Does KE=3/2nRT apply to all gases or only monatomic gases?

[ChloeYip]

"KE=3/2nRT=3/2kT" is applicable to only monatomic gas or all kind of gas?
What about " KE=1/2*fkT" ? (f: degree of freedom) Is it implies the same thing as "KE=3/2nRT=3/2kT" ?

Thank you.

 

[Phylosopher]

degrees of freedom change with temperature, the following scale shows how molecules degrees change:

Usually when solving a problem in undergraduate level, we assume that we are in room temp. So for mono. f=3 and diatomic particles f=5. Thus only translation and rotation freedom exist, but why mono. and dia. are different?

for mono. the shape is a sphere (symmetric in rotation), you can move it in x, y, and z but rotating it won't change anything so you have f=3+0=3. For dia. The shape is a line or rod (asymmetric in rotation) you can move it in x, y, and z direction but you can also rotate it in two angles (let's say: θ and φ) so you have f=3+2=5.

 

[ChloeYip]

thanks for your reply, but I do understand the concept of degree of freedom
however, from my notes, "KE=3/2nRT=3/2kT" is applicable to all kind of gas" made me doubtful.
why the formula need not consider degree of freedom? or I have misunderstanding to the notes?
thanks

 

[Hiranya Pasan]

Yes 3/2kT can be applied only for mono atomic gasses, Mono atomic gasses have only 3 degrees of freedom (X,Y,Z) direction, that's why K.E comes out with 3/2 factor (f=3), But when go to diatomic and poly atomic gasses those are having degrees of freedom more than 3 including rotational and vibration degrees of freedom ( f=5,7... )then K.E comes with factor 5/2,7/2... likewise

 

[ChloeYip]

So molecules like N2 should have KE=5/2kT, instead of 3/2kT, right?
However, when I read the post of https://answers.yahoo.com/question/index?qid=20110123011943AAjliwr , question 39 of it,
is it wrong for "Now use the equation for average molecular kinetic energy: Ek = (3/2)kT, where k is the Boltzmann constant (1.38×10ˉ²³ J·Kˉ¹) and T is the absolute temperature. "?
Thank you very much.

 

[Hiranya Pasan]

So molecules like N2 should have KE=5/2kT, instead of 3/2kT, right?
However, when I read the post of https://answers.yahoo.com/question/index?qid=20110123011943AAjliwr , question 39 of it,
is it wrong for "Now use the equation for average molecular kinetic energy: Ek = (3/2)kT, where k is the Boltzmann constant (1.38×10ˉ²³ J·Kˉ¹) and T is the absolute temperature. "?
Thank you very much.

Yes N2 has 5 degrees of freedom and KE=5/2kT
Yeah there must be a mistake unless they have neglected rotational and vibration effects

 

[ChloeYip]

So how to solve this problem?

my copy of the question is:
" What is the total translational kinetic energy in a test chamber filled with nitrogen (N2) at 2.16 * 10^5 Pa and 20.7°C? The dimensions of the chamber are 4.00 m * 5.70 m * 7.40 m. The ATOMIC weight of nitrogen is 28.0 g/mol, Avogadro’s number is 6.022 * 10^23 molecules/mol and the Boltzmann constant is 1.38* 10^-23 J/K."
And the answer is " 5.47 * 10^7 J"
Even I tried with KE=5/2kT , still I can't get the answer...

Thanks

 

[Hiranya Pasan]

So how to solve this problem?

my copy of the question is:
" What is the total translational kinetic energy in a test chamber filled with nitrogen (N2) at 2.16 °— 105 Pa and 20.7°C? The dimensions of the chamber are 4.00 m °— 5.70 m °— 7.40 m. The ATOMIC weight of nitrogen is 28.0 g/mol, Avogadro’s number is 6.022 °— 1023 molecules/mol and the Boltzmann constant is 1.38 °— 10-23 J/K."
And the answer is " 5.47 °— 10^7 J"
Even I tried with KE=5/2kT , still I can't get the answer...

Thanks

Make no mistake about, K.E= f/2kT is for per molecule or atom, the total K.E will be f/2Nkt, N is the number of atoms or molecules and also this question is about only transnational K.E, therefore you have to use 3/2Nkt (3 transnational degrees of freedom).
First you have to find the number of nitrogen molecules (N) using ideal gas equation. then use 3/2Nkt to calculate total transnational K.E

 

[ChloeYip]

I tried to calculate in the same way, i have got 25.644 *10^6 as the answer.
http://upload.lsforum.net/users/public/c39058IMG_3107k11.jpg

Is there any problem with my calculation? Thanks

 

[Hiranya Pasan]

I tried to calculate in the same way, i have got 25.644 *10^6 as the answer.
http://upload.lsforum.net/users/public/c39058IMG_3107k11.jpg

Is there any problem with my calculation? Thanks

use directly PV=NkT (or also you can us PV=nRT)
your calculation of N is incorrect; N=PV/kT I got N=8.9916 x 10^27 molecules

now using K.E=3/2NkT = 3/2*8.9916*10^27*1.38 x 10^-23*293.7= 5.4665x10^7 Joules

 

[ChloeYip]

I see. Thanks for the much simpler way for finding the number of molecules. I can get the answer now.
Thank you very much.

[Sorry, I found problem right after this post is post >.<]

 

[Phylosopher]

I tried to calculate in the same way, i have got 25.644 *10^6 as the answer.
http://upload.lsforum.net/users/public/c39058IMG_3107k11.jpg

Is there any problem with my calculation? Thanks

Hiranya Pasan did give you the right way to do it. But so you understand your mistakes, can you rethink about how you found the no. of moles?

 

[ChloeYip]

Wait~

K.E=3/2kT = 3/2*1.38 x 10^-23*293.7= 5.4665x10^7 Joules

You didn't use N,thus P at all and you found the answer! Why no need to multiply N? and the answer is not dependent on pressure?

 

[Hiranya Pasan]

Wait~

You didn't use N,thus P at all and you found the answer! Why no need to multiply N? and the answer is not dependent on pressure?

sorry I did a mistake when typing, you need to multiply by N, because to find Total translational energy

 

[ChloeYip]

Indeed, I just found that we can simply KE=3/2*pV :)
Thank you very much anyway~

 

[Hiranya Pasan]

Indeed, I just found that we can simply KE=3/2*pV :)

Yes, correct for this case, but KE=3/2*pV is not going to work every time, Energy is a function of Temperature only plus KE=3/2*pV works only for ideal gases, so better to go with f/2kT