Kap FL2 Questions about correct answers

[nightowl]

2 questions, for anyone who can explain: Help will be greatly appreciated

#59 physical sci: How does the work done by the automatic pitcher change as the angle of the barrel to the horizontal increases?
Answer: the work done does not change, because the air pressure behind the disk is unchanged
explanation: work energy theorom,work done equals change in KE
my q: what about Fdcostheta? How do you know to look at KE and not this, which is determined by horizontal distance?

#66 ps: CO2(g) + C(s) <--> 2 CO9(g)
when a system is stablized at 1200K a sample of helium is injected into the furnace. what should happen to the amt of carbon dioxide in the system?
answer: it should stay the same
exp: helium unreactive, dalton's law
my q: wouldn't adding a gas, even if it's unreactive, shift the rxn to the left based on Lechatier's principle?

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[SailCrazy]

I had *exactly* the same quesitions.

I'm going to try to track down a couple of professors today. If/when I get the answer, I'll post the info.

Good luck!

 

[SailCrazy]

I didn't get a chance to track down any professors, but I did speak with my Kaplan instrutor about these this evening. None of these explanations is going to be a solid theoretical "by the book" answer, but hopefully will be a good starting point...

#59 physical sci: How does the work done by the automatic pitcher change as the angle of the barrel to the horizontal increases?
Answer: the work done does not change, because the air pressure behind the disk is unchanged
explanation: work energy theorom,work done equals change in KE
my q: what about Fdcostheta? How do you know to look at KE and not this, which is determined by horizontal distance?

Work done relative to gravity is dependent on the angle. Total work, however (in both the X and Y direction) is relative to the velocity of the ball, which is not changing.

#66 ps: CO2(g) + C(s) <--> 2 CO9(g)
when a system is stablized at 1200K a sample of helium is injected into the furnace. what should happen to the amt of carbon dioxide in the system?
answer: it should stay the same
exp: helium unreactive, dalton's law
my q: wouldn't adding a gas, even if it's unreactive, shift the rxn to the left based on Lechatier's principle?

The reaction equilibria is relative to the partial pressures of each individual gas *within the reaction*. If instead of this example, the reaction was taking place in a cylinder with a compressible piston, for example, and the pressure was increased by compression of the piston only and therefore reduction of volume, then the partial pressure of each gas would change. This change in partial pressures would account for the increased total pressure and would effect equilibrium (a la Lechatlier).
When an inert gas is added, the total pressure does increase. However, this is because this additional gas has its own partial pressure which is added to those from the gasses that were already in the furnace, NOT because the partial pressures of the previously present gasses change. The partial pressures within the reaction do not change, so there is no change in CO2 present.


Anyone else have any thoughts? How are these explanations for a start?

 

[Tinker Creek]

2 questions, for anyone who can explain: Help will be greatly appreciated

#59 physical sci: How does the work done by the automatic pitcher change as the angle of the barrel to the horizontal increases?
Answer: the work done does not change, because the air pressure behind the disk is unchanged
explanation: work energy theorom,work done equals change in KE
my q: what about Fdcostheta? How do you know to look at KE and not this, which is determined by horizontal distance?

#66 ps: CO2(g) + C(s) <--> 2 CO9(g)
when a system is stablized at 1200K a sample of helium is injected into the furnace. what should happen to the amt of carbon dioxide in the system?
answer: it should stay the same
exp: helium unreactive, dalton's law
my q: wouldn't adding a gas, even if it's unreactive, shift the rxn to the left based on Lechatier's principle?

Regarding 66:
No, because inert gases don't change the equilibrium constant. I don't quite remember the reasoning behind it, but you could probably google it.

edit: ok, I see sailcrazy already explained it.