Quick test for 2012 candidates

[robot125]

Hey, i have a quick test for the class of 2012 and anybody else who wants to give it a shot..ready?? ok..

1)how do molecules in a gas exchange energy?

2) what effect, if any, does the exchange have on the maxwell-boltzman distribution curve?

3) give two changes that would cause the number of molecules with energy greater than activation energy to increase..

good luck!! they are very simple though...

Click the spoiler for the answers.

1) through collisions

2)it flattens it out

3)Temperature(increase), Catalyst

Edited April 22, 2012 by Summer Glau
added the spoiler

[LMaxwell]

How about a test for you: Is this your homework?

[robot125]

so that's why people aren't answering. No, its not my homework. however, seeing as there is no way to prove it, i guess i should delete this then. Can you help? i don't know how to. im relatively new to ibsurvival.

[LMaxwell]

If you say it's not your homework then mods could leave the test up (if you want). That way people can still do the questions to test their chemistry knowledge, fulfilling the original aim of this thread.

Hope this helped,

brofessional

p.s. As a suggestion, maybe put the answers to the test in a spoiler so that the candidates who do the test can check their answers.

[Award Winning Boss]

Alright, here are the answers to the questions:

1) through collisions

2)it flattens it out

3)Temperature(increase), Catalyst

I see what you did there... but I don't think you understood what brofessional meant.

you put a spoiler in your post by doing this [ spoiler ] (without the gaps) [ / spoiler ]

[omri]

I don't think they'd ever ask us how energy between gas particles are exchanged.

Second, it does not flatten out the curve. As some gain and some lose, but there will always be extremes (I think).

[robot125]

thank you for the clarification award winning boss. @omri, i actually took that question from a past paper so, they can. Oh, and the flattening out was kind of a guess on my part, i was actually hoping for some1 to correct me.

[Drake Glau]

Following the 2nd law of thermodynamics (keep this in a closed system btw) the molecules/particles of gas will EVENTUALLY all have the same energy. If you think of it this way then the curve would do the complete opposite of what you said. It would cause it to spike. The graph would also shift to the left a little since the high energy particles keeping the graph to the right would begin to go away causing the bulk (the spike) to move to a position of less energy.

Keep in mind the axes of this graph. If it were to flatten, then some particles would be gaining energy while other particles are reaching absolute zero. Thermodynamics will never agree with that

Edited April 25, 2012 by Drake Glau

[omri]

1 and 3 are pretty straight forward..

but number 2 is a tricky one, gah the IB always catch you somewhere.

@Drake Glau, are you sure about that?

Yes, according to the second law that should happen.. but it just sounds so strange.. would this be a very very long process?

Also, at any given temperature, there's an associated curve for the average kinetic energies.

Do you have any links you can refer to to back that up? (I'm a bit skeptical, sorry )

[Drake Glau]

http://www.webchem.net/notes/how_far/kinetics/maxwel2.gif

That's the curve/graph you were referring to in your question right?

If all the particles were to eventually contain the same energy (due to the 2nd law, and yes it would take forever but think of it as a process that is slowly changing the graph) then your "x" coordinates for your points would all begin to converge into on spot causing all your y coordinates to follow and spike in that area.

[omri]

Yep, that's the curve.

But why should they all attain the same kinetic energy?

All the collisions are elastic between the molecules. Why would they average out and all have the same kinetic energy?

I guess I just have a hard time applying the second law to this system.. because yes, that is what SHOULD happen,

Can you explain this in terms of collisions and not just in terms of the second law?

Thanks

[Drake Glau]

Particle A hits Particle B, some of the energy from A transfers to B if they are travelling in the "same direction." If they are travelling towards each other, they both lose some energy to heat/light/all those things that happen when things collide.

Now imagine if there were billions upon billions of these particles...

One transfers energy to another, who transfers some to another, who could possibly transfer some back to the original, or another particles. They all have "equal" probability of having a collision happen at every instant from every angle. So, thinking of the graphs movement as a process rather than going from graph A to graph B, the curve would 1) go to a state or lower energy due to collisions of opposite directions and 2) begin to spike up around that area as the particles continue to collide and normalize their energy to all be near equal (it is a curve...)

By spike, I am saying that those particles of high energy to start with can't just disappear, but they are no longer at that energy but are now at a lower energy and since # of particles is your y axis the graph would begin to move up due to the conservation of mass.

Cool, sort of related, topic that might make more sense is the thermal death of the universe. It follows the same idea. Entropy is ALWAYS increasing and particles are always colliding so in theory...eventually...in some non-quantifiable # of years...every particles will have the same energy and thus ALL work will stop and the universe will "die." It was something cool we learned in physics if you want to check it out since it kind of relates and might make more sense since it is on a scale of the universe.

[omri]

Particle A hits Particle B, some of the energy from A transfers to B if they are travelling in the "same direction." If they are travelling towards each other, they both lose some energy to heat/light/all those things that happen when things collide.

Now imagine if there were billions upon billions of these particles...

One transfers energy to another, who transfers some to another, who could possibly transfer some back to the original, or another particles. They all have "equal" probability of having a collision happen at every instant from every angle. So, thinking of the graphs movement as a process rather than going from graph A to graph B, the curve would 1) go to a state or lower energy due to collisions of opposite directions and 2) begin to spike up around that area as the particles continue to collide and normalize their energy to all be near equal (it is a curve...)

I think you misunderstand the concept of entropy.

Molecules of a gas in a close system can NOT lose energy. Collisions between molecules are completely elastic, that is, kinetic energy is conserved.

Energy is lost to heat in macroscopic colissions (between two bowling bowls for example).

I'll explain why there is no energy loss in a collision between two molecules.

Molecule A has kinetic energy and molecule B has kinetic energy. They both collide. According to you, some energy has been lost to thermal energy. But temperature is just a measure of the average kinetic energy of molecules. They cannot get "hot" like a bowling ball. When they get "hot" the meaning is that they gain kinetic energy.

Yes, perhaps they will all have the same energy eventually, but the system as a whole will not lose energy, it will not go to a lower energy level as you said.

Your statement about the heat death of the universe is correct. Eventually, all particles will have the same energy. This is because in ANY transfer of energy, some energy is always degraded (turned to thermal energy that cannot be done to do any useful work),

[Drake Glau]

Pretty sure you just answered your own question then