Calculating KE of Ideal Gases: Understanding its Independence from Molecule Mass

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The discussion centers on the kinetic energy (KE) of ideal gases, highlighting that the average kinetic energy is independent of the mass of the molecules. This is explained by the principle that, although heavier molecules have lower velocities, they still possess the same average kinetic energy as lighter molecules when in thermal equilibrium. The relationship between temperature and energy is emphasized, with temperature representing the mean energy per unit quantum, where the quanta are the molecules. In non-ideal gases, additional energy is associated with rotational modes, leading to different specific heat ratios. The concept of equipartition is introduced, stating that energy is equally distributed among available degrees of freedom, which varies depending on whether the gas is monoatomic or diatomic. Overall, the discussion clarifies how gas properties are defined by pressure and temperature, and how molecular mass influences velocity rather than average kinetic energy.
apache
when you equate the two formulas for ideal gases, one is evetually left with a formula to calculate the ke. of the ideal gas (3/2kt i think) how come the ke is independent of the mass of the molecule ?
 
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If you mix many balls with different mass in a box and shake the box, each ball will have the same average kinetic energy regardless difference in their mass.
 
wait... i got an idea, does the formula mean that although the ke. will be what the formula spits out, the velocities for heavier moleules (lets say of another ideal gas) would be lower, and thus the formula would be indepndent of mass.. ?
 
Yes.
 
A simple way to see it.

The gas properties are defined by the pressure and temperature. The temperature is basically the mean energy per unit quantum, in this case atoms or molecules are the "quanta" involved, (this isn't the standard definition of quantum) and the pressure is the mean energy per unit volume. The heavier molecules move slower to have the same energy as the lighter ones.

In non-ideal gasses some of the energy is tied up in rotational modes, which is why they have differing ratios of specific heat.
 
thanks guys,
i think i finally got it !
 
Mobility of elastic gases

Originally posted by apache
thanks guys,
i think i finally got it !

Hi apache,
I am attaching the table referred to below with the hope that the narrow field of this posting causes scrambling.

VERTICAL MOBILITIES of ATMOSPHERIC GASES

The Mobilities according to Graham’s law are listed in the table above as referenced to the mobility of N2, the major molecular constituent of the atmosphere. Positive factors indicate propensities of given gases to rise while negative factors indicate falling tendencies. Those labeled “gas” are usually Brownian gases in that their boiling points are above ambient atmospheric temperatures. For example when solid Iodine sublimes, a maroon cloud hovers nearby until dispersed and/or condensed. It should be remembered that volatility is a function of boiling point temperature as contrasted with mobility, which is the inverse square root of the molecular mass.
 

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Originally posted by Tyger
The temperature is basically the mean energy per unit quantum, in this case atoms or molecules are the "quanta" involved, (this isn't the standard definition of quantum) and the pressure is the mean energy per unit volume...

In non-ideal gasses some of the energy is tied up in rotational modes, which is why they have differing ratios of specific heat.

Standard postulate here is "In thermal equilibrium energy is equally distributed among all available degrees of freedom (equipartition)", so temperature T is defined in such way that each degree has in the average kT/2 amount of energy. If a molecule is monoatomic, it has 3 degrees only(x,y,z), thus <E>=3kT/2, if diatomic then it has two more (rotational) degrees, thus <E>=5kT/2, and so on.
 
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