Mean Free Path

[@aahilrafiq4850]

This is the simplest and Best explanation i found till now and i Finally understood , Thank You Professor

[@PhysicalChemistry]

You're welcome. Glad it helped

[@ayushverma4693]

Thankyou so much sir for this effort ,it helped me lot .I'm persuing ug course . Love from India🙏🙏🇮🇳

[@PhysicalChemistry]

Great, I'm glad to be of help

[@tejas28saini]

Thanks for the lecture really helped a lot 😊

[@arhamapon9282]

Sir you literally saved me Thank you very very much. Respect from Bangladesh .

[@PhysicalChemistry]

My pleasure. But all I did was present the information. You had to do the work of learning it

[@4m0d]

in my book the small part about relation between Vrms and Vrms relative wasn't mentioned and left me unsatisfied. thanks for the video.

[@kevinkamaumungai951]

many thanks from Kenya

[@PhysicalChemistry]

You're very welcome! And thank you for saying hi

[@medardjosephat1559]

thank you sir for your work kindly I would like to ask on the relationship above what does letter K implies an what is its value

[@PhysicalChemistry]

This k (lowercase) is the Boltzmann constant. It is the same as the gas constant, R, just in different units. The equation PV = NkT is the same as the equation PV = nRT that you may be more familiar with. More details here: kzbin.info/www/bejne/qXqZemaAj5mnhpY

[@crusader1234]

Excellent explanation Sir!! 12:26 Dear Professor , how rms velocity can be replaced by average velocity with square root 2 intact?

[@PhysicalChemistry]

The conclusion that v_rel = √2 v is just a statement about vectors. (It will be true for any vectors that point in all directions equally.) So it's equally true about v_rms and v_av and the magnitudes of any other vectors we care to discuss.

[@crusader1234]

Thanks a lot ! Your video really helped.❤

[@capybun]

Man you're so underrated

[@PhysicalChemistry]

Thanks. I much prefer that to being called overrated!

[@KazeNokorai]

Good for jee advanced

[@張翔喻-x9h]

the pressure unit is"pa"? why isn't it atm?

[@PhysicalChemistry]

Pascal (Pa) and atmospheres (atm) are both units of pressure. 1 atm = 101325 Pa. But if you use atm in the formula for mean free path, the units in the denominator don't cancel the units in the numerator. Your mean free path would end up in units of J m⁻² atm⁻¹. This **is** a unit of length, but not a useful one! If you use units of atm for pressure, and keep careful track of your units (as you always should), then you will end up with this ugly unit for your answer. You'll then need to add some unit conversions to get your answer into units of meters. I used units of Pa for the pressure so that the units would cancel easily, and result in a mean free path with units of meters.

[@張翔喻-x9h]

@@PhysicalChemistry thanks a lot 😍I just finish my third chemistry test yesterday ,I am so glad I saw your video before I did the mission impossible test ,he gave us a hole sheet of Zaa and mean free path ,20 questions in 2 hours 😭

[@angpeiying8108]

may l ask if wads the difference between the formula for mean free path taught in this video and the formula 1/square root 2ppid^2 and RT/square root 2pi d^2NaP note that Na is avagado's number

[@PhysicalChemistry]

The gas constant (R) is the same as Boltzmann's constant (k) multiplied by Avogodro's number (N_A): R = k N_A. So your equation with R / N_A will give the same result as the equation in this video, which just uses k instead. Your equation without a factor of kT or RT in the numerator is not correct. You can tell this because it doesn't have the right units. If you use that equation, your result won't have dimensions of length, so it can't possibly describe a mean free path.

[@solunke.pranav]

Reallyyyy helped me alot😇

[@ahmedalajlouni7173]

Thank you

[@truthphilic7938]

By Allah, I loved it.

[@PhysicalChemistry]

I'm glad to hear that

[@truthphilic7938]

Dear professor, Nobody explained me why we should take = 0 till you. You said why average of all dot products cancel out. By Allah, I loved it. But, Most of others say, "they aren't correlative, so zero" It seems so confusing. Are they explaining it correctly?

[@PhysicalChemistry]

Yes, both explanations are correct. If two vectors aren't correlated, then vᵢ·vⱼ will be negative exactly as often as it is positive, and so the average will be zero.

[@truthphilic7938]

@@PhysicalChemistry dear professor, I have a another question. Why it feels like / or we consider, molarity of solid and pure liquid as 'one' to do equilibrium math? like Ksp, Kc and Kw derivation.

[@PhysicalChemistry]

@@truthphilic7938 The way this is often explained is: the concentration of the pure solid or pure liquid is a constant; it doesn't change as the amount of the solid / liquid changes. So this constant concentration can just be included in the equilibrium constant. A more detailed explanation is that it's most correct to be using the activity of the solid / liquid instead of the concentration, and the activity of a pure solid or pure liquid (under standard state conditions) is equal to 1.

[@truthphilic7938]

@@PhysicalChemistry Thanks, professor. But I am not still clear about the concept. is it a concept of active mass(activity)? I just found a question answer in stackexchange. chemistry.stackexchange.com/questions/19001/why-is-active-mass-of-a-pure-solid-or-liquid-always-taken-as-unity

[@PhysicalChemistry]

@@truthphilic7938 Activity is the term we use. "Active mass" is not very common at all. The activity of a pure solid or liquid in its standard state is equal to 1 because those are the reference states against which the activity of a substance is measured. This is a more detailed subject than can be learned from comments. I would recommend reading about activity in a textbook. Here's the video of mine on the topic: kzbin.info/www/bejne/a2nKY3uAd9Gkn7M

[@a.h.6710]

I needed this 🥲Thank you

[@PhysicalChemistry]

My pleasure