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user116211
11:00 AM
@ACuriousMind, You remember the work function from yesterday's discussion?
 
user116211
As you know, $V,$ the potential energy was defined to be the negative of $U$ i.e., $V= -~U\,.$
 
user116211
But it is the case when $U$ doesn't depend explicitly on time $t\,.$
 
user116211
The relation doesn't hold for time-dependent $U\,.$
 
user116211
So, how should I define $V$ for $U:= U(q_1, q_2, \ldots, q_n; ~t )\,?$
 
ok I am running out of time, thus cross SE analysis stops here for today. But it seems that contrary to what I thought, while the phenomenon is very common across the internet as in there is always at least one example in any media, it seems that for SE, it is only concentrated enough on PSE and MSE to be able to allow me to pick them up via random sampling
 
user116211
11:04 AM
I dug the book but didn't find any relation between the two in the time-dependent case.
 
@DanielSank cc @Secret I think you massively overanalyzed what happened here. I think the question was just not particularly clearly written. I didn't have to crawl through the chat at all to understand what it really meant to ask, the comment from the OP was more than enough.
 
user116211
It is only written that $V=- ~U$ is true only for time-independent $U\,.$
 
@MAFIA36790 Who's saying that there is a $V$ for time-dependent $U$?
 
@MAFIA36790 \o
 
user116211
@ACuriousMind ah!
 
user116211
11:10 AM
But @ACuriousMind, can $V$ be defined for rheonomic cases?
 
user116211
He writes:
 
user116211
> $\delta V = \mathrm dV$ would no longer be true if $V$ were rheonomic depending through $t$ explicitly.
 
@ACuriousMind This is one facet of what the meta post is stressing, but there is also another important thing to brought focus on by the meta post is that often (at least in MSE, PSE), when the OP tries to clarify the question, eventually the comment flow stops as the answerers stops following up. The question is then effectively ignored no matter how many edits the OP followed. In a few lucky cases, ~2 months later, there will be a new answerer to answer the question, but often when the question is
 
user116211
where $\delta$ denotes virtual variation.
 
being erroresnously treated, people stopped trying to check how the question is fixed and thus the question is effectively being abandoned
 
user116211
11:12 AM
I mean he introduced $V$ first through this expression $V= -~U$ without giving separate definition of $V\,.$
 
user116211
Now, that $U$ is time-dependent and the relation doesn't hold true, how can I then define $V$ for rheonomic/time-dependent cases?
 
@Secret That only happens when the follow-up comment by OP don't actually make it clearer. In this case, OP made a comment to me explaining I misunderstood, but I was asleep at the time and you were acting in the whole chat discussion as if I had ignored that comment when I hadn't even seen it.
Slow down the horses here
@MAFIA36790 Why do you want to define $V$?
 
user116211
@ACuriousMind Because, as he said in the above quote 'if $V$ were rheonomic', there must be a general definition of $V$ which works for rheonomic cases also.
 
user116211
@ACuriousMind In fact, can it be defined for rheonomic cases?
 
@ACuriousMind No, I never said you ignored the comment, I am saying that the situation of that questions reminds of the more widespread situation I mentioned that I saw across MSE and PSE, even when the OP does end up clarifying the question. As you have said, you closed the question not because of duplicate, but because of being unclear. Daniel later realised that you are not the one who cast the duplicate, but the general point about the phenomenon remains
So our purpose of the metapost is to raise awareness that often questions get erroreously judged because answereers and closers sometimes overlooked the crucial part of the question
and the phenomenon of abandoned questions
 
11:18 AM
@MAFIA36790 I don't know, I don't have these parts of mechanics memorized. But from what you quoted, the sentence "$\delta V = \mathrm{d}V$ would no longer be true if $V$ were rheonomic" could equally well be read as an argument why rheonomic $V$ doesn't make sense. That's why I asked why you want to define a $V$ - what do you need it for?
@Secret I'm not sure how people making mistakes is news, or something that needs particular "awareness".
Users are human, and they are fallible. I think everyone is aware of that. Communicating such that everyone involved understands the same thing is difficult, and often fails. What's the point?
 
Well, for MSE and PSE, most questions either get closed or abandoned can be treated back to this, especially when a stream of OP comments followed that apparently get no responses for days or months. (and perhaps we are wrong) but we do noticed that when people answer things in general, they often missed the crucial point in the question that is crucial to kill the question, which is why our idea to raise the awareness of this specific mistake into view.
 
user116211
@ACuriousMind As why I need for, see, this whole thing started why the law of conservation of energy doesn't hold true for time-dependent $U$ and whether $V=-~U$ holds true for the time-dependent cases. Then digging up the unread chapters of the book, I came across that quote above. the next line confirmed that $V=-~U$ is true only for time-independent $U\,.$
 
Perhaps for you it is clear, as evidenced by most of your answers (provided ithe question is clear to you) you often managed to address the crucial point (and hence answer accepted and upvoted as expected), but it seems in general often a couple of answerers failed to do this
 
@Secret I think what I don't get is why you think there's a "specific" mistake here. The mistake is people missing the point of the question, which doesn't seem all that specific to me.
I agree most misunderstandings that happen happen due to this, but that's because it's a really rather broad class of mistakes
 
user116211
As that of conservation law, I saw that the total energy doesn't remain constant when the Lagrangian $L$ does depend on time. So, that query seems to get answered as well as the second one that $V=-~U$ doesn't hold true for rheonomic cases.
 
11:26 AM
@MAFIA36790 When you say "doesn't hold true", you must have a definition of $V$ other than that. What is it?
 
I think the specific mistake is that people often tries to address only the statement that forms the question, but actually to answer the question one need to address the relevant chain of logic that lead to the question, not just the question itself, which is something that most people overlooked
 
I'm not convinced that's something most people overlook.
 
user116211
@ACuriousMind Yes, that's my question! What is the other definition?
 
@MAFIA36790 Why do you think there is one?
(We're going in circles here!)
 
user116211
@ACuriousMind yeh; I'm also confused with that ;/
 
11:29 AM
What reason do you have to believe that there must be a meaningful general definition of $V$?
 
user116211
@ACuriousMind There must be some reasonable point why he said "if $V$ were rheonomic"... What does that mean? What is the definition of $V$ when it is rheonomic? He didn't touch those queries :(
 
user116211
In fact the whole point of potential energy is for the conservative cases....
 
@MAFIA36790 Without more context I can't really tell, but I already said that that sentence sounds to me like an argument why $V$ doesn't have a good definition. In my interpretation, he's saying that $\delta V = \mathrm{d}V$ is something we want from $V$, but if $V$ (as a function, however it may be defined) would be rheonomic, it can't fulfill that, so it's pointless to try and define $V$.
 
user116211
@ACuriousMind "... why $V$ doesn't have a good definition."- This answers my query.
 
I already said the same thing here:
15 mins ago, by ACuriousMind
@MAFIA36790 I don't know, I don't have these parts of mechanics memorized. But from what you quoted, the sentence "$\delta V = \mathrm{d}V$ would no longer be true if $V$ were rheonomic" could equally well be read as an argument why rheonomic $V$ doesn't make sense. That's why I asked why you want to define a $V$ - what do you need it for?
 
user116211
11:34 AM
@ACuriousMind ;P
 
...why are you starring those?
 
user116211
But he could have written that straight ;(
 
user116211
@ACuriousMind Sorry, haven't meant the second one.
 
user116211
Anyways, I would conclude that $V$ is defined only for the conservative time-independent/scleronomic cases and not for the other one as it doesn't make sense to define it for the rheonomic cases.
 
@ACuriousMind Well, perhaps "most" is too strong of a claim, thus not necessary convincing, but our experience said we tend to see one to a few cases occasionally. Anyway we thought that the meta post may help address the cases where the aforementioned issues arises. We will let the community to decide whether our aims are actually sound, so we just sit back crossing our fingers for now...
Uh, Danielsank, you sure you got enough sleep, you just went to bed a few hours ago?
but anyway...
 
11:41 AM
@Secret How will it help address those cases?
 
When making aware to the answerers that in order to answer those questions with a logic chain structure, one need to address the crucial point instead of the stated question, the answers will become less likely to miss the point hence allowing the OP's questions to be satisfactorily answered, and avoid questions closing as duplicates or other reasons because the related question answered the stated question but not the crucial point that makes them different
It also raised an important message that answerers need to follow up on the answers when OP clarify them unless it deviates to the point of becoming a separate question
 
I don't think the problem is people are not aware that sometimes questions have an underlying issue that's not the literal question itself. It's that they don't always see that issue, and all the "awareness" in the world can't help with that.
@Secret No, they don't. No one is obligated to respond to anything here.
 
@ACuriousMind and that issue is often mentioned by the OP in the comments but people are often too tired or (insert reason) to respond to the OP's clarification and result in questions being abandoned
@ACuriousMind This is true...
 
@Secret The idea is that all relevant information is in the question. If I have to read the comments to understand a question, it's a bad question. If there's relevant information in the comments, edit the question. Anyone may do this, but the burden is on the asker.
Besides that, I dispute your claim that this "abandonment" happens all that often. If the follow-up comments are clear, then questions are frequently modified and reopened. However, there are many cases where the follow-up comments are just as unclear as the original post - but in this case you can't blame the commenters/close voters for not responding.
Not every follow-up comment is a clarification.
 
The issue is that after that, often for those cases, no one will try to ask what is still unclear about the clarification, thus OPs often just stopped there and waiting for responses
Here's an MSE example:
3
Q: Why is : $ [ \mathbb{R} [x_1 , \dots , x_n ]^{ A_n } : \mathbb{R} [x_1 , \dots , x_n ]^{ S_n } ] = [ \mathfrak{S}_n : \mathfrak{A}_n ] $?

YoYoHow to show that : $$ [ \mathbb{R} [x_1 , \dots , x_n ]^{ \mathfrak{A}_n } : \mathbb{R} [x_1 , \dots , x_n ]^{ \mathfrak{S}_n } ]= [ \mathfrak{S}_n : \mathfrak{A}_n ] $$ such that : $ \mathfrak{S}_n $ : the symmetric group $ \mathfrak{A}_n $ : the alternating subgroup of $ \mathfrak{S}_n $. ...

But, $ \mathbb{R} [x_1 , \dots , x_n] $ is not a field, so we cannot use : $ [K \ : \ K^{ \Gamma} ] = | \Gamma | $, right ? — YoYo Sep 16 at 18:44
although you are right he/she should include that clarification in the edit and not rely on comments
 
11:52 AM
@Secret That question isn't closed, and has three upvotes. I don't see what it has to do with the discussion.
 
The question is 9 days ago, and the answerer Lubin, stopped responsing to Yoyo's clarification since Sep 16. That's what I mean by "abandoned"
The tricky thing is it does make use of your point that "No one is obligated to respond to anything here."
 
@Secret Lubin is not an answerer since that question doesn't have an answer. They presumably wrote only a comment precisely because they didn't want to invest the time to write an actual answer.
Why they didn't want to invest that time is nothing you could know, which is why I don't see what this has to do with the reason of "missing a node" that we were discussing.
 
ok, sorry, in that case, Lubin is someone who comment about the question, not an answerer (apologies for terminology). Well, Yoyo need that node he/she said to be addressed in order to consider the question itself addressed, at least that's what I think, perhaps it is wrong to judge what they are thinking based on limited info
However you do raised a point that DavidZ mentioned before regaridng comments and answer:
@ACuriousMind People should avoid answering questions in comments
 
@Secret Sure. But that comment isn't an answer, it's a hint at a general principle that will lead to the answer after applying unknown amounts of work.
If someone posted that comment as an answer, I would downvote it for not actually answering the question.
 
yup, in that case, the above MSE might be a poor example to illustrate the phenomenon
 
12:01 PM
Anyway, after all this discussion, I'm still not clear on what exactly your "abandonment" is, and what the meta post is supposed to do about it.
Which might be a sign that either I suck at reading comprehension or you need to make that a lot clearer in the post itself.
 
@ACuriousMind Well it seems there are 3 things: 1. People tend to have a habit of using comments as a partial answer to a question, *2. OP is confused and tend to ask these people for clarification, 3. After some time of back and forth commenting these comenters stopped responding and no new comments followed.:: Are we agree on the above obsevation?
and of course there is 3a: OP often fail to include the clarification in the comments in the actual question edit
 
@Secret Yes
I don't see what that has to do with the meta post, but yes.
 
Point *2 is where OP often state what the missing node is, and point 3a is a common thing that OP fail to do. Are we agree on these deductions?
 
Those are not deductions, because you haven't deduced them from anything, but yes.
 
Ok and finally: Point 3. and 3a. are the reasons why this issue is often left unresolved, and lead to question abandonment for those cases
 
12:09 PM
Also, it's "Do we agree", not "Are we agree".
@Secret Yes
 
Hence, the meta post is raising awareness of point 3, in response to point *2
 
I don't see what this has to do with the meta post though, as there your prime example is a question where, as I said, you didn't even give the commenter enough time to respond to even the first clarifying comment from OP
This pattern of commenters getting tired of OPs moving the goalposts in comments is completely disjoint from commenters not understanding the crucial point of a question. In this case, I admit that I simply misunderstood the question as written, but I don't see what this has to do with the partial answer-question-partial answer-question-silence pattern you're now talking about.
 
Ok I think it is like this (Sorry I have to make the format into a list) :
1. Question written
Issue 1:
2. Some answers, partial answers given which missed the crucial point
3. (Optional) Question get closed because of unclear, or duplicate (based on the partial answers) and other reasons
4. (Optional) OP edit the question
5. OP clarify question in comments
6. Commenters address the comments
7. (Optional) OP edit the question based on comments
8a. OP clarification is clear, question is saved from closing etc. and answered
So you are right there are actually two separate issues here, in that case, the meta post should be geared towards one of them to be more focused
which one will it be?
By the way, since we have use the term "partial answer" to refer to the commenters, the MSE post is suitable to illustrate Issue 2
In that case, I think 4 days is a unreasonably long time to respond to the clarification
So my point is, while Issue 1 and 2 are technically separated, they tend to occur together cc @DanielSank for more clarification details
@BernardMeurer @0celo7 can you help me test something? Can you screencap my previous message block. I want to check how it looks like on your computer
just screencap everything surrounded by this large grey block is ok
 
12:34 PM
@Secret Okay, so yes, these are two patterns that happen. Why do you think there needs to be something done about them?
 
Why am I doing this
 
@ACuriousMind Well, let's focus on PSE (although it is also common in MSE) because quite a lot of "marked as duplicate" is wrongly attributed because of Issue 1 as noted by Danielsank, and the situation is sometimes made worse by Issue 2
@0celo7 I want to check how linebroken messages look like on you guy's comps, in order to test something
 
@Secret You've lost me again, I'm afraid. Let's drop this for now and see if Daniel can explain this to me in another way later on.
 
Sure, he can explain it and the meta post better cc @DanielSank
 
12:40 PM
@BernardMeurer ok that looks fine, and not some weird jigsaw puzzle like mess, thanks
 
@ACuriousMind As of right now I don't think the proof will happen...there's a line in Petersen that makes zero sense. My advisor asked me "what does that even mean" and I had no answer :/
 
@Secret No worries :)
 
12:58 PM
@ACuriousMind It says: show that the evaluation map $Iso(M,g)\ni f\mapsto (f(p),df_p)$ is continuous. Note that $df_p:T_pM\to TM$ is linear so that convergence of the values of the evaluation map makes sense.
Problem is, it doesn't make sense.
Maybe if we give $TM^{T_pM}$ the compact-open topology, it does.
@ACuriousMind Any smart ideas?
 
user228700
1:35 PM
@JohnRennie: Hello again :-) I have a quick question. Are u free?
 
Yes, ask away!
 
@JohnRennie Did you see the picture I sent you of Daedalos? (laptop)
 
user228700
Okay! So. This is technically in the realms of chemistry but I asked at the CSE chat and people kept saying "this is plain physics" so I figured I'd ask here :P
 
user228700
@JohnRennie Are u familiar with Le Chatelier's principle..?
 
1:38 PM
Yes, though be very careful as it's just an empirical rule that sometimes fails.
 
user116211
in The Periodic Table, 2 hours ago, by MAFIA36790
@KaumudiHarikumar Remember Le Chateliar's Principle is just a rule of thumb. The reason for why it works depends on how the value of $\Delta_r G^\Theta $ and $K$ get affected by the process.
 
@MAFIA36790 great minds think alike :-)
 
user116211
@JohnRennie ;))
 
user228700
@JohnRennie It's confusing the crap out of me! :-( Let's take the statement "When volume is increased, the pressure decreases, so the system moves in that direction in which the pressure decreases".
 
user228700
@MAFIA36790 Oh my God, that is ur reaction to what he just said?! :o
 
1:41 PM
@KaumudiHarikumar I have to say that seems a statement so vague as to be meaningless.
 
user228700
@JohnRennie :'-( OK...This is what's given in my textbook sir.
 
user228700
I'm gonna burn these effed up textbooks someday!
 
user116211
@KaumudiHarikumar ;P
 
Suppose you have some reaction: $$ A + B + etc \rightarrow C+D+etc$$This reaction will have some quilibrium constant given by:
$$K = \frac{[C][D][etc]}{[A][B][etc]} $$
OK so far?
 
user228700
No raising them to the power of their stoichiometric coefficients..?
 
1:44 PM
Oops yes, quite right! In this case the powers are all one, but yes in general you need to include the powers.
 
user228700
Yeah, OK. Please go on...
 
If the change in the Gibbs free energy for the reaction is $\Delta G$ then the equilibrium constant is: $$K = exp(-\Delta G/RT)$$
At least it's something like that. I'm typing from memory and it was a long time ago.
The point is that how far the reaction goes depends on $\Delta G$.
The Gibbs free energy partly depends on the chemical energies involved but it also depends on enironmental factors like temperature and pressure.
 
user228700
@JohnRennie I have to say, your memory is outstanding!
 
user228700
@JohnRennie OK..?
 
Changing the temperature and pressure change $\Delta G$ so the change the equilibrium constant. Depending on how $\Delta G$ changes they can make the reaction go more to right side or more to the left side.
 
user228700
1:49 PM
Wait, changing pressure changes the equilibrium constant? What? I thought it depends solely on temperature..?
 
The equilbrium constant depends on the change in the Gibbs free energy
Anything that changes $\Delta G$ will change the equilbrium constant. Anything!
 
user116211
@JohnRennie You are a nerd; you can never be wrong ;P
 
@MAFIA36790 One of those two statements is true, but only one :-)
 
user228700
@JohnRennie And not on temperature..? Sir, you should seriously consider writing a book titled something along the lines of "ALL OF THE MISCONCEPTIONS OF SCIENCE!(for beginners)"
 
@KaumudiHarikumar Temperature will change $\Delta G$ as well, so temperature will also affect the equilibrium constant.
 
user116211
1:51 PM
However, @JohnRennie, that should not be $\Delta G$ but rather $\Delta_r G^\Theta,$ the standard reaction Gibbs energy; they are different things.
 
I'm using $\Delta G$ to mean the molar change in free energy.
 
user228700
@JohnRennie OK, great. We've been taught that it only depends on temperature. Wait, is this way advanced stuff? 'Cause you know, I don't really need to learn way advanced stuff...
 
user116211
@JohnRennie You defined $K$ using this, right?
 
I don't think it's especially advanced. You'd learn it in the first year of a chemistry degree ...
 
user228700
And BTW, I don't even know what the proper definition of ∆$G$ is so if it's really so important, I'll have to go away and read up on that. If you could just tell me what's wrong with the first statement I mentioned..?
 
user116211
1:55 PM
@KaumudiHarikumar Did you read my answer I gave to one of your queries? Wait....
 
user228700
@MAFIA36790 My bet is that I would've...
 
user116211
@KaumudiHarikumar Then you shouldn't have asked that....
 
@MAFIA36790 IIRC the standard Gibbs free energy is at 25C and 1 bar. But you're doing the reaction at different pressure and temp obviously the change isn't the standard change.
 
user228700
Okay :P Which query?
 
user116211
@JohnRennie yes, that's why they change at different temperatures.
 
1:57 PM
@KaumudiHarikumar Le Chatelier's principle is a rule of thumb based on how $\Delta G$ will change.
 
user228700
OK...
 
Suppose you have a reaction that increases the volume e.g. a gas dissociating into two componets.
 
user116211
2
A: What is the difference between the equilibrium position and the equilibrium constant?

MAFIA36790Equilibrium position: A reaction reaches equilibrium position when it has no further tendency to change; that is, the reaction does remain 'spontaneous in neither direction' . This happens when the reaction Gibbs energy becomes zero viz. $\Delta_r G= 0\;.$ Equilibrium constant: The reaction ...

 
user228700
@MAFIA36790 Yeah, no, I didn't read the whole thing, sorry.
 
user228700
@JohnRennie OK...
 
user116211
1:58 PM
2
A: Gibbs free energy change at equilibrium

MAFIA36790I'm really not getting why that post didn't answer OP's query. Nevertheless, I'm posting the following argument as an answer. Why is the Gibbs free energy change equal to zero at equilibrium? Well, it's not 'Gibbs free energy change' $\Delta G$ but rather Reaction Gibbs Energy $\Delta _...

 
As a rough guide if you increase the pressure you will increase the free energy of the right side more than you increase the free energy of the left side, so the $\Delta G$ for the reaction decreases.
 
@ACuriousMind In Yang-Mills theory, I've seen the gauge transformation $U$ defined as $U(x)=\mathrm{exp}(i\theta^a(x) T^a)$, but also together with the coupling constant as $U(x)=\mathrm{exp}(ig\theta^a(x) T^a)$. Is that just a matter of convention, or is there more to it? In the first variant, is the coupling constant absorbed in $\theta(x)$?
 
That means the equilibrium constant also decreases and it pushes the reaction to the left side.
So increasing the pressure tends to stop the reaction.
Le Chatelier would say that if the volume increases then restricting the volume (by applying pressure) will oppose the reaction
And that's mainly true, but it's because of the effect the pressure has on the free energy.
 
@Bass It's a matter of convention that depends on how the field Lagrangian is scaled, in particular, whether the $g$ appears as a $\frac{1}{g^2}$ in front of the entire Lagrangian or whether it appears as a $g^2$ prefactor in front of the four-gluon interaction term.
 
@ACuriousMind I see. So the answer to the second question is no, right? In the first variant, the coupling constant is not part of the exponent.
 
user228700
2:02 PM
@JohnRennie Yeah, OK, I don't understand any of that because I don't know what "free energy" is. Will it take a horribly long while to get my head around this after learning about free energy..?
 
@Bass I think that's right, yes
 
Thanks!
 
user228700
I mean, it's subjective and all, but do u reckon it will take me at least 2 hours to learn free energy first and get to this, now that u're familiar with me and the pace at which I learn..?
 
@KaumudiHarikumar Gibbs free energy is kind of the total energy released in a reaction including changes in pressure and temperature and entropy. Actually few of us care exactly what it is.
 
2:05 PM
There is an analogy to potential energy that chemistry uses called chemical potential, and the Gibbs free energy is related to this.
 
↑ omg that is horrifying
 
So it's kind of the change in potential energy when a reaction occurs.
 
user228700
@JohnRennie I'm afraid that doesn't make a lot of sense to me...energy change associated with change in pressure..?
 
@EmilioPisanty I found a proof that doesn't use horrible approximations.
 
2:07 PM
@EmilioPisanty What the...?
 
user228700
@JohnRennie Is that it..?
 
@0celo7 nope
 
Er yes, changing the pressure of a system obviously changes its energy. Isn't it obvious?
 
@EmilioPisanty did you do that?
my LaTeX never looks like that.
 
user228700
@JohnRennie Yeah, OK. Forgive me sir. I've been studying all day and it's 7:30 PM now, so my brain is a smidge...what's the word? Yeah, fried.
 
2:10 PM
Have a quick scan of the Wikipedia article on Gibbs free energy. But just a quick scan. Don't try to take it all in one go.
 
user228700
OK, so the proper definition is that the free energy is the change in potential energy when a reaction occurs..?
 
user116211
@KaumudiHarikumar kinda.
 
@ACuriousMind We're doing coherent states in QM, prof gave the hint "use BCH"
I still don't understand the power series nonsense
 
@0celo7 Oh, hell no
I was there cleaning up
 
user228700
@JohnRennie OK, gonna go do that. Brb.
 
2:11 PM
 
Wh
 
Never seen anything that terrible w.r.t. LaTeX on this site
or at least that I can recall
 
@EmilioPisanty Why didn't you remove the "that" in the title?
 
@EmilioPisanty Eek! :-)
 
@0celo7 oversight
 
user228700
2:16 PM
@JohnRennie I'm back...
 
More questions? :-)
 
user228700
@JohnRennie I'm waiting for the day u say "To hell with this kid!" :P
 
@JohnRennie I have a question can I ask you?
 
@Ramanujan yes of course
 
We know pressure is inversely proportional to velocity. Right
But f
 
2:20 PM
@Ramanujan hmm, I'm not sure what you mean by that. Can you give a bit more context?
 
But from formula of pressure=force / area.
 
user228700
@JohnRennie I'm not sure I fully understand this...
 
@KaumudiHarikumar Gibbs free energy?
 
We get pressure = mass×velocity/area ×time
 
user228700
Sort of. I meant that statement specifically.
 
2:21 PM
And hence pressure is directly proportional to velocity?
 
@Ramanujan think about direction, Force is vector but pressure isn't
 
@Ramanujan Your initial statement pressure is inversely proportional to velocity seems wrong to me. Where did you get that statement from?
 
It's my intuition.
 
I agree with you that pressure of a gas is the rate of change of momentum of the gas molecules per unit area. pressure it is proportional to the average molecular speed.
 
@JohnRennie do you believe that pressure is directly proportional to velocity?
 
2:24 PM
The pressure of a gas is (approximately) proportional to the velocity of the molecules that make it up.
 
user228700
@JohnRennie: Is there any way to understand Le Chatelier's principle without going into all this free energy business? I mean, what is wrong in stating "Volume decreases causes pressure decrease so system will move toward that direction in which pressure increases"..?
 
user228700
You said that that statement doesn't make sense. I was wondering why..?
 
Pressure is inversely proportional to velocity of a fluid if it's moving. You are correct when you say its directly proportional to velocity ( of molecules of fluid). You should differentiate between the two cases. Please clarify in what context you are asking.
 
@KaumudiHarikumar the statement is basically correct but it's worded in a slightly strange way, at least to my eyes.
The trouble with Le Chatelier's principle is it doesn't say why it works.
The reason it works is due to the effect of pressure, temperature etc on the Gibbs free energy.
 
user228700
OK...since that's the way it's written in my textbook, after reading it, I was wondering if there is no alternative way to think about it. Volume increases so system wants to decrease volume so if we were talking about, say a gas-liquid equilibrium, following this line of reasoning, won't it make sense to say that on increasing volume, the system moves toward the liquid state to decrease volume..?
 
2:29 PM
@JohnRennie but for moving fluid , velocity is inversely proportional to pressure. How can we show this from formula of pressure=force/area?
 
user228700
@JohnRennie Yeah, all of my textbooks explain it using "system does whatever it can to counteract the change/stress".
 
@Ramanujan You are thinking of Bernoulli's principle which is one type of pressure velocity relationship. The reationship between pressure and molecular speed in a gas is completely different.
@KaumudiHarikumar I suppose you could argue that the system is trying to minimise it's potential energy, which is kind of what Gibbs free energy is anyway ...
@KaumudiHarikumar So system does whatever it can to counteract the change/stress kind of means system is trying to minimise its energy
 
user228700
@JohnRennie Yeah, OK, that kinda makes sense...
 
user228700
What about what I said before about volume changing and all..?
 
@JohnRennie why can't be use only formula of pressure=force/ area to show pressure is inversely proportional to velocity? Why need Bernoulli theorem?
 
2:35 PM
@Ramanujan suppose I point a fire hose at you so the water from the fire hose hits you on the chest and exerts a pressure on you. If I increase the velocity of the water from the fire hose is the pressure on you going to increase or decrease?
@KaumudiHarikumar Yes?
 
user228700
@JohnRennie No, um, what are ur thoughts on that..?
 
@JohnRennie pressure increased because from more velocity water will come.
 
@Ramanujan Yes, so the pressure is proportional to the water velocity just like in a gas.
@KaumudiHarikumar You mean:
9 mins ago, by Kaumudi Harikumar
OK...since that's the way it's written in my textbook, after reading it, I was wondering if there is no alternative way to think about it. Volume increases so system wants to decrease volume so if we were talking about, say a gas-liquid equilibrium, following this line of reasoning, won't it make sense to say that on increasing volume, the system moves toward the liquid state to decrease volume..?
 
user228700
@JohnRennie Yup. Exactly.
 
@JohnRennie I think OP is asking about Hydrostatic pressure exerted on the wall.
 
2:40 PM
You say:
> on increasing volume, the system moves toward the liquid state to decrease volume
 
user228700
@JohnRennie Yeah..? Sigh. Tell me why I'm wrong, please..?
 
But I'm not sure what you mean by this.
I can control the pressure on my reacting system. I can't easily control it's volume.
In your example of a gas-liquid equilibrium then if I increase the pressure I would expect the system to move in a way that decreases the pressure e.g. it changes from a gas to a liquid.
By doing this its molar volume decreases
... if that's what you meant
 
user228700
Remember I told u that we're taught that the system tries to counteract the stress applied? In this case, the "stress" is that you change the weight on the piston or something, so the gas expands and its volume increases. So, to counteract this increase, the system tries to decrease volume by changing some of its gas molecules to liquid.
 
@JohnRennie thanks for answering I will be studying more about Bernoulli theorem
 
user116211
@Ramanujan You can edit this within 2 minutes.
 
2:44 PM
> you change the weight on the piston or something, so the gas expands and its volume increases
Do you mean I increase or decrease the weight on the piston?
 
user228700
@JohnRennie Yup.
 
Liquefaction of gases is achieved by expansions
 
If I increase the weight the pressure goes up and the volume goes down. The system will try to counteract this by turning from gas into liquid i.e. by decreasing its molar volume.
Look at it this way ...
Assume we are above the boiling point of the liquid, so the material would normally be in the gas form rather than the liquid form.
If a mole of the material were to change from gas to liquid this would cost energy, so it normally doesn't happen.
 
user228700
OK, no. We remove some of the weight on the piston so the volume increases. To decrease the volume to counteract the stress, some of the gas turns into liquid. But looking at the same situation in terms of pressure, we increase volume, pressure decreases. To increase pressure, the liquid converts to gas. Which one of these is correct? I KNOW that I'm doing something wrong.
 
user228700
I just need u to help me identify what that is.
 
2:51 PM
@KaumudiHarikumar if you decrease the pressure the liquid will turn into gas.
That's why the boiling point of liquids falls as you decrease the pressure.
 
user228700
@JohnRennie That's the correct one, u mean..?
 
Yes.
 
user228700
OK, why is the other one wrong then..?
 
If you decrease the pressure the equilibrium shifts towards the gas state i.e. some liquid turns to gas.
The statement To decrease the volume to counteract the stress, some of the gas turns into liquid doesn't make sense to me.
 
user228700
We're increasing volume so why can't we look at in terms of volume itself and say that gas will turn into liquid to decrease volume, 'cause liquid has lesser volume..? Gah!
 
2:55 PM
That's like saying if you pull ona spring the spring will try to shrink to oppose the stress
 
user228700
Remember the counteracting stress thing..? That's what I'm applying...is the statement "volume of liquid is less than volume of gas" wrong..?
 
user228700
@JohnRennie OK..?
 
in The Periodic Table, 2 hours ago, by orthocresol
That is like saying that if you push a door, it opens towards you
 
You're saying if we pull on, i.e. expand, a system it will respond by trying to decrease its volume. That seems obviously wrong. If we pull on a system it responds by expanding.
 
user228700
@Loong Yeah, I was gonna post that lol. Everybody's saying the same thing :-(
 
2:58 PM
what the shit, I have to have a security clearance to apply to entry level internship
 
user228700
@JohnRennie It responds by expanding and then what happens?
 
>then what happens?
It expands.
 
user228700
:P
 
It expands by liquid turning into gas
 
user228700
OK, I think I understand what I'm doing wrong. God, I could kill my brain right now for being so effing dumb.
 

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