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Topic: Ideal gas - energy of molecules  (Read 3457 times)

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Offline cliverlong

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Ideal gas - energy of molecules
« on: May 08, 2008, 10:59:19 AM »
Hi,


I will try and remember question accurately  :)

Question:

We have two ideal gases, A and B. The conditions of each: volume, temperature and pressure are identical. There are equal number of moles of each gas.

The mass of molecules in B are twice those of mass of molecules in A.

However, average kinetic energy of molecules is same in both gases.

Why is this?


Suggested answer

The kinetic energy of a particular molecule is 1/2 mv2. Now since the PVT conditions are the same and there are the same number of molecules of gas in each, the average kinetic energy of the molecules must be the same in both gases. Since mass of particles in B mB are twice those of A mA, the speed of the particles in B vB must be proportionally lower than those in A vA so that average kinetic energy is same.


1/2 mAvA2 = 1/2 2mAvB2

giving

vB = vA / sqrt(2)

My problem

I'm not very convinced by my answer. Is there some standard "equation" that is a bit "stronger" to support my argument?


Clive

Offline Borek

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Re: Ideal gas - energy of molecules
« Reply #1 on: May 08, 2008, 11:22:06 AM »
E = n/2 kT

http://en.wikipedia.org/wiki/Boltzmann_constant

Perhaps someone will give you better link, I have a black hole between ears ATM  >:(
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Offline Yggdrasil

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Re: Ideal gas - energy of molecules
« Reply #2 on: May 08, 2008, 01:58:21 PM »
Suggested answer

The kinetic energy of a particular molecule is 1/2 mv2. Now since the PVT conditions are the same and there are the same number of molecules of gas in each, the average kinetic energy of the molecules must be the same in both gases. Since mass of particles in B mB are twice those of A mA, the speed of the particles in B vB must be proportionally lower than those in A vA so that average kinetic energy is same.


1/2 mAvA2 = 1/2 2mAvB2

giving

vB = vA / sqrt(2)

My problem

I'm not very convinced by my answer. Is there some standard "equation" that is a bit "stronger" to support my argument?


Clive

This answer is exactly correct and I do not think it requires a standard equation to make it stronger.  In fact, in most cases, reasoning such as yours is used to derive some of the standard equations from first principles.  For example, if you combine your reasoning with the equation that Borek posted, you can write an equation for the "average" (more specifically the root mean squared) speed of the molecules in a gas (see http://en.wikipedia.org/wiki/Kinetic_theory#RMS_speeds_of_molecules).

Offline cliverlong

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Re: Ideal gas - energy of molecules
« Reply #3 on: May 08, 2008, 04:15:51 PM »
  For example, if you combine your reasoning with the equation that Borek posted, you can write an equation for the "average" (more specifically the root mean squared) speed of the molecules in a gas (see http://en.wikipedia.org/wiki/Kinetic_theory#RMS_speeds_of_molecules).

Great.

Thanks both of you


Clive

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