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Hey /sci/. I had this very puzzling assignment regarding thermodynamics, it's not in English originally so I translated it as well as I could.
An empty pot is left on a stove. The cross-section of the pot is a circle. On top of the pot there is a tightly shut ceramic lid with a mass of 0,420kg. The height of the pot is 0,22m. Let’s assume that the air inside the pot is at equilibrium with the environment. How much thermal energy does the air inside the pot has to receive before air starts escaping from under the lid?
The internal energy for a two-atom gas is [math]\frac{5}{2}[/math] [math]k_B[/math] [math]T[/math] where [math]k_B[/math] is the Boltzmann constant [math]\frac{R}{Na}[/math]
The answer I got is 2,4 Joules, which sounds like a ridiculously low amount. It's the correct answer according to the assignment, but I can't wrap my head around the fact that this could be answered without knowing the radius of the pot.
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Bump, did I just stump this board?
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Its been a while, but if you think about the condition that gas from the inside escapes, that doesnt really specify how much gas is escaping.
So the lid really only has to be lifted enough to let a single molecule through, wich wouldnt be a lot of expansion from the gas.
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>>8632975
>The answer I got is 2,4 Joules
How did you get to this?
My answer is dependent upon ambient temperature, pressure and the area of the lid.
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>>8633163
Never mind, forgot to multiply by mass.
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>>8633163
Kinda funny, isnt it? How only the height of the pot somehow affects the energy needed. Also, we don't know the number of molecules N but since the internal energy given by 5/2 kbT is per molecule, it just works this way.
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>>8633187
Also the answer is 2,3 Joules rounded, I messed that one up in the OP.
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Well, since you are not considering the area of the lid, it becomes a problem in 1 dimension. So you don't need the radius.
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>>8632975
0
Just lower the ouside air temp or air pressure.
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>>8633187
Think of it this way. If you had a much wider pot with a lid weighing the same, then you'd need less pressure increase (thus less heat transfer) to raise the lid, but you'd have to raise the pressure of more mass since the pot is larger. These two effects cancel out since the equations governing them are linear.
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>>8633187
You are wrong. You need the relation between surfaces, and since the surface is the same, the relation is 1. You need to know radius if surfaces differ on area.
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>>8632975
>but I can't wrap my head around the fact that this could be answered without knowing the radius of the pot
You did the sums and made it to the end without the radius.
Clearly you didn't need it.
Wouldn't be the first time.
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>>8633205
Interpretation is that pressure is equal around all the volume.
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