The h Bar

Kyle Kanos

00:01

The LP (from "long playing" or "long play") is an analog sound storage medium, a vinyl record format characterized by a speed of 33 1⁄3 rpm, a 12 or 10 inch (30 or 25 cm) diameter, and use of the "microgroove" groove specification. Introduced by Columbia in 1948, it was soon adopted as a new standard by the entire record industry. Apart from a few relatively minor refinements and the important later addition of stereophonic sound, it has remained the standard format for vinyl albums. == Format advantages == At the time the LP was introduced, nearly all phonograph records for home use were made...

?

ACuriousMind

@rob heh

Bernard Meurer

@KyleKanos Just got a new Talking Heads LP

rob

@ACuriousMind If I've told that joke once, I've told it an infinite number of times

heather

@DanielSank, judging from the way I did on the problems, I did not learn calculus in a day. Rather, I learned enough to think I knew it in a day. I'll keep working on it. =)

@rob, I don't understand...=)

Kyle Kanos

@BernardMeurer I don't know who Talking Heads is/are

Bernard Meurer

00:03

@KyleKanos What?

Dude

rob

The Banach–Tarski Paradox

►

Bernard Meurer

@KyleKanos wat

Kyle Kanos

@BernardMeurer Not unsurprising given my choice of music

rob

@heather Worth the time.

Bernard Meurer

"Life During Wartime" Talking Heads LIVE

►

@KyleKanos Psycho Killer? No?

Kyle Kanos

00:04

I know the movie Psycho

Sorta.

Never watched it, but I'm aware of its existence

Bernard Meurer

How old were you in the 80's?

Talking Heads - PsychoKiller

►

Kyle Kanos

@BernardMeurer Between -3 and +7

Bernard Meurer

bah

Okay, I thought you were older

So yeah

Give them a listen :)

Kyle Kanos

I got married at 24. Had my first kid at 25

heather

@rob, okay, I'll watch it =) thanks

Kyle Kanos

00:06

That is what makes me seem older

Bernard Meurer

@KyleKanos Are you tall and gorgeous?

Kyle Kanos

@BernardMeurer 5'9" isn't tall, so only half-yes there

Bernard Meurer

@KyleKanos Hmmmm Ivy League school?

Kyle Kanos

No

Emilio Pisanty

00:41

Bla bla bla bla bla bla... — Marty Green 8 mins ago

↑ wow, discussion got constructive fast.

Bernard Meurer

@EmilioPisanty Where's the bla bla comment?

ACuriousMind

A grue ate it.

Bernard Meurer

@ACuriousMind You deleted it?

ACuriousMind

@BernardMeurer Duh.

Bernard Meurer

@ACuriousMind Why you delete everything man :/

ACuriousMind

00:56

...because that comment was the very definition of "not constructive"?

Bernard Meurer

@ACuriousMind Meh, it wasn't offensive, I don't see why deleting it

ACuriousMind

Because it served no purpose. Having to dig through comment threads to find the few that actually relate to the post that's being commented on and have something relevant to say is rather annoying, and detracts from the Q&A model.

If you look at the available flags for comments, you'll see that "rude/offensive" is only one among "not constructive", "too chatty" and "obsolete".

Bernard Meurer

@ACuriousMind Troll

Emilio Pisanty

@ACuriousMind I think it served a pretty clear purpose, I don't think it should have been deleted.

Bernard Meurer

^

ACuriousMind

01:00

@BernardMeurer ?

Emilio Pisanty

It shows quite clearly that the OP has no answer for the clear and concrete criticisms offered.

Bernard Meurer

The Ukranian lady that cleans stuff here in the residence keeps coming in my room when I'm still sleeping, and then goes away

I'm going to slap a poster of Stalin on my door, see if that helps

ebay.com/itm/Vintage-oil-painting-man-portrait-Joseph-Stalin-/…

This might just be the funniest Stalin ever

Wtf is that even

ACuriousMind

@EmilioPisanty I don't see how it's better to leave it there, since no response effectively says the same.

Bernard Meurer

@ACuriousMind 0 and None are not the same thing

Emilio Pisanty

@ACuriousMind ^ that

Bernard Meurer

01:04

in other terms, $\{\}\neq \{0\}$

ACuriousMind

@EmilioPisanty How does it impact my judgement of the post itself? One judges a post on whether or not the presented arguments are sufficient to support what it says, not on whether the user who wrote it might have a better way of explaining it that they didn't tell us. It would impact my judgement of the user, but we try very hard to not make things about users

Emilio Pisanty

@ACuriousMind If nothing else, it marks the discussion in the comments as concluded.

I do see your argument, though.

Not that OP will be very pleased at all, I imagine, with the disappearance of the comment.

ACuriousMind

@EmilioPisanty That's not your cross to bear (and I would advise not to engage should he try to fault you for it).

Emilio Pisanty

@ACuriousMind No, I've had plenty of particularly souring interactions with OP, and I have no appetite for more of those. I posted that comment, and one here, because I think it is important that that argument is re-stated, but I hope I won't have to engage in any more discussions.

ACuriousMind

Another one of these questions?

Emilio Pisanty

01:13

@ACuriousMind yup

fresh off the press

ACuriousMind

And how is it different to the Oct 5 one?

It's the same question, just the order of the hydrogen example and the tungsten filament are reversed.

Emilio Pisanty

@ACuriousMind Not quite sure. It makes explicit references to hydrogen atoms, which the 5/10 one doesn't (much)

ACuriousMind

> Let me clarify: in a hydrogen atom in the ground state, I would say there are NO charge density fluctuations. But if there is a superposition of the ground state and a 2p state, I would say the charge distribution fluctuates sinusoidally at the difference frequency of the two states. THAT is the kind of charge density fluctuation I am asking about for in the tungsten filament. Does it or doesnt' it?

from the "edit" on the Oct 5 question.

Bernard Meurer

Can a GameBoy Color play original GameBoy games?

Emilio Pisanty

@BernardMeurer yes

@ACuriousMind If you want to close as a dupe I won't object

Bernard Meurer

01:17

@EmilioPisanty Asking because I want to get one just to play Tetris, and the original ones are far too expensive

50€ is too much

ACuriousMind

...the originals are more expensive that the color one by now?

Huh.

Bernard Meurer

@ACuriousMind Yep, rarer

Emilio Pisanty

I won't vote to close as a duplicate because it'd be an unilateral closure and OP would take exceptionally badly to that.

ACuriousMind

@EmilioPisanty Oh, right, you got that gold badge, too

Bernard Meurer

The real money will go into finding an original tetris

best game ever made

Emilio Pisanty

01:20

I'm not convinced that the formal text of the question is quite close enough to dupe-close. However, it's pretty clear to me that it's the exact same underlying question, which is only just waiting to boil over in the comments (and eventually as an incorrect self-answer saying what OP already believes).

(I hope I'm wrong, btw - but I see no signs of this go-round being any different.)

ACuriousMind

I don't see that playing groundhog day with this question will be productive. I closed it and asked OP to clarify how this question is different because I can't tell after reading both. The sentences are different, but I would write the exact same answer to both.

Argon

I am supposed to figure out $[p_x^2, x^3]$, but I am not clear as to how to do this exactly, Should I start off with $[p_x, x]=-i\hbar$ and work back using $[A,BC]=[A,B]C+B[A,C]$ and $[AB,C] = [A,C]B+A[B,C]$?

Emilio Pisanty

@ACuriousMind fair enough

ACuriousMind

@Argon Yes, that's most likely what you're supposed to do.

Argon

@ACuriousMind Thanks.

Emilio Pisanty

01:23

@Argon you're also supposed to suffer through the algebra and then be grateful for the suffering =P

Argon

@EmilioPisanty Well, the first part is working, so I guess I'm halfway there.

Emilio Pisanty

The only such commutator really worth calculating is $[p,f(x)]$.

ACuriousMind

I guess you could, in this case, also compute the classical Poisson bracket and just slap an $\mathrm{i}\hbar$ on the result, but that's a bad habit to get into

Emilio Pisanty

@ACuriousMind is that guaranteed to return the right answer? if so, when?

ACuriousMind

@EmilioPisanty It's not guaranteed, that's why it's a bad habit.

Groenewold and van Howe showed that there is no map from classical to quantum observables that turns the Poisson bracket into the commutator for all observables

If you want such a map, you must deform the Poisson bracket into the Moyal bracket

Emilio Pisanty

01:27

@ACuriousMind there might still be one for some reduced but broad-enough-to-be-useful class of operators

@ACuriousMind Moyal brackets are cheating

ACuriousMind

But, in this case, I think it would give the correct result because there's no ordering ambiguity, the two observables are simple powers of p and x.

Hm, you might still need to be careful, better not do it :D

Emilio Pisanty

Yeah, it's worth asking @Argon once s/he's finished with the full calculation - does it match the Poisson bracket?

actually, no, wait

ACuriousMind

@EmilioPisanty Yes, but you already get into problems if you try to apply the canonical Poisson->commutator replacement with polar coordinates.

Emilio Pisanty

The Poisson bracket is $\{p^2,x^3\}=-6px^2$, and that's not even hermitian

Argon

What are Poisson brackets?

Emilio Pisanty

01:32

@ACuriousMind the procedure doesn't even make sense

@Argon They're a structure in hamiltonian classical mechanics that is directly analogous to the commutator in qm

Argon

@EmilioPisanty Ah, thanks. I didn't take classical mechanics, so I miss out on a lot of these analogies.

Emilio Pisanty

@Argon If you want anything resembling a reasonable understanding of QM and its underpinnings, it is important to understand classical analytical mechanics all the way up to Hamilton's equations.

Argon

@EmilioPisanty Well, I don't think (quantum) physics will take up much of my future; I am learning electrical and computer engineering. Maybe one day...!

DanielSank

@ACuriousMind isn't there a trick for computing commutators by thinking of them as derivatives?

And isn't that trick 100% kosher?

ACuriousMind

@EmilioPisanty I guess you should...Weyl order that to have a chance at getting the quantum result?

@DanielSank Do you mean the $[p,f(x)] = \frac{\partial f}{\partial x}$ (modulo factors of $\mathrm{i}\hbar$) that Emilio already alluded to?

Emilio Pisanty

01:36

@DanielSank $[p,f(x)]$ is. to go beyond that and stay kosher, as far as I know, you need to do the nested commutators down from whatever power of $p$ you have

@ACuriousMind yeah, but is that guaranteed to get you the right answer?

sounds pretty risky to me

ACuriousMind

@EmilioPisanty No, I don't think it's guaranteed.

Emilio Pisanty

@ACuriousMind anyways, hopefully @Argon can fill us in on what the proper QM answer was

ACuriousMind

But I think that Weyl ordering might work for brackets of the form $\{f(p),g(x)\}$. I seem to very faintly remember something to that effect, but I can't tell you where from

Argon

@EmilioPisanty If my math ever works out. I keep getting zero.

Bernard Meurer

3 months into uni, I now understand 1 in every 100 words you guys use :P

Getting there!

Emilio Pisanty

01:43

@Argon no, that is pretty much certainly wrong.

Argon

So I have $[p_x, x^2] = [p_x, x]x + x[p_x, x] = -2ix\hbar$

Then $[p_x, x^3] = [p_x, x^2]x + x[p_x, x^2]$

Is $xx=x$?

Emilio Pisanty

@Argon what? no.

$x\,x = x^2$.

Argon

Ah woops, that was obvious. Sorry.

@EmilioPisanty -2 or -4?

Looks to me like $-4i\hbar x^2$

Emilio Pisanty

@Argon yeah, but that can't be right, it needs to come out to $-3i\hbar x^2$, I think

your identity for $[p,x^3]$ is wrong

it should read $[p,x^3] = [p,x^2]x + x^2[p,x]$

which does give $-3i\hbar x^2$.

Argon

@EmilioPisanty Good point. Thanks.

ACuriousMind

01:49

Hm, no, I might be thinking of the Moyal bracket again since the product of weyl-ordered $f(x,p)_W g(x,p)_W$ is the Weyl order of their Moyal product, $(f{\star_\hbar} g)(x,p)_W$.

Emilio Pisanty

then $[p^2,x^3]=p[p,x^3]+[p,x^3]p=-3i\hbar(px^2+x^2p)$

So probably the Weyl ordering of the Poisson bracket does get it right?

if I'm getting the Weyl ordering right

@Argon That's a very easy mistake to make with these ones. You avoid it by making sure that the cancellations do work out every time you make an expansion of that type. It really is worth doing that, every time.

Get into the habit early

or well, ok, if you're going off into engineering maybe it's not that important

but it can save your life in an examination

Argon

@EmilioPisanty Hehehe thanks. Can I do something further with $p_x x^2$ and $x^2 p_x$?

Emilio Pisanty

@Argon nope

Argon

Ok. Too bad there isn't a less tedious way to get though this.

Emilio Pisanty

@Argon well, you can, if you insist, and you'll get something like $[p^2,x^3]=-6i\hbar (px^2-i\hbar x)$.

However, any semblance of simplicity that that gives you is 100% misleadingness and 0% simplicity

$px^2$ is not hermitian, and neither is $i \hbar x$.

$px^2+x^2p$ is explicitly hermitian, and that is how you should keep it

@Argon in terms of tedium.... well, yes.

In practice, you're only faced with those relatively rarely on research-grade calculations (in my experience)

Argon

01:56

@EmilioPisanty Ok, I imagine they want me to keep it like this then.

Emilio Pisanty

and when you do, you just buckle down and suck up the tedium

@Argon yes, very much so

DanielSank

@ACuriousMind Yep.

Argon

@EmilioPisanty (You should see the rest of this assignment!) But thanks, ill keep pushing through this.

Emilio Pisanty

@Argon Wherever you are, a large part of building proficiency at the calculations and of getting an intuitive feel for how they work in practice, and for the fundamental structures that underlie them, will always be some significant amount of symbol-pushing

so just keep pushing 'em symbols

it builds character something something

Argon

@EmilioPisanty vignette1.wikia.nocookie.net/candh/images/8/8e/…

Argon

02:20

Why doesn't $H\psi(\vec x, t)=E\psi$ violate energy-time uncertainty? It seems like $\psi$ is in a definite state of energy at a certain time.

ACuriousMind

@Argon The energy-time uncertainty relation probably doesn't mean what you think it means. Have a look at this answer for its correct meaning.

user228700

Ello everyone :-)

user228700

I've a super quick question. When we write $x^y$, $y$ is called the exponent. What's $x$ called?

Argon

@Kaumudi Base?

user228700

Is there another term, perhaps..?

Argon

02:30

Base (exponentiation)

In exponentiation, the base is the number b in an expression of the form bn. == Related terms == The number n is called the exponent and the expression is known formally as exponentiation of b by n or the exponential of n with base b. It is more commonly expressed as "the nth power of b", "b to the nth power" or "b to the power n". For example, the fourth power of 10 is 10,000 because 104 = 10 × 10 × 10 × 10 = 10,000. The term power strictly refers to the entire expression, but is sometimes used to refer to the exponent. == Roots == When the nth power of b equals a number a, or a = bn, then b is...

user228700

Alright, thank you :-) [I thought there was another term. Maybe just my brain screwing around again]

Ramanujan

y has another term…power and index(not so shure)

user228700

@Ramanujan Dude. Stick to 1 room :-P

user228700

(To answer the same question, ie.)

ACuriousMind

@Kaumudi That's what you get for asking the same question in several rooms :P

user228700

02:35

Gots to do what I gots to do to get the question answered, somehow :-P

ACuriousMind

The courteous thing to do would be to ask it first in one room, and then try the other when you didn't get your answer in a reasonable timeframe, otherwise you get several people giving the same answers

Ramanujan

@ACuriousMind you mean iam reason of it?

ACuriousMind

Reason for what?

Ramanujan

To make kaumudi ask in other rooms

ACuriousMind

No, that's not at all what I was saying

I didn't say anything about you, really

user228700

04:05

@ACuriousMind Hm. Will decide the length of the reasonable timeframe and then do this from now in, sorry.

Anonymous

04:21

Can someone check whether my solution to the question is correct or not ? physics.stackexchange.com/a/293338/135977

Anonymous

0

A: In a time varying magnetic field why is the current flowing through a conducting wire connecting two points of a closed conducting loop zero?

Superposition Theorem: The total current in any part of a linear circuit equals the algebraic sum of the currents produced by each source separately. Let resistance per unit angle subtended at the center (in radians) of conducting loop be $\lambda$. Total EMF ($E_{total}$ ) around a circular...

Secret

05:38

Nothing other than having everyone saying it is impossible or not meaningful,. Initially tried to proof what they all said as a no go theorem, but failed as counterexamples continued to arise. The project is then generalised to the more general notion of zero divisors. You might say it is nothing other than I want to plug that hole in division, and wheels are not good enough because there is no legitmate zero inverse.
MY perfectionism personality means if someone said that something is impossible, then I will proof it to be impossible no matter how witty you are. If counterexamples arises, …

koolman

@JohnRennie can you help me in a chemistry problem(concept)

John Rennie

I can try. What's the problem?

koolman

-2

Q: graph of positive azeotrope

In the above graph what does the line AB signifies and why there are two curves ? Please help me , its getting too much confusing .

vapor-pressure

John Rennie

@koolman Have you looked at:

Azeotrope

An azeotrope or a constant boiling mixture is a mixture of two or more liquids whose proportions cannot be altered by simple distillation. This happens because when an azeotrope is boiled, the vapour has the same proportions of constituents as the unboiled mixture. Because their composition is unchanged by distillation, azeotropes are also called (especially in older texts) constant boiling mixtures. The word azeotrope is derived from the Greek words ζέειν (boil) and τρόπος (turning) combined with the prefix α- (no) to give the overall meaning, "no change on boiling". The term "azeotrope" was coined...

koolman

Yeah

Not able to understand phase diagram

@JohnRennie

Anonymous

05:52

@koolman See this link chemguide.co.uk/physical/phaseeqia/nonideal.html It explains the graphs in detail

koolman

Ok

Anonymous

And before that read this chemguide.co.uk/physical/phaseeqia/idealpd.html#top

Anonymous

You will get the whole picture. Even I had a problem initially, but that website helped a lot.

Anonymous

You can ask if you have further doubts :)

koolman

Yeah thank you

user228700

06:01

@JohnRennie@S007: Morning :-)

John Rennie

Morning :-)

Anonymous

@Kaumudi yala yolo :-D

koolman

In this the blue line shows boiling point if liquid(mixture)

John Rennie

@koolman Yes

koolman

At that temperature whole liquid will convert to vapour

John Rennie

06:04

No, it's more complicated than that ...

koolman

Please explain me @JohnRennie

John Rennie

Take your diagram and suppose we start with a 50/50 mixture of A and B

koolman

What is 50/50

John Rennie

50% A and 50% B

koolman

Ok

John Rennie

06:07

The composition shown by the vrtical green line.

koolman

Yes

John Rennie

Because A boils at a lower temperature than B the vapour will contain more A than B. Where the horizontal green line meets the magenta curve shows the composition of the vapour. OK so far?

koolman

I think B boils at lower temp

John Rennie

Oops, yes, sorry I misread the text on the scale. Yes, the vapour contains more B than A.

With that correction does my statement make sense?

user228700

Hey @S007: Do we have magic numbers in our syllabus?

user228700

06:10

(Answer honestly pls :-P)

koolman

Ok @JohnRennie

John Rennie

The Magic Numbers

The Magic Numbers are an English pop rock band comprising two pairs of brothers and sisters from Hanwell. The group was formed in 2002, releasing their debut album titled The Magic Numbers on 13 June 2005. Their follow-up album, Those the Brokes was released on 6 November 2006, The Runaway was released on 6 June 2010, and their most recent album Alias was released on 18 August 2014. The Magic Numbers consists of Romeo Stodart (lead guitar, vocals), his sister Michele (bass guitar, vocals, keyboard), Angela Gannon (melodica, percussion, glockenspiel, vocals) and her brother Sean Gannon (drums)....

koolman

@Kaumudi which syllabus

user228700

:-P Sure, yes, that's ^ exactly what I mean.

user228700

@koolman JEE adv.

John Rennie

06:12

@koolman so because the vapour contains more B than A that means as the liquid boils it loses more B than A and the composition of the liquid changes - the proportion of A increases.

Anonymous

@Kaumudi No. Atleast our teacher never taught it to us in class or I didn't find them in the etoos lectures :-P

Anonymous

Magic number (physics)

In nuclear physics, a magic number is a number of nucleons (either protons or neutrons, separately) such that they are arranged into complete shells within the atomic nucleus. The seven most widely recognized magic numbers as of 2007 are 2, 8, 20, 28, 50, 82, and 126 (sequence A018226 in the OEIS). Atomic nuclei consisting of such a magic number of nucleons have a higher average binding energy per nucleon than one would expect based upon predictions such as the semi-empirical mass formula and are hence more stable against nuclear decay. The unusual stability of isotopes having magic numbers means...

user228700

@S007 Alright, thanks :-)

John Rennie

@koolman The composition of the liquid moves along the blue curve to the right as shown by the black arrow.

user228700

Your teachers at school are good?

koolman

06:13

@JohnRennie yeah

@Kaumudi we don't

Anonymous

@Kaumudi Well, okay okay. Not that good though. For Jee I have to study myself only....

user228700

@S007 I see.

John Rennie

So suppose out mixture starts out boiling at 100C, then as it boils away vapour it will get gradually more A than B and its boiling point will increase. Until finally it's almost pure A and its boiling point will be the boiling point of pure A.

Anonymous

@Kaumudi You go for any tuitions ?

user228700

@S007 Nope. Tuitions, really? Dood, I'm done with my boards!

Anonymous

06:16

@Kaumudi For jee you are studying alone ?

koolman

@JohnRennie for example at 50/50 compositon boiling point is T°C . What will happenif temp. is more than that .

user228700

@S007 Yes. Well, not so alone now I've found this place :-)

user116211

@JohnRennie halp!

user228700

@JohnRennie: Please clone yourself to help all of the students!! We will try to pay for the machine :-P

user116211

06:18

How did they get the differential operator?

Anonymous

@Kaumudi That's good. But the problem most people face without tuitions (including me) is lack of knowledge of what exactly to study and what not to study :-P

Anonymous

@Kaumudi True :-P

John Rennie

Suppose you start with a 50/50 mixture at room temp and heat it. At the temperature T it will start boiling. If you now increase the heat two things happen: (1) the boiling gets more intense and (2) the temperature increases only slowly as you move the system along the blue curve.

user228700

@S007 Well, I have great materials, so.

Anonymous

@Kaumudi Great then :)...and we have John Rennie to help us also :)

Anonymous

06:20

@Ramanujan helllo :)

koolman

@JohnRennie but when boiling starts the liquid composition should also change

John Rennie

@koolman Yes, and that's why the system moves along the blue curve to the right, because the proportion of A to B increases as B is preferentially lost in the vapour.

user228700

:-) @S007: Oh dude. I just remembered learning magic numbers (just the numbers) for BITSAT.

John Rennie

@MAFIA36790 isn't that just a coordinate transformation?

Anonymous

@Kaumudi I don't think it ever came in the history of jee or cbse or icse or bitsat.... I don't want to stuff my mind with unecessary things :-P

koolman

06:22

@JohnRennie and when we can say whole liquid is converted to vapour

@Kaumudi might be the syllabus of bitsat is different

user228700

@S007 Well, it's given in Arihant's book for BITSAT so I'd think it's important to at least know about these numbers.

John Rennie

As you keep boiling the system loses more an more B and the liquid gets gradually closer to being pur A. That is, it moves along the blue curve to the right axis.

user228700

@koolman Not that much.

Anonymous

@Kaumudi Okay...i'll just read through it sometime later :)

John Rennie

@koolman When the liquid becomes pure A then it will completely boils away to vapour.

Anonymous

06:24

@koolman bitsat is just ncert syllabus...much easier than jee :P

koolman

@JohnRennie ok thank you very much

@S007 yeah

Ramanujan

@S007 hello :)

Anonymous

@Ramanujan sup ?

Ramanujan

@S007 means?

Anonymous

@Ramanujan sup= what's up ? :-P no school today ?

Ramanujan

06:28

I didn't go, learning myself today

Anonymous

Me too.I bunked school today :-P

Ramanujan

For what?

Anonymous

To study at home...at school I don't get to learn much anyway :-P

Ramanujan

I bunked because 1.5 hour for lunch and evening there will be free period to practice for tomorrow exam

@S007 you are in 2nd year right?

Anonymous

Same here

Anonymous

06:30

yes

Anonymous

12 th

Ramanujan

CBSE?

Anonymous

icse

Ramanujan

OK,iam from telangana state board

user116211

@JohnRennie The only relation that was given is $\mathbf r\cdot \mathbf s ~=~ \zeta\,.$ I'm not getting how the coordinates are transformed then.

John Rennie

06:35

I don't think I understand the notation ...

user116211

Which one?

user116211

$\mathbf s$ is a unit vector perpendicular to the plane.

user116211

$\mathbf r$ is the position vector of a certain point.

John Rennie

Any of it. If $\zeta$ is a scalar then how can you have $d/d\zeta$

user116211

They said $\mathrm{O\zeta}$ is one of the new set of Cartesian axes along the direction of $\mathbf s\,.$

koolman

06:48

chemguide.co.uk/physical/phaseeqia/nonideal.html

What you can get (although it isn't very useful!) from the mixture is pure water. As ethanol rich vapour is given off from the liquid boiling in the distillation flask, it will eventually lose all the ethanol to leave just water.

What does this para means

@JohnRennie @S007

How can we get pure water

Anonymous

See the graph and the explanation below it. It clearly written.

Anonymous

you start with 95.6 percent mixture

koolman

@S007 i cannot uderstand the last paragraph

How we can get pure water

Anonymous

you start with a mixture > 95.6 percent ethanol

Anonymous

06:53

okay?

Anonymous

then boil

koolman

Okay

Anonymous

the vapour pressure is given by the pink line

Anonymous

on the right hand side

koolman

Yes

And i think vapour composition of water will increase

Anonymous

06:55

So what happens when you boil > 95.6 percent mixture according to the graph and then condense it again ?

koolman

We will reach at constant boiling point

Anonymous

correct

Anonymous

now try the same procedure

Anonymous

when conc <95.6 percent

Anonymous

on the left hand side of the graph

Anonymous

06:57

what happens when you boil it and then recondense ?

koolman

Again will reach at constant boiling point

Anonymous

"You have hit a barrier. It is impossible to get pure ethanol by distiling any mixture of ethanol and water containing less than 95.6% of ethanol."

Anonymous

So now find your mistake

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