- #1

LastTimelord

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So, at 1 kelvin, how fast do the molecules actually move? Is it a direct, or exponential increase to 2 kelvin?

This might be a stupid question. I'm not a scientist.

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- Thread starter LastTimelord
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- #1

LastTimelord

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So, at 1 kelvin, how fast do the molecules actually move? Is it a direct, or exponential increase to 2 kelvin?

This might be a stupid question. I'm not a scientist.

- #2

xlines

- 96

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So, at 1 kelvin, how fast do the molecules actually move? Is it a direct, or exponential increase to 2 kelvin?

This might be a stupid question. I'm not a scientist.

Density does not matter. Relating macroscopical phenomena of thermodynamic equilibrium to atomic level events was subject to much research and is legitimate scientific question. Quick was to relate temperature and velocity of a particle is through equipartition - let say you have pointlike atoms. For every degree of freedom - number of independent ways particle can move - you get 1/2kT of kinetic energy. Such particle has 3 degrees of freedom, so

3*1/2kT = 1/2 mv

where k is Boltzmann constant, T is temperature in kelvins, m is mass of a particle and v is the thing you are interested in - velocity of a particle.

- #3

cmos

- 367

- 1

Regardless, it's clear from the equation that velocity increases as the square root of temperature. So increase the temperature by a factor of 100, the velocity increases by a factor of 10.

- #4

LastTimelord

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This equation is very helpful, but what would the unit of velocity be in this case?

- #5

Mute

Homework Helper

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Density does not matter. Relating macroscopical phenomena of thermodynamic equilibrium to atomic level events was subject to much research and is legitimate scientific question. Quick was to relate temperature and velocity of a particle is through equipartition - let say you have pointlike atoms. For every degree of freedom - number of independent ways particle can move - you get 1/2kT of kinetic energy. Such particle has 3 degrees of freedom, so

3*1/2kT = 1/2 mv^{2}

where k is Boltzmann constant, T is temperature in kelvins, m is mass of a particle and v is the thing you are interested in - velocity of a particle.

Just to be very clear, the equipartition theorem does not give you the velocity of a single particle, it gives you the root mean square (rms) velocity of the atoms of an ideal gas in thermal equilibrium at temperature T. In this case, where all of the atoms are the same (i.e., have the same mass and are classical particles), it follows that this is the rms velocity of a single atom, but that atom could at times be going much faster or much slower than the rms velocity.

That is to say, temperature is related to the

Also, it's worth noting that when one asks how the rms velocity increases with temperature, the word "increases" is a bit misleading. The equipartition theorem relates, for example, the rms velocities and temperatures of two systems in equilibrium, but it only relates the rms velocities and temperatures of a single system at different points in time when the system is in thermal equilibrium at those two points in time.

That is, you can use it relate the rms velocity of a system in equilibrium at temperature ##T_1## at time ##t_1## compared to its rms velocity at a later time ##t_2## and temperature ##T_2##, but only so long as you have waited for the system to settle back down into equilibrium. The equipartition theorem does not hold while you are heating the system from temperature ##T_1## to ##T_2##; i.e., while the temperature is actually increasing - hence why I say "increasing" is a misleading word to use.

So, you can relate the rms velocities and temperatures of two different systems in equilibrium, or you can relate the rms velocities and temperatures of a single system that is in equilibrium at different temperatures at different points in time, but you can't do that while you are actually physically increasing the temperature.

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