Raoult's Law

For an ideal solution, the partial pressure of a component above the solution is directly proportional to the concentration of that component. This relationship is known as Raoult's Law.

[@junioralxse1041]

I am disappointed that I didn't find a channel before I did. thanks for the tutorial. I have a chemical thermodynamics exam tomorrow !!

[@PhysicalChemistry]

Glad you found it eventually, and hopefully you did well on your exam.

[@charleyroe8995]

Great review for my P-chem test this week! Very helpful video

[@PhysicalChemistry]

Glad it helped. Good luck on your test!

[@mortezakhoshbin]

what beautiful you say Pchem.thanks alot prof.stuart

[@PhysicalChemistry]

Welcome!

[@captainamericawhyso5917]

This is so well explained! Thank you sir

[@PhysicalChemistry]

You're quite welcome. I'm glad you liked it

[@user-uu1we7db2i]

super helpful omg ty so much!

[@PhysicalChemistry]

You're very welcome

[@esoto1386]

4:24 why didn't you write the total number of moles (na+nb) on the left side of the equation (delta-Gmix) when multiplying?

[@PhysicalChemistry]

Good question. I **did** multiply by (n_A + n_B). But I just used that to convert the molar (intensive) Gibbs energy (ΔḠ with a line on top) to the extensive Gibbs energy (ΔG without a line on top). ΔG = (n_A + n_B) ΔḠ In case you want more info on molar properties or intensive / extensive properties, see here: kzbin.info/www/bejne/q2i5gJhqlMegh7c

[@ishaan9017]

3:05..Sir can we directly replace the Boltzman's constant with the Gas constant?

[@PhysicalChemistry]

Yes! If you multiply Boltzmann's constant by Avogadro's number (to get a value per mole), you'll see that it's actually the exact same number as the gas constant. There's more detail here: kzbin.info/www/bejne/qXqZemaAj5mnhpY

[@ishaan9017]

@@PhysicalChemistry Sir you mean to say that the expression Delta(S mix)= -k(XA lnXA+XBlnXB) is entropy of mixing per molecule and if we replace k(boltzman's constant) with Gas constant(R) it represents molar entropy of mixing. Am I correct on this Sir?

[@PhysicalChemistry]

@@ishaan9017 Yes, that's exactly right

[@ishaan9017]

@@PhysicalChemistry Thanks so much Sir