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Volume of 1 mole of Gas at STP

 
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Sep24-11, 08:27 PM   #1
 

Volume of 1 mole of Gas at STP


Can anyone give an explanation (other than algebraically) as to why one mole of a gas under Standard Temperature and Pressure and behaves like an ideal gas will always have the same volume (22.4 L)?
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Sep24-11, 09:40 PM   #2
 
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pV=nRT for an ideal gas, so if temperature and pressure are constant, then so is volume.

The reason the volume is the same for any ideal gas is because an ideal gas uses the assumption that the molecules are much smaller than the volume of the gas, and they collide elastically and there are no intermolecular forces, so for these reasons, the type of molecule won't affect the equation above.
Sep25-11, 04:24 PM   #3
 
Quote by JSGandora View Post
Can anyone give an explanation (other than algebraically) as to why one mole of a gas under Standard Temperature and Pressure and behaves like an ideal gas will always have the same volume (22.4 L)?
A somewhat elliptical answer is that because a mole of Gas A will have exactly the same number of molecules (Avogadro's Number) as a mole of Gas B.

Moreover, for an Ideal Gas, you don't even have to stick with NTP. No matter what the temperature and no matter what the pressure, the number of molecules in a fixed volume will be exactly the same for any gas or any mixture of non-reacting gases.
Sep26-11, 08:00 AM   #4
 
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Volume of 1 mole of Gas at STP


Quote by klimatos View Post
Moreover, for an Ideal Gas, you don't even have to stick with NTP. No matter what the temperature and no matter what the pressure, the number of molecules in a fixed volume will be exactly the same for any gas or any mixture of non-reacting gases.
I don't think you really meant what you said there. The density of a gas, or anything else for that matter, is not fixed.

In the case of a gas: n/V = P/RT

AM
Sep26-11, 02:59 PM   #5
 
Quote by Andrew Mason View Post
I don't think you really meant what you said there. The density of a gas, or anything else for that matter, is not fixed.

In the case of a gas: n/V = P/RT

AM
Andrew,

I can see how my wording might be interpreted in a way that I did not intend. What I meant was that for any specific temperature (T) and for any specific pressure (P), the number of molecules per cubic meter (n) is the same for any gas or any mixture of non-reacting gases.

n = P/kT
Sep26-11, 06:47 PM   #6
 
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Quote by klimatos View Post
Andrew,

I can see how my wording might be interpreted in a way that I did not intend. What I meant was that for any specific temperature (T) and for any specific pressure (P), the number of molecules per cubic meter (n) is the same for any gas or any mixture of non-reacting gases.

n = P/kT
Right. That is true only for any ideal gas, which I think is what you were trying to say.

AM
Nov29-12, 08:19 AM   #7
 
I understand that the number of molecules in an isolated system will remain the same, no matter what you expand the volume to, raise the temperature to etc.
But how can the equation N=P/kT (I'm going to assume you meant uppercase N for number of molecules there, rather than lowercase n for moles) be derived from PV=NkT; completely disregarding V?
Nov29-12, 09:09 AM   #8
 
Quote by Gavandeshaq View Post
I understand that the number of molecules in an isolated system will remain the same, no matter what you expand the volume to, raise the temperature to etc.
But how can the equation N=P/kT (I'm going to assume you meant uppercase N for number of molecules there, rather than lowercase n for moles) be derived from PV=NkT; completely disregarding V?
klimatos has redefined the quantity "n" and is using it to denote "number of molecules per cubic meter" rather than "number of molecules" or "number of moles" alone.

So when he writes n=P/kT you should read it as n/V = P/kt.

Obviously that's algebraicly equivalent to PV=NkT.
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