Confused about the Boltzman distribution curve and ideal gas law

[larrylu]

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Ok so from Datbootcamp and my own memory I was pretty sure that the average kinetic energy of all ideal gas at the same temperature is the same, or average kinetic energy of all ideal gas is only proportional to temperature.

The question on Destroyer(2014 version) , gen chem #57. Choice C states that "All molecules of a gas is proportional to the absolute temperature" is false. And choice A " Kr has a lower average speed than Ne at the same temperature".

Can anyone explain this to me please??

 

[BrazilianRider]

Is it talking about ideal gasses or real gasses? That's all I can think of. Also, maybe it's referring to Rate of Effusion?

 

[FeralisExtremum]

The average KE of all ideal gas at the same temperature is the same. Keep in mind that is the average, not necessarily the KE of every single gas particle. If you were to actually graph the KE of every single molecule of an ideal gas at a certain temperature you would see a distribution almost like a bell curve: some with the average KE, some lower, some higher, etc. This curve (when graphing the speed of the ideal gas particles - distribution is the same) is called a Boltzmann distribution curve. So it is false to say that all molecules of a gas are proportional to the absolute temperature, because they are only all proportional on average. It is this specific wording that makes the choice incorrect.

I don't have the 2014 version of Destroyer but I assume you mean to say that choice A "Kr has a lower average speed than Ne at the same temperature" is true, which it is. Remember that as you stated, the average KE of all ideal gas at the same temperature is the same. The formula for KE that we're using here is KE = 1/2 * mV^2 (or more specifically for this case, KEavg = 1/2 * m* Vavg^2). If two ideal gases have the same KE on average, but have different masses (as Kr and Ne do), then the one with the higher mass would have to have a lower average velocity for their average KE's to be equal.

 

[larrylu]

The average KE of all ideal gas at the same temperature is the same. Keep in mind that is the average, not necessarily the KE of every single gas particle. If you were to actually graph the KE of every single molecule of an ideal gas at a certain temperature you would see a distribution almost like a bell curve: some with the average KE, some lower, some higher, etc. This curve (when graphing the speed of the ideal gas particles - distribution is the same) is called a Boltzmann distribution curve. So it is false to say that all molecules of a gas are proportional to the absolute temperature, because they are only all proportional on average. It is this specific wording that makes the choice incorrect.

I don't have the 2014 version of Destroyer but I assume you mean to say that choice A "Kr has a lower average speed than Ne at the same temperature" is true, which it is. Remember that as you stated, the average KE of all ideal gas at the same temperature is the same. The formula for KE that we're using here is KE = 1/2 * mV^2 (or more specifically for this case, KEavg = 1/2 * m* Vavg^2). If two ideal gases have the same KE on average, but have different masses (as Kr and Ne do), then the one with the higher mass would have to have a lower average velocity for their average KE's to be equal.

Thank you so much man. It's so clear now.