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Topic: thermodynamics monotomic gas  (Read 1524 times)

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Offline p.ella

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thermodynamics monotomic gas
« on: March 27, 2013, 10:10:29 PM »
GRADE 12 CHEMISTRY: Why would a monotomic gas like Helium require less heat than a diatomic gas like hydrogen to undergo the same temperature increase?

I think it has something to do with bonds and rotational/translational/vibrational energy of molecules but don't know exactly help

Thank you everyone!


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« Last Edit: March 28, 2013, 04:19:33 AM by Dan »

Offline Corribus

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Re: thermodynamics monotomic gas
« Reply #1 on: March 28, 2013, 10:02:57 AM »
Temperature is basically a measure of the average kinetic energy of molecules in a system.  When you put heat into a system, the temperature rises because that heat is absorbed by molecules, which, among other things, makes thems move around faster.  However, as you've noted, different systems will raise their temperature to different degrees for the same amount of heat energy applied.  This relationship is called the heat capacity (or specific heat capacity, when expressed on a per mass basis).

So basically what you're asking is: what determines the specific heat capacity of a substance?

Let's answer it using the specific case you've mentioned.  First answer a very simple question: What is the primary difference between a monatomic gas like helium and a diatomic gas like hydrogen? And with that in mind: if you apply heat to a monatomic gas like Helium, the temperature will rise a certain amount because a large portion of that heat will go into increasing the kinetic energy of isolated Helium atoms.  If you apply the same amount of heat to di-hydrogen gas, why would you expect the average kinetic energy to be less than it would be for the same mass of helium? 

Think about it, see if you can come up with an answer.  Post it here and I will tell you if you have got the right answer.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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