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Total translational kinetic energy

  • #1
[SOLVED] Total translational kinetic energy

1.A 0.03m3 vessel contains helium (monatomic) gas at 0.0C and 1.00 atm. The total translational kinetic energy of the gas molecules is (in KJ).

2. 3/2KbT


3. pV=nRT

where p is the pressure, V is the volume, n is the number of molecules present, R is the gas constant (8.31J/(mol*K)), and T is the temperature in Kelvins (273K = 0ºC)
The other equation is that the average translational kinetic energy K of a single molecule is

K = (3RT)/(2N)

where R and T are from the first equation and N is Avogadro's number (6.022E23).
Just sub in numbers:
K = (3 * 8.31 * 273) / ( 2 * 6.022E23) = 5.651E-21 Joules. This is the kinetic energy of one atom of helium at 0º C. Change the first equation around to get n = (pV) / (RT) and then multiply 5.65E-21 by n


The answer in the book shows its suppose to be 4.5kJ, but I'm yet to get that, I'm getting a way off number. Thanks for any help given :+)
 

Answers and Replies

  • #2
hage567
Homework Helper
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2. 3/2KbT
This doesn't look right. It should be [tex] K = \frac{3Nk_BT}{2}[/tex] where N is the number of molecules in the vessel.

K = (3RT)/(2N)
Again, you are missing the N in the numerator. This equation is just the one above with a substitution for k_B.

If you have the N in the numerator, you will have a ratio of [tex]\frac{N}{N_A}[/tex] which is the number of moles of gas, n.

So just start over with the right equation and you should come out to the right answer.
 
  • #3
Awesome, Thanks for showing me where I made my mistake :+)
 

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