At 0 Kelvin, all molecular motion has stopped completely as this is the minimum temperature possible and since temperature is related to the average kinetic energy of the particle….if the temperature is zero, so is the KE, thus no motion.
There is nothing particularly special about the temperatures 0 °C and 0 °F, other than water freezing at 0 °C but this is irrelevant right now. At both of these temperatures, the molecules are not at a minimum KE, they can be cooled much further.
Absolute zero (0 K) converted onto the Celsius scale is -273.15 °C. -273 °C is very close to absolute zero, but not quite there; it is only a few hundredths of a degree off. At this temperature, molecular motion is still possible, but just barely.
What is the ‘minimum’ possible motion of a molecule? No molecule can go slower than ‘zero [length units per second]’ (You can’t go slower than stopped). The temperature corresponding to zero molecular motion (the slowest, minuminum, possible speed) is 0 Kelvin. At any temperature greater than 0 K, the molecules can always be slowed / cooled further.
2007-03-10 03:51:05
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answer #1
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answered by mrjeffy321 7
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A. The Kelvin scale starts at absolute zero where all molecular motion has stopped. On the other hand, celcius has zero at the freezing point of water and 100 degrees at the boiling point of water
2007-03-10 03:32:04
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answer #2
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answered by Zefram 2
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0 K. 0 Kelvin is the lowest temperature possible, so the motion of the molecules must by the least at that temperature.
2007-03-10 03:33:25
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answer #3
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answered by physandchemteach 7
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gas particles move slowest in the lowest temperature due to the fact that particles require kinetic energy inorder to move as kinetic energy increases the motion of the particle increases as well. in this case the answer is D
(hint: use the conversions between the different units inorder to obtain the smallest temperature)
An easy question hope u can handle it by now sweetie :)
2007-03-10 04:16:39
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answer #4
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answered by Pharmalolli 5
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A, absolute zero.
2007-03-10 03:53:07
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answer #5
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answered by Anonymous
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