BET Isotherm - Discussion

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  • Опубликовано: 16 окт 2024
  • The BET model for adsorption gives the same prediction as the Langmuir model, when the pressure is very low. As the pressure approaches the saturated vapor pressure, though, the BET model predicts that an infinite amount of adsorbate will adsorb onto the surface, as the gas condenses.

Комментарии • 16

  • @rameshkalagnanam2946
    @rameshkalagnanam2946 2 года назад +1

    Prof. Stuart - you videos are absolutely a class of their own! You are a very effective teacher.

  • @MacClips
    @MacClips 2 года назад +1

    PhD student here working with this equation in a seminar in two weeks. These videos are a god send and you make this so much more understandable so thank you!

    • @PhysicalChemistry
      @PhysicalChemistry  2 года назад +1

      Great! Once you're an expert, post your own tutorial, and we can all learn more from you

  • @amoledraikoon4432
    @amoledraikoon4432 Год назад +2

    Does "Teller" here refers to Theoretical Physicist "Edward Teller"?

    • @PhysicalChemistry
      @PhysicalChemistry  Год назад

      Yes, that's right. The same Edward Teller who is known as the father of the H-bomb, and who appears in the movie Oppenheimer.

    • @amoledraikoon4432
      @amoledraikoon4432 Год назад +2

      @@PhysicalChemistry Yes Sir! That's the very reason I asked it....I also watched Oppenheimer, but was not completely sure if it was the same Teller from BET model, hence found your awesome video..by the way, the movie was like a dream come true, saw many great scientists whose theories we study today 😃

  • @neroscr7
    @neroscr7 Год назад +1

    Would you be able to explain why more gas adsorbs onto a material as pressure increases? This appears to be the general trend shown on all isotherms, regardless of type.

    • @PhysicalChemistry
      @PhysicalChemistry  Год назад +1

      Think of it as an example of Le Chatelier's Principle. The process we're considering when molecule A adsorbs from the gas onto a surface is:
      A(g) -> A(ads)
      Any time you increase the amount of the reactant (gaseous A) by increasing the pressure, the equilibrium will shift to reduce that increase, thus consuming some A from the gas phase and causing more to absorb onto the surface

    • @neroscr7
      @neroscr7 Год назад +1

      @@PhysicalChemistry amazing, thank you

  • @xinyueliu5498
    @xinyueliu5498 2 года назад +1

    what about c=1, maybe a liquid water surface + water vapor? then theta = P/P*/(1-P/P*). In the physical picture, water vapor only adsorbs to the water surface when p/p*=1. but why theta is still a positive number? (btw your courses are fantastic!!)

    • @PhysicalChemistry
      @PhysicalChemistry  2 года назад

      Even if c=1, as in your example, that doesn't guarantee that K=1.
      (The quantity K has dimensions of inverse pressure. Even if it equals one in some particular set of units, that is only an accident; it will not equal one in another set of units.)
      The surface coverage will always be between zero and infinity, so it must always be a positive (or at least non-negative) number.

  • @xinyueliu5498
    @xinyueliu5498 2 года назад +1

    does K' here also has a similar dependence on temperature, just like the langmuir model? i am curious if we can also write down the K in an exp() form. thanks!

    • @xinyueliu5498
      @xinyueliu5498 2 года назад +1

      in the general form, adsorption amount = c(p/p*)/(1-p/p*)/(1-p/p*+cp/p*), i cannot tell how does the temperature affect the adsorption. Assuming the relative humidity (or relative pressure p/p*) is maintained the constant, only c should show some temperature dependence. I saw that the c is like c=exp(epsilon0-epsilonL)/kT. then we should have the similar discussion as in the "Langmuir Isotherm"? Thank you!

    • @PhysicalChemistry
      @PhysicalChemistry  2 года назад

      Yes, K' definitely has some temperature dependence. It is the equilibrium constant for the surface + gas ⇌ adsorbed molecule reaction. So it is very similar to K from the Langmuir model.

    • @PhysicalChemistry
      @PhysicalChemistry  2 года назад

      Yes, you're right