@RishiNandhaVanchi India really is underdeveloped in terms of pure science sector, you should probably give SATs and try to get opportunities outside. I wanted to go for science but so far is not very feasible. ..

I am having one doubt in weight percent standard state...Please see the quote,"If deviations from Henry's law occur below 1 wt %,then the procedure is to extrapolate the Henry's law line upto 1 wt% and conceive that as standard state,which will be hyptothetical."But it does not matter for thermodynamic treatment."

Now,

Firstly,why are we assuming a negative deviation here.I even am not motivated how to judge between positive and negative deviation secondly,Could someone explain the quote with reference to the diagram provided.I do not understand a single thing between the quote and the picture.

@RishiNandhaVanchi I think that's a distinction only made in India, in the UK there is a just a BSc, and I don't know enough about the Indian university system to comment.

Hello @JohnRennie sir

@PrateekMourya hi :-)

Sir Tommorow' question on beam nitrogen

Yes ... ?

I treated it another way

I want to confirm

If we take the system

And we calculate after how much time the cycle of its particular behaviour repeat

Then we find that immediately after hitting the wall the cycle repeats

But in case of box cycle repeats after the molecule tracel to its original position

Therefore method of taking time period is different in both scenario

Is it write?

There isn't a cycle or a time period in the nitrogen beam question. The nitrogen molecules come in from infinity, hit the wall and bounce, then fly away to infinity. Each nitrogen molecule hits the wall only once.

Yes for entire system of hitting and leaving

The cycle of hitting repeats afyer certain time

No

I don't know why i am not able to accept the method of calculating momentum change oer unti time

Each nitrogen molecule experiences a momentum change $\Delta p$ when it bounces. If $N$ nitrogen atoms bounce per second, then the momentum change per second is $N\Delta p$.

And the momentum change per second is just the force.

Whereas a single atom experience 2mv change in momentum in 2l/v time then momentum change per unti time(unitsry method is force?

In case of box

Ok got it

@JohnRennie Can you help me about the question I just posted.

@RajorshiKoyal what question? Can you post a link to it?

Coincidentally @user586228 has a doubt similar to mine....

Infact It is totally same to the question I posted on Chem Stackexchange

I lifted his question and

posted on chem SE so that someone may answer..

Actually I find Chem SE to be very quiet..

@RajorshiKoyal which edition of Gaskell's book do you have?

@JohnRennie Do u use Gaskell?

I mean the thermodynamics book

?

Actually it is sort of written by Prof HS Ray in an abridged version

Who is a legendary metallurgist from India..

I am not familiar with the book. I can look around for a copy, but it would be good to get the same edition that you're using so our books match.

So if you tell me what edition you're using I'll look for that edition.

Ok in that case Gasekll is better

Fourth

The name of the book is introduction to thermodynamics of materials by Gaskell

(Fourth Edition)

This one?

Yes..

@JohnRennie Can you do me one favour?

@RajorshiKoyal yes ... ?

books.google.co.in/books/about/…

Can you help me find this book

I am desparately looking for this one...

@JohnRennie Did u get the book

May be u wull not get the exact quote

But u will get the concept for sure.

Do u nedd to know the name of the chapter?

@RajorshiKoyal what page is the diagram on in Gaskell's book?

page 252

That doesn't match the image and text in your question.

I need to drop out now to do some work. I'll be back in about an hour.

ok...Tbh the image is taken from an abriged version..

It is better to check for the topic alternative standard states..

I am seeking for an explanation to the topic

@JohnRennie Check chapter 13

Alternative standard states

subsection...

check page 475

@JohnRennie Precisely this is what I am talking about

Section 13.3 is the most dreaded section.. for me....

@RajorshiKoyal give me a moment to open the book and read that section.

ok

OK, I read as far as page 475 and it all seems to make sense.

ok

Please explain what you understood

@JohnRennie Firstly,why that deviation?

@RajorshiKoyal it's generally true that solutions are non-ideal. It's rare to find a binary system that is even approximately ideal over the whole range of concentrations.

ok

ok explain line by line but why a negative deviation?

So we should expect that the curve will deviate from a straight line. How much it deviates from a straight line will depend on the system.

@RajorshiKoyal By a negative deviation do you mean that the activity is less than expected?

I mean there is another curve which shows a symmetrical upward thingy

That is what I am worried about..

@RajorshiKoyal I'm not sure what that means.

This is what means by positive and negative deviation

OK

So what are you asking?

Why negative deviation

The figure you posted above explains that. It happens when A and B have a tendency to attract each other. In extreme cases a compound AB may be formed.

but is is a solid dissolved in a liquid right?

Yes or no?

What are A and B in this case?

Please explain this to me @JohnRennie

The curve could be a solid dissolved in a liquid or a liquid dissolved in a liquid. I guess in this case they mean B to be a solid and A to be a liquid, because they are making the point that we can define a Henry standard state for B that isn't the same as the standard state for solid B.

But you can get curves like that for liquid-liquid systems as well. I think ethanol-water has a negative deviation like that.

When to expect solid and when to expect liquid

I don't think there is any way to tell just from looking at the curve. It could be either.

why ethanol water has negative deviation

I don't know exactly why it happens. It happens because ethanol and water attract each other, presumably due to some form of hydrogen bonding, but I don't know the details.

why is it called negative deviation plz explain

Because the activity is lower than expected from an ideal solution.

raooultian or henrian which one?

Raoultian

Ideal solutions obey Raoult's law over the entire composition range.

And non ideal

Could you elaborate slighly what it means

In an ideal solution the two components do not have any specific interactions with each other. The enthalpy of mixing is zero across the entire composition range and the mixing process is driven only by the entropy of mixing.

We get non-ideal behaviour when the two components do interact and the enthalpy of mixing is not zero.

@JohnRennie Hi!

@YouKnowMe hi :-)

@JohnRennie Do you know Mathematica?

Only in a very basic way.

Usually I have to Google to find out how to do stuff in Mathematica, or just used someone else's notebook.

@JohnRennie Basic is enough :-)

Plot[{2 x, -3 x, 3 x, -2 x}, {x, -9, 9}, PlotLegends → "Expressions"]

Why this gives error?

> Plot::nonopt: Options expected (instead of PlotLegends[RightArrow]Expressions) beyond position 2 in Plot[{2 x,-3 x,3 x,-2 x},{x,-9,9},PlotLegends[RightArrow]Expressions]. An option must be a rule or a list of rules.

Let me try it ...

It'll take me a moment as I have Mathematica installed on a different PC that isn't on at the moment.

Ok :D

OK, I get the same error. Give me a moment to Google it ...

If you remove the , PlotLegends → "Expressions" it works, so that's the problem ...

Yes, tried that :p

Hmm, according to the documentation it should work. What version of Mathematica are you using? I have v10.

12.2

This works:

Plot[{Sin[x], Cos[x]}, {x, 0, 2 \[Pi]}, PlotLegends -> "Legends"]

Oops, I meant:

Plot[{Sin[x], Cos[x]}, {x, 0, 2 \[Pi]}, PlotLegends -> "Expressions"]

"Legends" just prints the word "Legends" :-)

@JohnRennie One more question $a=\gamma x$

Right

But here they are taking a=X

Please juftify this....

justify*

@YouKnowMe OK, now I'm officially confused. Now this is working:

Plot[{2 x, -3 x, 3 x, -2 x}, {x, -9, 9}, PlotLegends -> "Expressions"]

Yeah

But this looks identical...

There's a difference in the spacing:

@JohnRennie I didn't know that the spacing matter :-O

Aha!

I think it's the arrow

Delete the arrow and retype it, and it works.

Boom!

:-)

@JohnRennie Thanks :-) One more doubt, Where can I ask Mathematica error, Mathematica SE, or Stackoverflow?

I would probably ask on the Mathematica SE.

I doubt Stack Overflow will know much about Mathematica.

@JohnRennie Ok, I had just seen a question.

Ah, OK. Still I would ask on Mathematica SE first. That seems an obvious place to ask.

:-)

Maybe ask on SO if you don't get an answer on the Mathematica SE ...

@JohnRennie You would be available tomorrow, now?

@YouKnowMe I need to go in a few minutes. I will be back later - probably around 4 p.m. UK time (9:30 p.m. Indian time) or I will be around tomorrow morning as usual.

:-)

Hello @JohnRennie sir

@PrateekMourya hi :-)

@JohnRennie Hi! Are you free?

@YouKnowMe yes :-)

@JohnRennie Why does this give error?

I assume that \:f00f is meant to be [

What are these pathetic numbers? :-O

Unicode?

Try:

Yes this works

I was wondering why that doesn't work...

You managed to get some weird characters into you text somehow.

Yes, that's what I got when I copied and pasted your text. I deleted the brackets and retyped them and it was fine.

Where is the text coming from? Are you copying and pasting it from a web site or some other document?

Uhm.. .nb converted to .pdf by Mathematica .

And you are then copying and pasting from the pdf back into Mathematica?

Yes

was*

The characters must be getting converted to similar looking UNICODE characters in the pdf, then when you paste them back into MM it doesn't know what to make of the UNICODE.

Ohk

Good night(Evening)! :-)