Igneous Petrology Series: Lesson 6 - Binary eutectic phase diagrams

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  • Опубликовано: 13 сен 2024

Комментарии • 15

  • @tirthadeysarkar1128
    @tirthadeysarkar1128 3 года назад +11

    diopside liquidus and anorthite liquidus is wrong placed here, the area which pointed as liquid+anorthite is basically liquid+diopside area and vise versa, but your explanation is too good

  • @Oceanuniverse_.1
    @Oceanuniverse_.1 6 месяцев назад

    I'm truly grateful ... Thank you so much ( watching you from Jordan 🇯🇴 )

  • @raniamasha9623
    @raniamasha9623 3 года назад +4

    This is so helpful.Thank you so much

    • @WillDSmith7
      @WillDSmith7  3 года назад +1

      Thank you for your comment ☺️

  • @TrendsinGeology108
    @TrendsinGeology108 3 года назад +1

    That was Rapid yet Cool explanation.....good job .......

  • @jackglover3632
    @jackglover3632 2 года назад

    Very helpful but I don’t understand where a few of the numbers came from, for example the percentages in the equilibrium crystallisation.

  • @eliasminkim
    @eliasminkim 2 года назад

    this is divine

  • @muhdlawan7931
    @muhdlawan7931 2 года назад

    Please I want to know why absent of anorthite will hinder the melting of Diopside until at 1553 degree Celsius?

  • @lexluthor1642
    @lexluthor1642 3 года назад +1

    Could you please tell me why is the melting temperature higher for individual phases than it is for the mixtures? I know it has something to do with dG=VdP-SdT but I don't fully understand why that is.

    • @WillDSmith7
      @WillDSmith7  3 года назад +5

      Hey! It is essentially all about energy and entropy. If everything is the same (i.e. all one phase) the bonds will be stronger, entropy will be lower, and therefore the energy required to disrupt the substance will be higher. If you start adding impurities, you're increasing the entropy and thus lowerong the amounts of required to melt it. As with anything, I'm sure there is exceptions to the rule, but I'm yet to come across one in geology! Hope that helps!

  • @noahkang3910
    @noahkang3910 2 года назад

    Hello! I’m having a bit of trouble understanding what you mean when you say you remove one of the products from the system during fractional crystallization/melting. How does a product get removed from a system and what would this look like?

    • @xxxxxx-ef3lj
      @xxxxxx-ef3lj Год назад +1

      One way is that the solid products, once formed, will sink to the bottom because of density difference with the liquid.
      For example, once Olivine crystallizes out of a melt, the melt becomes more silicic than it was prior to the crystallization of Olivine, and olivine, being mafic and therefore more dense than the silicic melt, sinks to the bottom. If the liquid migrates to another location leaving these olivine crystals, the crystals are said to have been "removed" from the system (the liquid).

  • @simonmadileng9097
    @simonmadileng9097 2 года назад

    how does the crystal equilibria change in the Di+An system if we have an open system and crystals are being isolated during cooling

    • @WillDSmith7
      @WillDSmith7  2 года назад +1

      Hi Simon, interesting question. Firstly, it depends on the open system behaviour. For example, if we have an evolved magma that then interacts with a less evolved magma - we would produce less evolved minerals (its how we explain reverse zoning in cumulus minerals). However, if we have perfect fractional crystallisation, then the crystals would be removed from the system immediately and be unaffected. This is very rare in natural systems and crystals will always want to equilibrate with the magma they interact with (we call this process trapped liquid shift). Hopefully this answers your question.

    • @simonmadileng9097
      @simonmadileng9097 2 года назад

      @@WillDSmith7 Thank you, got that