Berkely Review Question-Electricity

[September24]

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"Suppose protons were boiled off of a filament instead of electrons and they considered to obey the ideal gas law. Assume that temeprature remains the same, the KE of ejected protons compared to KE of ejected protons would be"

A- greater, due to the mass of proton
B- smaller, due to mass of proton
C-, smaller due to the magnitude of charge of protons
D- Same

Answer is D (the same)

Berkeley says that KE is related to 3/2kT so if temperature does not change, KE does not either. I thought KE also equals 1/2MV^2 so shouldn't a proton have more KE?

 

[monkeyvokes]

"Suppose protons were boiled off of a filament instead of electrons and they considered to obey the ideal gas law. Assume that temeprature remains the same, the KE of ejected protons compared to KE of ejected protons would be"

A- greater, due to the mass of proton
B- smaller, due to mass of proton
C-, smaller due to the magnitude of charge of protons
D- Same

Answer is D (the same)

Berkeley says that KE is related to 3/2kT so if temperature does not change, KE does not either. I thought KE also equals 1/2MV^2 so shouldn't a proton have more KE?

Yeah, I agree with you and hope someone can provide more insight. When boiling charged particles off of a filament (as with a cathone ray), their motion is because of an electric field. The magnitude of force applied should just be due to their charge, and I would think that a larger mass would result in less acceleration.
The process of boiling off charged particles is called thermionic emission. I found this information:

So a larger mass should confer smaller velocity and less kinetic energy. Hopefully someone can clarify our confusion.

ref: http://www.physics.csbsju.edu/370/thermionic.pdf

 

[sps27]

I have come across this one before too, but I think when they mention gas laws etc., realize they are talking about KE as a fn of temperature as is understood under gas laws. The concept of temp comes from average KE of particles and an increase in temp is related to increase in KE, so if the temp remains same, the average KE will be the same.

 

[monkeyvokes]

I have come across this one before too, but I think when they mention gas laws etc., realize they are talking about KE as a fn of temperature as is understood under gas laws. The concept of temp comes from average KE of particles and an increase in temp is related to increase in KE, so if the temp remains same, the average KE will be the same.

ohh right, the point is that the particles are to be treated as a gas under the ideal gas law. Now the passage makes sense, thanks

 

[September24]

Oh okay. For gases, KE=3/2RT (or kT) and not 1/2MV^2?

KE is sort of confusing between a solid, liquid and gas.

For fluids, increasing velocity decreases temperature and KE since fluid velocity reduces random motion.

For gases, KE is realted to temperature only?

It's sort of confusing.

 

[monkeyvokes]

Oh okay. For gases, KE=3/2RT (or kT) and not 1/2MV^2?

KE is sort of confusing between a solid, liquid and gas.

For fluids, increasing velocity decreases temperature and KE since fluid velocity reduces random motion.

For gases, KE is realted to temperature only?

It's sort of confusing.

As stated in EK chem book:
four chars of ideal gas not in a real gas:
1. zero volume gas molecules
2. gas molecules exert no force other than repulsive forces due to collisions
3. completely elastic collisions
4. the average kinetic energy is directly proportional to temperature

 

[sps27]

Oh okay. For gases, KE=3/2RT (or kT) and not 1/2MV^2?

KE is sort of confusing between a solid, liquid and gas.

For fluids, increasing velocity decreases temperature and KE since fluid velocity reduces random motion.

For gases, KE is realted to temperature only?

It's sort of confusing.

Yes the key words here are gas laws and so we have to think, kinetic molecular theory of gases.

As goes to the R v/s k issue. KE = 3/2(kt) where k = boltzman constant. The ideal gas law PV = nRT can also be written as PV = NkT. So judging by the two eqn u can see that nR = Nk so k = nR / N. So to make matters simple, lets assume 1 mole so in that case n = 1 and N = no of molecules per mole = Avogadros number. So you can say, that Boltzman constant (k) = Universal Gas Constant (R) / Avogadros number. This is all not very important but I hope u can understand the relation between Boltzman constant (k) and Universal Gas constant (R).


n=no of moles
R=universal gas constant
N= Number of molecules
k= boltzman constant