Activity Coefficient

[@chemtoniumhabling1182]

Just wanted to drop a comment saying that I've very much enjoyed these videoes. Thanks a lot! They have really cleared up difficult to understand concepts in the physical chemistry course I'm taking!

[@PhysicalChemistry]

Great, thanks for the comment. For the trickier concepts, it definitely helps to hear a couple of different explanations.

[@rameshkalagnanam2946]

Prof. Stuart - you videos are absolutely a class of their own! You are a very effective teacher. I will make it a point to go back to all videos I have seen and post this exact comment!

[@PhysicalChemistry]

Thanks, I appreciate the support!

[@whythelongface64]

Sir, this is really helpful. I will forever be grateful for the clear lesson. My college made physical chemistry seem more difficult than it is. Your videos are Godsend ❤ 🙏

[@beerpan9176]

This is way too helpful for me as a geologist who is trying to understand the thermodynamic concept of mineral (solid-solution) and magma at equilibrium.

[@PhysicalChemistry]

Wow, magma equilibrium sounds like a complicated system

[@sophah6866]

Thanks so much for taking the time to create these videos, they are really well explained.

[@PhysicalChemistry]

You're welcome; thanks for taking the time to comment

[@barriehemming1189]

thank you for the upload you actually made this make sense in my head.

[@rameshkalagnanam2946]

Prof. Stuart - also as someone mentioned in a comment in another of your videos, given the sheer size of the population, you will have tremendous success in reaching out to school and college students within India. The medium of instruction in most parts of the country is English and your videos will be very effective in supplementing whatever they learn from their institutions. For most part, the teachers there (of course, there are exceptions) simply read what is in the text book without explaining the underlying concepts effectively. This way, the student seldom develops a mastery over the topic. Please find a way to increase your visibility in the Indian online educational environment (I dislike using the word "market" - gives everything a commercial flavor).

[@PhysicalChemistry]

That's very kind of you. Like you, I am not inclined towards "marketing". But I'm very happy if students or instructors find them to be beneficial -- that's why I posted them, of course.

[@nawaurielpunabantu9141]

Clear and easy to understand, great video

[@PhysicalChemistry]

Great to hear!

[@Mehdiranjb]

🙏🙏

[@MusafirHoonYaro]

Prof Stuart, I have a question. I will try my best to articulate it clearly. I see the graphs that are being drawn to depict non-ideal behaviour or deviation from Raoult's Law. In order to actually draw the graphs we need points to plot them. How are the points obtained for the 3 behaviours (above Raoult's Law, as per Raoult's Law and below Raoult's Law) actually obtained in order to be able to plot them? This is where I am having some difficulty and I am unable to understand how in the first place we are able to envision these kind of behaviours. If you have covered this in another video, could you kindly point me to it? If not, could you kindly post a video describing the answer to this question. A very BIG THANK YOU in advance.

[@PhysicalChemistry]

The points on the graph are partial pressures. So you would just prepare solutions at various different concentrations x_A, and then measure the partial pressure of the gas above the solution when it is in equilibrium with its vapor. You could do this, for example, by measuring the total pressure and analyzing the composition of the vapor.

[@MusafirHoonYaro]

@@PhysicalChemistry Thank you, Prof. Stuart, for the clarification.

[@rinsim]

Is there any video on this channel about fugacity and how to calculate fugacity and/or activity from a cubic Equation of State?

[@PhysicalChemistry]

Sadly, not yet. That's on my to-do list.

[@murugeshpalanisamy774]

Professor, Would Activity Coefficient be less than one for positive deviation as per your explanation? If we take xA = 0.6, for positive deviation from the graph, the actual aA comes to be less than 0.6 which evaluates activity coefficient to be less than 1. Am I misconceiving some of the concepts? Thanks.

[@PhysicalChemistry]

No, that's not quite right. If x_A is 0.6, and the solution exhibits positive deviations from ideality, then the partial pressure will be larger than 0.6 ⨉ P*, so the activity will be greater than 0.6.

[@snigdhas2959]

Very helpful ! Thank you

[@PhysicalChemistry]

Great! Glad it helped

[@Ganirage]

the vapor pressure of an aqueous solution of 0.5m kno3 at 100°c is 749.1 torr while the vapour pressure of pure water at the same temperature is 760 0 torr calculate the activity of water in kno3 solution

[@Ganirage]

Sir can you help to solve this kind of numericals

[@PhysicalChemistry]

I'll leave that to you, but this video and/or the previous one (kzbin.info/www/bejne/a2nKY3uAd9Gkn7M) are good ones to teach yourself what you need to know

[@GulzarAhmad-sw1kh]

a(H2O) ~ 0.985 assuming KNO3 to be non-volatile. OK?

[@qqn4531]

👍

[@hmm..1560]

Are you writing backwards?

[@PhysicalChemistry]

No, it's reversed digitally. Here's some more info: kzbin.info/www/bejne/j57Ze4mhrq-VgsU