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jkasdhk
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The critical temperature is the temperature at and above which vapor of the substance cannot be liquefied, no matter how much pressure is applied. WHY?????
 one year ago
 one year ago
jkasdhk Group Title
The critical temperature is the temperature at and above which vapor of the substance cannot be liquefied, no matter how much pressure is applied. WHY?????
 one year ago
 one year ago

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jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
@shubhamsrg
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
Because the kinetic energy at critical temperature becomes too much, in comparison to the internal attractive forces, hence, attractive forces are unable to dominate, regardless of the applied pressure.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
Glad i could help. :)
 one year ago

jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
i have another question...
 one year ago

jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
Suggest two reasons why the noble gases become less ideal in their behavior down the group?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
Only good reason I can think off is Increase in size, don't know the 2nd reason.
 one year ago

jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
is this correct that noble gases get heavier down the group?
 one year ago

jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
ok, so as they get heavier, their grip on the electrons weakens, they become more capable of losing an electron and therefore become less ideal... can this be the another reason??
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
This is the first reason only I'd say.
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.1
since the atomic radius of the gas increases as it goes down the period, it causes there to be a decrease in the effect that the nucleus's positive charge has on the valence electrons ( more electrons in between, and more distance ). This causes the valence electrons to become less tightly bound to the atom, which translates to a deviation from the noble gases tendencies to remain nonreactive.
 one year ago

jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
@zaara ok this is the 1st reason.. What about second?
 one year ago

jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
i also knew this one :) but dont know what will be the other...
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.1
it might be the ionization energy,it decreases down the period so that removal of electron becomes easy.. so they tend for a less ideal behavior.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
Indeed.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
Though everything is related to increase in size.
 one year ago

Zaara Group TitleBest ResponseYou've already chosen the best response.1
right! size matters... that will be the major reason bt when going for a 2nd option, hope i thing ionization energy is commentable...
 one year ago

Carl_Pham Group TitleBest ResponseYou've already chosen the best response.1
The reason noble gases become less ideal going down the column is that (1) the atoms become bigger, and (2) the London dispersion forces between them get stronger, since those forces are proportional to the number of electrons in an atom. You'll recall the essential assumptions of the ideal gas model are that (1) the particles have zero size, so they do not collide (collisions would increase the pressure, relative to the ideal model); and (2) they do not experience attractive forces, which would reduce the pressure relative to the ideal model. You can see that the ideal gases as you go down Group 8A increasingly violate both (1) and (2).
 one year ago

Carl_Pham Group TitleBest ResponseYou've already chosen the best response.1
Also, you description of the meaning of the critical point is flawed. It's not that above the critical T the "gas" cannot be "liquified"  it's that above the critical T there is no meaningful distinction between liquid and gas. You might just as well say it's a liquid that cannot be turned into a gas. As for why that's true: that's requires a fairly subtle understanding of the nature of phase transitions, and what the real differences are between a liquid and a gas. A little more than you want right now, I think.
 one year ago

jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
alright... i understand now.. thanks
 one year ago

jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
i have another question: Why compressibility factor is unity at Boyle's temperature?
 one year ago

jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
i know the answer: If compressibility factor = 1 then degree of nonideality= z1 = 11 = 0 this means that the gas is ideal.... But this is not the REASON....!
 one year ago

Carl_Pham Group TitleBest ResponseYou've already chosen the best response.1
Because the Boyle temperature is *defined* to be where Z = 1, that's all. You might as well ask why 0C is where ice freezes  becaus that's how 0C was defined, that's all. What you might be asking is why such a temperature exists at all. Why isn't Z > 1 or Z < 1 always? The answer to that is that the repulsive forces that would tend to make Z > 1 have no important temperature dependence. They are there regardless of the temperature, and don't change much, because they come essentially from the fact that two bits of matter can't occupy the same space. That is, particles collide. Nothing much happens when the temperature is higher or lower, except that particles collide harder. Not so for the atractive forces that tend to make Z < 1. These forces become much more important at lower temperatures. This is because attractive forces have a finite range and finite strength. They're like a little pool of molasses surrounding each atom: if another atom runs through that pool, it slows down as it struggle through the molasses. The degree to which these attractive forces matter is determined, roughly speaking, by how long, on average, particles spend running through molasses. If they are traveling slowly (temperature is low) they spend a lot more time dragging themselves through molasses, and the pressure will be significantly lowered. But if they are traveling quite fast (T is high) they will fly through the molasses very quickly, and it will have very little effect on the pressure. So, in summary, at very high temperatures, repulsive forces must dominate, and Z > 1. At very low temperatures, attractive forces must dominate, and Z < 1 (we know at a low enough temperature the atoms or molecules will simply stick together). Since Z > 1 at some high T and Z < 1 at some low T, it follows logically there must be some intermediate T at which Z = 1 exactly. We define this as the Boyle temperature. I should add that for a model system in which there *are* no attractive forces, Z > 1 always and the Boyle temperature is 0. There are no real systems of this nature, although things like He come close.
 one year ago

jkasdhk Group TitleBest ResponseYou've already chosen the best response.0
Thanks a lot!! :)
 one year ago
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