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pop000 · Apr 25, 2011, 04:23 AM

In the picture you can see Three different tanks contain Three different gases.
The temperature are equal to 25 Degrees Celsius
In all the Three tanks, and it not changing.

I asked to calculate the kinetic energy of the Three different gases.
So I thought to use this formula:3/2*RT this formula is for ideal gas but how can I know if what I got here is an ideal gas ?

Or maybe I need to use in another formula?
Thanks.

Unknown008 · Apr 25, 2011, 08:28 AM

Well, in the lessons I had, I was told to assume that the gases I worked with to have ideal behaviour.

pop000 · Apr 25, 2011, 08:59 AM

OK so if I assume that this gases are ideal do I correct in my formula ?
Thank you.

Unknown008 · Apr 25, 2011, 09:27 AM

Well... you used the wrong constant.

$PV = nRT = NkT$

$U = \frac32nRT = \frac32NkT = \frac12 Nm\bar{c^2}$

$E = \frac12 m\bar{c^2} = \frac32 kT = \frac32 \frac{n}{N}RT$

$\frac{n}{N} = \frac{1}{N_A}$

pop000 · Apr 25, 2011, 10:17 AM

well look here :http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html I think I correct in my formula no ? :)
btw: N = number of molecules is not given in my question info, also n. so I can't use in your formula yes?
thank you again.

Unknown008 · Apr 25, 2011, 10:25 AM

Your site is correct, but what you typed is not :p

See in my post, all the equivalents of E, you'll see the one you can use :)

But still, you don't need n, or N, since you know N_A ;)

pop000 · Apr 25, 2011, 12:07 PM

well so I can use this formula:E=3/2kT (kinetic energy of gas molecule) k=Boltzmann constant T=Kelvin temperature.
so I got the correct formula now? Lol.

sorry for making it to look so hard :) thanks.

Unknown008 · Apr 25, 2011, 12:23 PM

Yes, now, this is correct :)

pop000 · Apr 26, 2011, 12:30 PM

Hi.
Thank you very much for help but if you says that is will be correct to use this formula E=3/2kT and we know that k and T in this case are equal in all the 3 gases so I get that the kinetic energy of each gas here are equal ?

Is it possible ?

pop000 · Apr 26, 2011, 11:28 PM

Lol never mind I did it :)

Thanks.

Unknown008 · Apr 27, 2011, 09:30 AM

Yes that's right. As long as T is constant, the KE of the particles will be constant.

Well, the average KE that is.

Remember that KE of molecules of gas is only affected by their temperature.

pop000 · Apr 28, 2011, 04:08 AM

OK so this what I got:3/2*1.38*10^-23J*K*298K=6.1686^-21 so 6.1686^-21is the average KE of the 3 gases ?

Unknown008 · Apr 28, 2011, 09:04 AM

Yes, I get that too :)

pop000 · Apr 28, 2011, 10:44 AM

YeA thanks you for your patience :)