you understand the whole thing

user228700

So argue that the temperature drops when $\Delta H$ is the same. That's where I'm coming from and JR spent a lot of time beating that over my head.

dmckee

1:47 AM

@BernardoMeurer Fair enough. I hit accept on the review. At this rate I should get some reviewing badges on Stack Overflow by ::checks watch:: about 2027.

Kenshin

@Kaumudi.H ok let's assume Delta H is the same

@Kaumudi.H I can derive T will drop

user228700

Okay.

Kenshin

@Kaumudi.H so we agree that -Delta S is more positive in the impure

user228700

Yeah.

Kenshin

and we agree that for the process to occur, Delta G must be -ve in both

user228700

1:48 AM

Yup.

Bernardo Meurer

@dmckee Ah, thanks for that :)

Kenshin

and we agree Delta H doesn't matter (same in both)

so let's for simplicity just say Delta H is 0 (but will work for any number)

then Delta G = - TDelta S

now -Delta S is more positive in the impure

therefore

T must be lower in the impure

Bernardo Meurer

@dmckee Hopefully SO will still be around in 2027

Kenshin

to ensure that T* -Delta S is balanced

That is, Because the -DeltaS component is higher, the T component must be lower

in order for the process to remain spontanous

user228700

Let's forget about that negative sign for a moment-it needs to be present for $\Delta G$ to be negative anyway. $\Delta S$ decreases, yeah?

user228700

1:53 AM

So to balance that decrease, the temperature must rise, not fall. This only works in our favor when we define $\Delta S$ to be the other way around, which is what the O.P did, which is also wrong (I think).

user228700

So Idk how to argue that the temperature drops.

Kenshin

gimme a sec

the -ve's ar egetting confusing let's put some numbers to it

Let's say Delta H = -10 in both (-ve because heat is lost during freezing)

user228700

No, let's revisit this in about 4.5 hours. JR will also be around and 3 are better than 2.

Kenshin

For the pure let's say Delta S = Ss - Sl = 5

and fo rthe impure let's say Delta S = Ss - Sl = 4 (less because Sl is higher)

user228700

Please? It will be more productive.

Kenshin

1:58 AM

so for the pure we have,

@Kaumudi.H this will be the best way fo ryou to see why T decreases

but yeah we can wait for John but if you choose a -ve value of G and solve for T you will see why T must be less

this is basically what the solution is showing on chemistry.SE

user228700

Hmm, okay.

user228700

I'll try that on my own and yes, let's please wait for @JohnRennie. Thanks! _/\ _

Kenshin

2:13 AM

np

user228700

Hey, dyou know anything about the medical condition "edema"?

Kenshin

yes

it can be caused by reduced albumin in the blood (protein defieincy)_

user228700

Does an excess of salt in one's diet cause it?

Kenshin

there are several causes

if you have excess salt intake your blood pressure will increase

user228700

'Cause I found this on a website called "medical daily":

user228700

2:16 AM

> "Excessive salt in the diet can cause a symptom known as edema. As reported by Medical News Today, edema is characterized by swelling, particularly in the hands, arms, ankles, legs, and feet..."

Kenshin

yep, it can be caused by (1) reduced cardiac output

(2) reduced blood proteins (blood seeps out of blood vessels by osmosis)

user228700

No, no, I'm only interested in this particular cause (salt)

Kenshin

when you eat salt, your body compensates with thirst

user228700

Edema = Puffing up of cells. Presumably, this would be caused by insufficient intake of salt, right? In this case, the cells in our body are surrounded by a hypotonic solution, which causes the water to flow into the cell, resulting in their swelling, yes?

Kenshin

because the mechanism for controlling blood volume (thirst etc.) is by detecting the salt concentration in the blood

@Kaumudi.H edema isn't puffing up of cells

edema is when the blood leaves the blood vessels into the surrounding tissue

so it is actually between the cells

in the interstiitial space

user228700

2:18 AM

Ohh.

Kenshin

there are conditions where cells puff up and lyse (and yes this is ude to osmosis) but this isn't oedema\

user228700

@Kenshin OK..?

Kenshin

and usually you won't see any visible signs of cells puffing up

user228700

Alright, so why does an excess in salt intake cause edema?

Kenshin

through hypertension I believe

your body compensates for the hypernatremia (increased salt) by retaining water

this increased water then seeps into the surrounding tissues due to hydrostatic pressure

user228700

2:20 AM

Uhh, okay.

user228700

Thanks! :-)

Kenshin

So not due to osmosis

user228700

Yeah.

Kenshin

but if you have low albumin in the blood, you do get edema due to osmosis

user228700

No pralem, I am not interested in learning about that :-P

Kenshin

2:21 AM

(because albumin is there to keep blood in the blood vessles through osmosis. if you loose albumin, blood seaps out)

lol

why u hate bio so much?

user228700

No, no, I don't hate it. I have a gazillion things to retain in my mind till June of next year, u see. I don't want to trouble my brain with information I could learn later as well.

Kenshin

ah good point

user228700

:-) 'Kay, thanks!

user228700

Actually, no, I'm not done yet. Can u please help me some more, with Osmosis?

Kenshin

yea

user228700

2:25 AM

Here:

user228700

user228700

Additional information that might help: some artificial SPMs are made out of $Cu_2[Fe{(CN)}_6]$

Kenshin

wow point (ii) at the top is just wrong

user228700

@Kenshin :-P Dood, it has been established already that my textbooks suck pretty bad. I always need to check with people here.

Kenshin

yeah

even the wording for the solved example is bad

user228700

2:30 AM

Yeah, in that there is no proper wording, really.

Kenshin

"in deep blue colour will obtain"

user228700

x'D Yeah. Never mind the wording.

Kenshin

i dunno this one, it depends on what the SPM allows through

user228700

Can u help me to figure out why a deep blue coloring would be obtained if the membrane was fully permeable?

Kenshin

well copper-ammonia complex is deep blue

user228700

2:33 AM

I guess the ions form that complex.

Kenshin

you can't figure that one out

you just have to know it

Now if Cu reacts with NH3 to form the copper-ammonia complex, it will form the deep blue

user228700

Yeah, OK. Simple enough, I guess. I was confusing this with the material of the SPM itself.

Kenshin

yea

user228700

Alright, thanks again!

Kenshin

np

G. Bergeron

2:52 AM

It seems there is no perfect solution to work desk ergonomy...

Bernardo Meurer

@G.Bergeron Yes there is, it's called retirement

G. Bergeron

@BernardoMeurer I don't want that! But I'd like a a setup where I could just lay on my back with a screen on the ceiling and half keyboards under each hand...

Bernardo Meurer

I don't want a chem exam monday

Life's tough

user116211

-1

Q: How fast could Earth spin and sustain life?

When dinosaurs roamed the Earth was the Earth rotating faster creating less gravity to allow for their size? How fast could have the Earth spin and be slowed down by an asteroid to the current rotational speed not completely wiping all life out? How much would 100 kg weigh then?

homework-and-exercises forces gravity centrifugal-force asteroids

user116211

off-topic?

user116211

3:05 AM

Hey @bernardo! What's your plan for Christmas?

Bernardo Meurer

@MAFIA36790 Going to Brazil, hanging out with my Grandma

You?

user116211

aha!

Bernardo Meurer

(Not sure whether or not you celebrate it)

user116211

@BernardoMeurer I have to take my last exam in 23; and yes, we celebrate ;)

user116211

No major plans though.

Bernardo Meurer

3:10 AM

Cool :)

Good luck on the exam!

user116211

@BernardoMeurer Why the hell in the world is the chem course not ending? I never got why your university is giving course on chem for your branch ;/

user116211

@BernardoMeurer Thanks!

Bernardo Meurer

It ends Monday

and I have no clue man

it's a nightmare

user116211

@BernardoMeurer good

user116211

@BernardoMeurer I know

G. Bergeron

3:15 AM

@Secret "Theorem 1: (M,∘)(M,∘) is a semigroup iff ∀k,l∈S,∃a∀k,l∈S,∃a, such that aTli=TkiaTli=Tki. Equivalently. ∀a,b∈S∀a,b∈S, (ab)S=aTbi(ab)S=aTbi. " This is not true. Your proof is assuming any element can be decomposed into a product of two elements... Secondly, the converse is not true! In your matrix notation: $((ab)S)_i=T_{(ab)i}\neq a T_{bi}$.

@Secret Finally, it's much better than last time, but the notation is still not consistent throughout your post and extremely convoluted to describe what ends up being the left and right action of the magma on itself. I had a really hard time following your argument.

@BernardoMeurer It is gonna be like that until it starts becoming quite the opposite, once realize the power and the solidity in it's arguments it has. Unless, you opt out before...

Bernardo Meurer

@G.Bergeron My problem isn't with the argument, it's with the fact that I can't do it and it really frustrates me

even when it's something that I feel like I know well I fail to prove simple things

Like the Trace example earlier

user116211

I know sometimes it is itchy; but this is mathematics.

user116211

In transplanting theorem, there are two parts to be proved: existence and uniqueness.

user116211

Although both the parts are more or less same; but this is the rigorous way to do it; even though it is very evident at the beginning.

user116211

Initially, it took me over an hour to comprehend why the author bothered to do the more or less same thing twice; but this is the way it should be done.

G. Bergeron

3:30 AM

@BernardoMeurer Get yourself exposed to it more is my best recommandation. This is to learn why an "evident" is not all evident in terms of only what you supposed. Get to see all the nasty counter exemples in math that leads to the more rigorous formal definitions. e.g. : What is the problem with naive set theory, why is a vector space not always isomorphic to the dual of it's dual, why are smooth functions not necessarily analytic, why do you need branch cuts in complex analysis...

... why can't you always permute the order of operations like limits, infinite sums and integrals?

What is meant in math when an object is said to be "well-defined"?

Bernardo Meurer

3:43 AM

Wat? A vector space is not isomorphic to the dual of it's dual?

ACuriousMind

@BernardoMeurer It is iff it is finite-dimensional

Bernardo Meurer

@ACuriousMind Why on earth are you still up?

And also, I always forget that infinite dimensional spaces are a thing

@ACuriousMind Go like my answer, I'll buy you a beer for no particular reason if you do

user116211

@BernardoMeurer I can give proofs that he doesn't sleep at night.

dmckee

4:16 AM

@ACuriousMind In other words infinities are jerks: they go around messing up all kind of beautiful stuff for no good reason. Bunch of punks.

user116211

Almost all introductory linear algebra books make their treatises on finite dimensional vector spaces. It's actually implied from the very beginning; most often we get introduced to infinite dimensional vector spaces when studying functional analysis.

DanielSank

@BernardoMeurer they're really cool

user116211

@DanielSank o/

DanielSank

\o

Bernardo Meurer

@DanielSank Go review my edit and answer :P

It's 4:33 AM, I got at university at 9:00 AM

user116211

4:32 AM

BTW, @dmckee remember the Max-Born question I asked and you commented? I re-read the Feynman lectures where I could remember there was some sort of this discussion; well it is better put and discussed there.

user116211

@BernardoMeurer Sleep, sleep.

user116211

Max-Born is getting duller and duller, hmm :|

user116211

I want to continue reading it as there are so much elements in it but it's way too dull ;/

Bernardo Meurer

@MAFIA36790 Studying Chem

user116211

@BernardoMeurer You have to :{

DanielSank

4:40 AM

@BernardoMeurer still with family.

@MAFIA36790 the infinite dimensional case is mostly the same.

user116211

@DanielSank okay; apart from some physics y touch on Hilbert Space, I still am not too much accustomed to the infinite-dimensional case.

DanielSank

5:46 AM

@MAFIA36790 It's not that complicated.

Consider the set of functions $\mathbb{R} \rightarrow \mathbb{R}$.

If you add two such functions, you get another one, and you can multiply by scalars.

Therefore, it's a vector space.

No big deal.

Consider one of these functions $f$.

user116211

good.

DanielSank

Suppose you want to express this as a sum over a basis.

You know what that means?

user116211

@DanielSank sure.

user228700

This is the definition for the term "field of view" (optics) given in my textbook:

user228700

> "The region in which the observer's eye must be present in order to view the image"

DanielSank

5:58 AM

Ok, so you know that if I say a vector $v$ is $(1, 5, 3)$, that's just one possible representation of $v$, I.e. in a particular basis.

user228700

OK, nvm.

user116211

@DanielSank so far good.

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