The h Bar

user228700

00:03

Hi everyone :-) I've a quick question about electric polarization density.

user228700

heather

@Kaumudi, hello!

user228700

@heather Hi :-)

user228700

In the above picture, I don't quite understand why we say $\Delta p$ and $\Delta V$ instead of just $p$ (or $\mu$ or whatever) and just $V$.

user228700

If we don't include the $\Delta$, we don't get equation 1 :-/

user228700

00:35

@DanielSank: Hi :-) In response to this answer (for the question "Do metals have infinite permittivity?"):

user228700

3

A: Dielectric constant or permittivity of metals

In electromagnetism, absolute permittivity is the measure of the resistance that is encountered when forming an electric field in a medium. In other words, permittivity is a measure of how an electric field affects, and is affected by, a dielectric medium. Yes, metals have infinite permitti...

user228700

You said this:

DanielSank

@Kaumudi meh, the Delta is just a choice of notation, I think.

user228700

> 'Woah, what?! Metals do not have infinite permittivity when you are dealing with signals at nonzero frequency, and this definitely matters in applications."

DanielSank

@Kaumudi true

user228700

00:36

Okay. I don't understand what u meant...

DanielSank

Permittivity depends on frequency.

user228700

What frequency? .__.

user228700

I found this:

user228700

> "Metal's dielectric constant depends on external electromagnetic field. For a low frequency field, metal's dielectric constant is a complex, with a very high imaginary part. As the field frequency increases, its dielectric constant can become a negative value, meaning metal reflects the electromagnetic wave in this frequency domain. When the frequency becomes even higher, metal's dielectric constant becomes a positive value, meaning metal acts as a "normal" dielectric medium."

user228700

Alright, I definitely don't need to delve deep into that :-P

DanielSank

00:40

So you want me to think about how to explain this, or not?

user228700

@DanielSank If it was just a choice a notation, we wouldn't get equation (1) in that differential form and that's the trouble! :-/

user228700

@DanielSank No, thank you! :-)

DanielSank

@Kaumudi why not?

user228700

Um, OK, I actually dunno. How would we obtain it?

DanielSank

If I have an equation of ratios, why can't I go to a small limit?

Emilio Pisanty

00:42

@Kaumudi You're essentially getting to the fact that for a low enough volume $\Delta V$, the electric dipole moment of the charges inside that volume is proportional to $\Delta V$, with some proportionality constant which you denote $\mathbf P$ and you call the polarization

Thus $\Delta \mathbf p = \mathbf P \Delta V$ (for small enough volumes)

You use $\Delta \mathbf p$ to emphasize the fact that it is a small quantity that depends linearly on the small volume $\Delta V$.

DanielSank

^ that is very well said.

It took a long time for me to understand that.

user228700

@EmilioPisanty Ohh, okay.

user228700

So the polarization density is a constant for the given material?

DanielSank

@Kaumudi no

Emilio Pisanty

now, as to why you expect $\Delta \mathbf p$ to depend linearly on $\Delta V$... hang on, I need to sneak off to a corner to scratch my head for a bit

@Kaumudi Not really.

First off - you don't say polarization density.

you just say polarization

wait, no

sometimes you do

DanielSank

00:47

@EmilioPisanty uh, because the charges are all roughly sitting in parabolic potential minima?

Emilio Pisanty

Polarization density

In classical electromagnetism, polarization density (or electric polarization, or simply polarization) is the vector field that expresses the density of permanent or induced electric dipole moments in a dielectric material. When a dielectric is placed in an external electric field, its molecules gain electric dipole moment and the dielectric is said to be polarized. The electric dipole moment induced per unit volume of the dielectric material is called the electric polarization of the dielectric. Polarization density also describes how a material responds to an applied electric field as well as...

most often you just say polarization as far as I usually see

user228700

@EmilioPisanty I thought the two terms are synonymous.

Emilio Pisanty

@Kaumudi yes

@DanielSank wait, what?

user228700

Okay...

DanielSank

@EmilioPisanty no?

Emilio Pisanty

00:48

@Kaumudi anyways, the polarization just tells you how polarized the material is

@DanielSank what minima?

DanielSank

@EmilioPisanty electrostatic.

Oh I was explaining why P should be proportional to E.

user228700

Alright, so polarization depends on the electric field applied and all, according to this:

Emilio Pisanty

@Kaumudi however, most materials (and in particular most dielectrics) will change their polarization depending on conditions

@Kaumudi exactly

for most materials

DanielSank

Actually, I have no idea what Eq. (1) means.

Emilio Pisanty

but you also have things called permanent electrets, which have a frozen polarization that does not depend on the outside conditions

Electret

Electret (formed of elektr- from "electricity" and -et from "magnet") is a dielectric material that has a quasi-permanent electric charge or dipole polarisation. An electret generates internal and external electric fields, and is the electrostatic equivalent of a permanent magnet. Oliver Heaviside coined this term in 1885. Materials with electret properties were, however, already studied since the early 18th century. One particular example is the electrophorus, a device consisting of a slab with electret properties and a separate metal plate. The electrophorus was originally invented by Johan Carl...

and you can also have materials that don't respond linearly to the applied electric field, like e.g. $P_z = \chi E_z +\chi^{(3)} E_z^3$.

user228700

00:52

@EmilioPisanty Oh, I see.

user228700

Hang on, so the electric susceptibility of most materials is a constant?

user228700

What is constant, basically? :-P

DanielSank

@Kaumudi yeah that's roughly true.

Although it can be frequency dependent.

user228700

OK...

DanielSank

::ducks::

user228700

00:55

@DanielSank :-P Yeah, I don't think I need to worry about frequencies right now.

Emilio Pisanty

@DanielSank I'm happy to chuck stuff at you if you're feeling intellectually guilty right now

DanielSank

Please don't.

Emilio Pisanty

@Kaumudi yeah, that's mostly true

DanielSank

Goodness, we should learn more pedestrian linear response before we learn the 3D electromagnetic version!

Physics curricula are quite un optimized.

Emilio Pisanty

A more complete statement goes like: the electric susceptibility of a linear dielectric, at a constant driving frequency, is constant.

user228700

00:57

Alright. I'm a little bit confused, still, because the way u've explained it, I don't get eqn. (1) either...

DanielSank

Oh actually I get it now.

user228700

@EmilioPisanty Yeeah, no, I dunno what "constant driving frequency" is, but that doesn't matter :-P

DanielSank

The polarization density is the total dipole moment over the volume.

Emilio Pisanty

It might still depend on the position (unless the material is homogeneous) and on the direction of the electric field (unless the material is isotropic).

DanielSank

@Kaumudi right, that has to do with frequency.

Emilio Pisanty

00:58

@Kaumudi sorry, that was a typo

@Kaumudi frankly, I think that equation (1) is mostly rubbish

what does $\frac{d}{dV}$ even mean?

user228700

@EmilioPisanty o.O

user228700

OK..?

Emilio Pisanty

@Kaumudi That means: for a driving field of the form $E=E_0\cos(\omega t)$, where $E_0$ is time-independent.

DanielSank

@EmilioPisanty it's not rubbish, it's just not ask that useful.

Emilio Pisanty

and where you don't intend to compare results for different $\omega$s

DanielSank

01:00

@EmilioPisanty it's not rubbish, it's just not all that useful.

Emilio Pisanty

@DanielSank OK, I won't ask that useful

:-P

DanielSank

@EmilioPisanty you don't have response at another frequency if the response is time invariant.

user228700

@EmilioPisanty Uhh, I dunno, how s'thing changes with volume? .__.

Emilio Pisanty

@Kaumudi so the thing with that (1) is: tell me what you want to use it for, and I'll tell you what it means.

(which is actually a good thing to ask about any equation that someone is confused about, in general)

user228700

@EmilioPisanty I actually dunno the answer to that question so I'll get back to you when I do..? :-P

Emilio Pisanty

01:02

@Kaumudi yeah, but "volume" is not a variable

@Kaumudi sure. For now, I would tell you to take it as follows: $\mathbf P = \frac{d\mathbf p}{dV}$ is a shorthand for the statement that $\Delta \mathbf p = \mathbf P \Delta V$ for small enough $\Delta V$.

I pretty much guarantee you that whatever the usage ends up being, it will be consistent with that interpretation.

user228700

@EmilioPisanty Hmm. Alright, I will get back to you once I actually find some use for that equation (if I ever do)

user228700

Thanks, guys! :-)

Emilio Pisanty

@Kaumudi honestly, there's like an 80% chance that the authors themselves don't know what $\frac{d}{dV}$ means, and they were just using that in a handwavy way because they thought it looked pretty even if it doesn't actually mean anything that you can make rigorous.

DanielSank

@EmilioPisanty what, dude, it just means the dipole moment is the volume integral of $dp/dV$.

Right?

Emilio Pisanty

@DanielSank that's what I said above, essentially

the thing is, volume is not a variable

DanielSank

01:09

@EmilioPisanty I see, you object to /dV because wtf does that mean?

Emilio Pisanty

@DanielSank yeah

or how would you write it down? $$\frac{d\mathbf p}{dV} = \lim_{V_2\to V_1}\frac{\mathbf p_2-\mathbf p_1}{V_2-V_1}$$?

That's just flapping in the wind.

DanielSank

I see your point

Emilio Pisanty

You just say: there exists a vector field $\mathbf P(\mathbf r)$, called the polarization (density), such that the bound-charge electric dipole moment $\mathbf p_V$ in a given volume $V$ is given by $$\mathbf p_V=\int_V \mathbf P(\mathbf r)\mathrm d^3\mathbf r.$$

and you're done.

DanielSank

Yeah

Emilio Pisanty

(CC @Kaumudi on the above - that's what $\mathbf P$ really means.)

user228700

01:25

@EmilioPisanty I will read it all in a bit. Thanks very much :-)

Emilio Pisanty

> Nigerian astronaut lost in space needs $3m to get home boingboing.net/2016/02/12/nigerian-astronaut-lost-in-spa.html

user228700

I just realized that tomorrow (today) is Thanksgiving! Happy Thanksgiving, everybody! :-)

David Z

02:01

In going over the recent meta post, I found this question which was closed as homework-like but really should not have been, IMO. @JohnRennie @dmckee can you explain?

Same goes for this one (@JohnRennie @MAFIA36790 @Qmechanic)

and this one (@MAFIA36790 @JohnRennie @heather)

and probably this one (same people)

DanielSank

@Kaumudi thanks.

David Z

and maybe this one

I don't mean to come out and tell the close voters on those questions they were wrong. But I think those particular questions (especially the first 3) are obviously enough good homework-like questions that the fact they were closed indicates a significant difference between how the close voters interpret the homework policy and how I think it's supposed to be interpreted. That I would like to understand.

dmckee

02:16

@DavidZ Frankly, I"m not sure I read it closely enough. Between the title and the first few paragraphs I assume I had formed a judgement before getting to the end.

My fault, I suppose.

heather

@DavidZ, the first one you reference me in, I can't see because it's deleted. Would you mind giving a screenshot?

David Z

It happens.

@heather sure, here. They're all in the meta post too.

@dmckee I'd kind of like to reopen those questions, but not without at least understanding why people voted to close them in the first place.

Then in the longer term, if there is a difference of understanding regarding the homework policy, that's something that should inform how we replace it.

heather

@DavidZ, thanks. Reading through them, I still stand by my vote. Perhaps I misunderstand the homework policy (actually, that's the most probable reason here) but they seem to fall afoul of the conceptual part of the homework policy. And honestly, you say those should be reopened, but I've seen many a "homework" question that had that same level of effort/conceptual-ness and were closed. So I was also going off of what I've seen of the policy being implemented.

dmckee

I don't think that is a great question, but don't delay on my account. Not sure if you should wait to hear from John on that one too.

heather

I do, however, disagree with the close votes on the first two and voted to reopen there.

Finally, your very last one seems to be clearly a "check-my-work" type of problem, and those really aren't allowed either.

So I don't think the last one should be reopened.

David Z

02:21

@heather In the Lagrangian one "Can I just treat the left part as one mass, even though the masses are not still?" is the conceptual question. So I would say, anyway.

heather

@DavidZ, I guess that could maybe be conceptual enough; it's still kind of borderline in my mind.

David Z

@heather My thought on the last one is that it is not actually asking us to check the work. It shows the work, but doesn't ask us to check it. What it does ask is "... since the particle is moving diagonally, I don't know if just taking the height is sufficient."

@heather ah, well I would say it's exactly the sort of thing I had in mind when using the word "conceptual" in the homework policy. That particular phrase, at least, is a textbook example of what I mean by a conceptual question.

heather

@DavidZ, okay, well the reopening of the Lagrangian question makes sense now; I think I must have skimmed past that, and I misunderstood there what exactly was wanted of a conceptual question. However, I think I still kind of disagree with the last one being on-topic.

now that's all just me.

I probably (as I said earlier) misunderstand exactly how the close reason should be wielded.

@DavidZ, @dmckee, you know what might be interesting: an experiment where we replaced the homework close reason with a few new close reasons for, say, a week, and see what happens, and then maybe evolve those close reasons based on the results.

David Z

@heather OK, that's good to know. This can inform our discussion of the new policy - I guess we should have better explanations of whatever the key terms are. As for the last one, why do you still disagree? Would you still disagree if the work shown (everything from "To find the amount of work..." onwards) were edited out?

@heather Well, that's kind of what we're doing - I mean, even if we did it temporarily, we'd have to figure out which new close reasons to use, and that is what we're doing. Close reasons are not meant to be changed lightly, in any case.

heather

@DavidZ, yes and no: "What I want to know is if my work below is correct" is what he is asking, and then he provides a doubt for his work. If he changed the question to remove that and just ask if taking the height was sufficient, then yeah, and then provide the work as a note. When I read through, I saw that phrase "What I want to know is if my work below is correct" and was like, okay, that's the question. VTC.

David Z

02:30

Ah, I see what you mean. I think that part didn't entirely register with me. In that case I agree; the question is close to being on topic, but in its current form it needs an edit to make it on topic.

heather

@DavidZ I suppose; the discussions on meta have been going on for years, right?

David Z

@heather about one year, maybe a little more, IIRC

heather

@DavidZ, oh =P Not years plural =)

@DavidZ not to be nitpicky, that's just how I read it.

3

Q: Pully and Inclined Ramp

100 kg box is held in place on a ramp that rises at 30° above the horizontal. There is a massless rope joint to the box that makes a 22° angle above the surface of the ramp. Coefficients of friction between the box and the surface of the ramp are μk = 0.40 and μs = .60. The pulley has a mass of 1...

Finally, with this one ^

why is that on-topic?

David Z

That's another one that is close to being on topic, but I could see it not quite being there... but the conceptual question there seems to be "I've gotta imagine that the intention of Mass B is that the speck of dust would represent just about nothing, and the point of it was to explain that the system would be in motion and I should just account for it as the same mass from before, but I don't know."

That seems to be the core of what the question is asking.

Of course the stuff after that is a distraction and should be removed, and I could see that being justification for keeping it on hold until that edit is made, but that other stuff is pretty clearly not the core of the question.

heather

@DavidZ, okay, I think I missed that. That does seem like more of a conceptual question.

David Z

02:42

Yeah... of course, I do agree the question would benefit very much from an edit that cuts out a lot of its content.

I'll wait a bit for more feedback but at some point soon I think we can make the edits to some of these questions and reopen them, as discussed.

user116211

03:28

@DavidZ This is basically check-my-work problem; OP is asking why a certain method he followed failed; I don't think it's a question I would vote to reopen.

user116211

@DavidZ If OP only asked the second question, I would not have voted to close it; but in its form, it's a homework question.

user116211

Of course, this is my opinion; if someone feels they should be reopened, just vote to reopen them; I'll not intervene in the reopen queue concerning these posts then.

David Z

03:52

@MAFIA36790 I would argue quite strongly that asking why a method fails is not check-my-work. (It matters for future questions)

@MAFIA36790 which second question do you mean?

user116211

@DavidZ Check Gert's answer' he broke the questions into two parts.

David Z

Oh, you mean A and B?

user116211

@DavidZ If the question were how the method works, it would be conceptual; or why doesn't this method apply here still looks a good conceptual query; but simply why I failed by applying this method is more or less synonymous to check-my-work problem.

user116211

@DavidZ yes.

David Z

Yeah, everything relating to part A isn't a question and should be edited out. I don't think that itself makes the question as a whole off topic as per our policy, but I'm not too sad about it being on hold in its current form since it does need such a significant edit to be the best question it can be.

(Personally, I might have just made the edit rather than VTC'ing, but either is okay)

user116211

03:57

@DavidZ The edit should have come from OP, not some external user, IMO.

David Z

@MAFIA36790 Well, even though the edit would change a large fraction of the question's text, it doesn't affect what is really being asked. It'd just be removing irrelevant fluff. That sort of edit is okay for anybody to make, I'd say.

user116211

I very much agree with this.

user116211

There is a good answer, that's why OP, having got the answer, didn't attempt to edit the post; so yeh.

user116211

What? AdBlock support is closed? Why?

user116211

Till 28Nov; okay.

user116211

04:03

@DavidZ, do you guys celebrate Thanksgiving in China?

user116211

I don't even know what the attitude of people in China is towards Thanksgiving.

David Z

@MAFIA36790 Chinese people don't, though Americans living in China might

It's not a Chinese holiday, and not even one that is recognized by Chinese people in general. Unlike Christmas and Halloween.

user116211

ahh!

user228700

I've a quick question again!

user228700

> "The nuclear force is not a central force."

user228700

04:09

What does that ^ really mean?

user228700

> "The force depends on whether the spins of the nucleons are parallel or antiparallel, as it has a non-central or tensor component. This part of the force does not conserve orbital angular momentum, which under the action of central forces is conserved."

user228700

Okay, perhaps that ^ is a bit advanced and I should just leave it at "The nuclear force is noncentral" for now :-P

user228700

Ah, nvm, I think I found the answer.

DanielSank

04:40

Hi, everybody.

user116211

o/

Anonymous

@Kaumudi It is basically due to interaction between nucleons. The force does not act from a central point or towards a central point.

user228700

04:58

This is the definition of a central force:

user228700

> "A central force is a force that points from the particle directly towards (or directly away from) a fixed point in space, the center, and whose magnitude only depends on the distance of the object to the center."

user228700

Now I dunno if the nuclear force acts along the line joining two nucleons or not, but I do know that its magnitude definitely doesn't depend solely upon the distance b/w the two nucleons-it also depends on their spin quantum number.

Anonymous

@Kaumudi Yes the nuclear force depends on a lot of factors like spin also. It is not as simple as electrostatic forces. The potential formula for nuclear force is something like $$V(r) = - \dfrac{4}{3} \dfrac{\alpha_s(r) \hbar c}{r} + kr$$

Anonymous

From the study of the spectrum of quarkonium (bound system of quark and antiquark) and the comparison with positronium one finds as potential for the strong force.

Anonymous

where the constant $k$ determines the field energy per unit length and is called string tension. For short distances this resembles the Coulomb law, while for large distances the $kr$ factor dominates (confinement). It is important to note that the coupling αsαs also depends on the distance between the quarks.

Anonymous

05:05

Anyway you need not go into the details. Just remember that it is not a central force :).

Anonymous

32

Q: Is there an equation for the strong nuclear force?

The equation describing the force due to gravity is $$F = G \frac{m_1 m_2}{r^2}.$$ Similarly the force due to the electrostatic force is $$F = k \frac{q_1 q_2}{r^2}.$$ Is there a similar equation that describes the force due to the strong nuclear force? What are the equivalent of masses/charg...

particle-physics forces quantum-chromodynamics

user228700

Oh God, why did u bother to go into that much detail?!

user228700

Thanks :-)

user116211

06:27

Morning @JohnRennie.

John Rennie

Morning everyone

user116211

So, what are you doing in Thanksgiving @JohnRennie?

John Rennie

We don't observe Thanksgiving in the UK.

user116211

:(

John Rennie

Perhaps we should be giving thanks for ridding ourselves of those pesky pilgrims :-)

@dmckee Curiously only four of the eight downvotes were reversed. Oh well. The functioning of the serial downvote detection script is one of those fundamental mysteries :-)

user228700

07:04

@JohnRennie: Ello Sir! :-) Can u please ping me if u're free to talk about those phase diagrams?

John Rennie

@Kaumudi I'm free for 23 minutes :-)

user228700

Dyou think 23 mins. will be enough to talk about phase diagrams? Dyou remember how terrible it was the first time? :-P

John Rennie

Well ask your question and let's see ...

You now have 21 minutes :-)

user228700

Damn! Okay, well, this time, it's the T xy diagram that I'm finding a little difficult to understand.

John Rennie

OK

user228700

07:11

Here:

user228700

John Rennie

OK, that should be simple enough. What's the question?

user228700

The pressure is kept constant this time and the temperature is varied.

user228700

@JohnRennie The question, erm, I'm unable to get a feel for these diagrams. Is there a source that explains these really well? My textbook, as always, sucks, and I haven't found any good videos for my level.

user228700

It's very frustrating-I've been trying to understand these for the past two/three days! :'-(

John Rennie

07:16

I agree that the subject can be confusing at first. Most of us old timers are comfortable with it only because with time you get used to anything.

I don't know of any good resource - I'm rather out of touch with what educational media are available.

However once you get the hang of these diagrams they are simple enough.

Maybe you could ask something, anything, and we can use that as a jumping off point to discuss the diagram.

For example, it often seems as if the space between the two lines is a mysterious forbidden zone. Does it seem that way to you?

user228700

@JohnRennie It's the place (:-P I mean the temperatute conditions) where the liquid and the vapour coexist and that's all I know about this region.

John Rennie

Well suppose you tried to prepare liquid with the composition and temperature shown by the red dot. What would happen?

user228700

Uhh, one moment...

user228700

What dyou mean "What would happen" ?

John Rennie

Suppose we took a 50/50 mixture of A and B and instantaneously heated it to the temperature of the red dot then stood back and watched. What would happen to our superheated liquid?

user228700

07:26

Some of it would definitely turn into vapour, I assume...

John Rennie

The key thing to understand about these systems is that if we have a vapour and liquid in equilibrium the compositions of the vapour and liquid are not the same.

In this case B is more volatile than A i.e. boils at a lower temperature.

user228700

Yes, the vapour phase will be rich in the more volatile substance, where as the liquid phase will be rich in the less volatile substance...

John Rennie

So as a general rule if we have a vapour and liquid in equilibrium then the vapour contains more of B and the liquid contains more of A.

So what happens is or liquid starts to boil, and the vapour that gets boiled off contains more B. But that means the liquid contains more A, and therefore the boiling point of the liquid increases as it loses B.

Eventually so much B has boiled away that the boiling point of the liquid that is left becomes equal to the temperature, and at that point the liquid stops boiling. The system now looks like this:

Where the green dot shows the liquid and the blue dot shows the vapour.

user228700

> "The boiling point of the liquid that is left becomes equal to the temperature, and at that point the liquid stops boiling"

John Rennie

Yes?

user228700

07:33

I don't understand this in that OK, A is the liquid that is left and being less volatile, it has a higher boiling point. What dyou mean when u say "The boiling point becomes equal to the temperature"?

John Rennie

Let's make it concrete. Suppose B boils at 20C and A boils at 100C. And suppose a 50/50 mixture boils at 40C.

Is this OK to start with?

user228700

OK wait.

user228700

When u say "A 50/50 mixture boils at 40°C", u mean the whole bulk of the liquid? The whole mixture? How to determine when that happens?

John Rennie

I mean if we start with a 50/50 mixture of liquid at 0C and gradually raise the temperature then at 40C the liquid starts boiling. By boiling I mean that if we put more heat in the temperature doesn't go up any more, we just get more bubbles of vapour. Just the same as water does at 100C.

user228700

Okay. So both liquids boil at the same temperature... .___.

user228700

07:41

(^ that's s'thing my puny brain is finding difficult to fully digest)

John Rennie

@Kaumudi remember that these are miscible liquids. The mixture boils at 40C. It isn't possible to say both liquids boil at the same temperature because we have a mixture not separate liquids.

In a mixture we can't say what is the boiling pointof B because we can't just boil the B part of the mixture.

user228700

I get that, but...okay.

user228700

@JohnRennie Yeah.

John Rennie

OK, if we have the 50/50 liquid at 40C the Raoults law (again!) tells us what the composition of the vapour will be.

And it will have more B than A because B is more volatile i.e. has the higher vapour pressure.

In fact we can get the composition of the vapour from our diagram. Give me a moment to redraw it ...

user228700

@JohnRennie And yet, here, we speak of the two different liquids...

John Rennie

07:49

@Kaumudi where do we speak of two different liquids?

user228700

@JohnRennie I think I get this...

user228700

@JohnRennie Never mind, my brain is just being a moron.

Anonymous

@Kaumudi I think you are still having a confusion there. Here the two liquids are miscible, think like the molecules of the less volatile liquid (having more inter molecular attractive forces) are somewhat holding back the molecules of the more volatile liquid (having lower inter molecular forces of attraction) from vaporizing. This way, whenever two liquids are miscible always the BP lies in between the BP of the two different liquids which are mixed.

DanielSank

@MAFIA36790 \o

user228700

@S007 OK, this makes sense.

John Rennie

07:55

I'm not sure where we've got to with this ...

user228700

Why not? .__. U were telling me the boiling points of the individual liquids and of the mixture...

John Rennie

Ok if we have a 50/50 mixture at 0C and start heating it then it boils at 40C. This is shown by the green dot on my diagram. OK so far?

user228700

OK, I've a question.

user228700

@JohnRennie Yeah. Before proceeding, may I ask a quick question?

John Rennie

Yes, the whole point of this is to try and probe which aspects of it you need clarifying.

I just made up this scenario as a starting point for discussion.

So at any point you should feel free to ask questions or indeed change the direction of the discussion completely.

user228700

08:01

OK. So we mix the more volatile B and the less volatile A. 50 mols of A and 50 mols of B.

user228700

@JohnRennie Yaay, OK :-P

John Rennie

@Kaumudi yes

user228700

And now we have a mixture, a proper mixture so that every molecule of A is attracting/attracted by every molecule of B. Is this correct?

John Rennie

Yes

user228700

OK, now how does the difference in volatility factor in here?

John Rennie

08:04

Are you asking how at the molecular scale we get this dependence of the boiling point on the composition?

user228700

I...dunno. OK, I'm asking about why the mixture has a boiling point in b/w that of the two liquids.

Anonymous

@Kaumudi Intermolecular Force between A-A>A-B>B-B (where A is less volatile than B). In the mixture A lessens the ability of B to vapourize and B in turn increases the ability of A to vapourize. As the whole the whole mixture attains a uniform BP between the BP of A and B...

John Rennie

@Kaumudi I'm going to have to talk about free energy again. Are you ready for this :-)

user228700

::braces herself::

John Rennie

Gibbs free energy has two components. The enthalpy, $H$, which is basically heat content, and the entropy, $S$.

The equation is $$G = H - TS$$

user228700

08:10

::shudders:: Is there no other way to explain..?

John Rennie

Note the minus sign in $-TS$. That means as the entropy increases the free energy decreases.

@Kaumudi It's simpler than you think. Certainly simpler than you fear :-)

user228700

Yeah, the term free energy certainly does give me nightmares.

user228700

@JohnRennie Right...

John Rennie

If you dilute anything you increase its entropy. In an arm waving way, this is because there are more ways to arrange a dilute stuff in space than a concentrated stuff - because the molecules in the dilute system have more room to move.

user228700

Oh @DanielSank: Sir John and Kaumudi's Physics Gallery is back!

John Rennie

08:13

Does that make sense in an arm waving sort of way?

user228700

@JohnRennie Not really :-/

John Rennie

Hmmm

user228700

Gimme a moment...

user228700

Alright.

user228700

(Sort of :-/)

user228700

08:17

What I don't understand (but somehow, is intuitively obvious) is why A-A>A-B>B-B. Particularly what A-B is doing in the middle.

John Rennie

Well let me rush ahead, and we'll see what happens.

user228700

::prepares for the worst::

John Rennie

As you heat a liquid its free energy rises. When the free energy has risen enough to become equal to the free energy of the vapour the liquid boils.

Anonymous

@Kaumudi More volatile the substance lesser is the inter molecular force. That is why A-A>A-B>B-B.

John Rennie

But if you take pure B and dilute it with A then the dilution causes the free energy of B to decrease.

This means you have to heat to a higher temperature to get the free energy up to the point where it matches the vapour.

user228700

08:21

@JohnRennie This makes sense but also doesn't on account of how I don't understand why dilution with A causes the free energy of B to decrease.

John Rennie

@S007 I don't think that's a good argument. We would get this effect even if A and B were completely non-interacting.

@Kaumudi because it increases the entropy of B. Increasing the entropy lowers the free energy.

user228700

OK, why is it that the force b/w A and B is greater than that b/w two Bs but less than that b/w two As? (Assuming A_2 and B_2)

John Rennie

@Kaumudi I would ignore that argument as i don't think it's valid.

user228700

@JohnRennie Wokay .__.

John Rennie

We know B molecules attract B molecules otherwise B wouldn't form a liquid. And likewise for A. But there is no reason to suppose any particular interaction between A and B molecules.

user228700

08:24

Why not? OK, what does them being miscible mean?

John Rennie

If we mix A and b there will be a free energy of mixing. And that free energy will be $$\Delta G = \Delta H - T\Delta S $$

The enthalpy $\Delta H$ is effectively the internal energy and is dominated by the interactions between the molecules.

The entropy $\Delta S$ depends on the arrangement of the molecules.

user228700

OK...

John Rennie

When you mix the pure liquids you vastly increase the number of possible arrangements of the molecules because A an B molecules can mix in a vast number of different random ways.

So there is a large positive entropy of mixing: $\Delta S \gg 0$

user228700

When u say A and B, u mean how they were molecules of A_2 and B_2 and now they're just a bunch of As and Bs?

John Rennie

@Kaumudi No, I don't mean the molecules dissociate. I'm treating the molecules as elementary objects. If it makes things simpler assume A and B are like noble gases i.e. in the liquids (and the mixture) we just have A atoms and B atoms.

Suppose we ignore the interactions between the molecules completely, i.e. $\Delta H = 0$, then we get $\Delta G = -T\Delta S$.

user228700

08:30

@JohnRennie Oh, alright...

John Rennie

So the positive entropy of mixing means the free energy change is negative.

user228700

So the only interactions even in pure liquid state are van der Waals forces?

John Rennie

And all system want to reduce their free energy, so a negative free energy means the process is spontaneous i.e. the liquids mix spontaneously just because of the entropy of mixing.

Anonymous

@JohnRennie If A and B were completely non-interacting I do not think they would be miscible. Miscibility is the cause of interaction as far as I know...

John Rennie

@Kaumudi the details don't matter. I'm making general comments that don't depoend on the exact details of what the liquids are and how they interact.

@S007 wrong, wrong, wrong. It's the entropy of mixing that drives the process and that doesn't depend on the interactions.

@S007 if A and B actively repel each other than that effect can be great enough to overcome the entropy of mixing and we get systems like oil and water.

user228700

08:34

@JohnRennie Let me get this straight-when we have two "miscible" liquids constituting a mixture, they aren't interacting?!?!

John Rennie

@Kaumudi I didn't say they aren't interacting, I said they don't have to interact in order to mix.

user228700

.___. OK.

user228700

Can u like, give me one example of when they're miscible and still not interacting?

Anonymous

@JohnRennie But in systems (where they repel) like that the liquids will be immiscible like in oil-water mixture...wouldn't it ?

user228700

@S007 Yeah, that's what he meant (I think). 'Cause the mixing doesn't happen 'cause $\Delta G$ is positive...

Anonymous

08:37

Can you give example of any mixture where the components do not exert attractive force on each other, but due to entropy being favourable they mix? @JohnRennie I think for mixing attractive force is necessary between A and B...I could'nt understand "It's the entropy of mixing that drives the process and that doesn't depend on the interactions"

John Rennie

@S007 if A and B repel each other enough to overcome the $-T\Delta S$ term then yes they won't mix.

user228700

@JohnRennie What Dyou mean by "overcome" that term? I realize $\Delta G$ must be negative for the mixing to happen...and if they repel, what happens to the entropy? Does it actually decrease?!

Anonymous

delta G for mixing i think is a result of inter molecular attractions between A and B. If there is no intermolecular attraction I do not think the liquids mix

user228700

@S007 He means that when we mix A and B, if entropy increases, then the process becomes thermodynamically favorable...

John Rennie

We're getting well off the subject, but i don't think that matters as this is all useful discussion. @Kaumudi are you still happy with the way the discussion is going.

Anonymous

08:41

Ok I will ask my doubts later. @Kaumudi You get your doubts cleared first. I will wait

user228700

@JohnRennie I realize this but I sort of feel like I need to know about the entropy and attraction and all, to proceed happily :-P Is that not OK..?

user228700

Because this has stirred up enough of a confusion in my mind that I sort of need to clear that up to move ahead with ease. Is that OK..?

John Rennie

Where we came in is why a 50/50 mixture of A and B boils at a higher temperature than pure B.

user228700

Yes.

John Rennie

And it's because in the 50/50 mixture the entropy of B is greater than it is in pure B.

That makes it harder to boil off molecules/atoms of B because the entropy change on boiling is lower.

user228700

08:45

@JohnRennie OK, this makes sense, I guess...

John Rennie

This is a thermodynamic argument so it's a high level/large scale/macroscopic argument.

user228700

U mean macroscopic?

John Rennie

You started by asking about what happens at the molecular/atomic scale and that's harder to explain because entropy starts getting hard to explain in these temrs.

user228700

@JohnRennie I see...

John Rennie

I would have to go away and think about how to phrase the argument in microscopic terms.

user228700

08:48

Okay... And u'll talk about the miscibility and entropy and all, later?

John Rennie

Do you want to go back to the phase diagram we started with, or have we gone beyond that?

user228700

Can u pls give me 5 mins?

user228700

_/\ _

user228700

..?

John Rennie

Take as long as you want. In the mean time I'm going to make more coffee.

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