# Homework Help: Average kinetic energy of particles of an ideal gas

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1. May 22, 2015

### Sipko

1. The problem statement, all variables and given/known data
Determine the average value of the kinetic energy of the particles of an ideal gas at 0.0 C and at 100 C (b) What is the kinetic energy per mole of an Ideal gas at these temperatures.

I took the above right out of the pdf we got from our professor.
I know that:
The gas constant is 8.3145 = R
T = Temperature

What got me confused here is that this question seems to be a two-parter. So according to this task I have to find the average kinetic energy for particles AND moles. Is there a difference? Or am I done already?

2. Relevant equations

Ek = 3/2RT --- As far as I can tell.

3. The attempt at a solution
So far I got down to this:

0.0 C = 273K;
Ek= 3/2 x 8.3145 x 273k =3404.79 J/mol = 3.405 kJ/mol
100 C = 373K
Ek= 3/2 x 8.3145 x 373k =4651.96 J/mol = 4.651 kJ/mol

2. May 22, 2015

### Orodruin

Staff Emeritus
Yes, there is a difference. There is significantly more kinetic energy in a mole than in the average particle. A mole consists of $N_A$ particles.

3. May 22, 2015

### Sipko

And there I was hoping it was a typo. But tell me, is my attempt correct? As far as I understand I found the kinetic energy of a mole right?

4. May 23, 2015

### Orodruin

Staff Emeritus
Yes, and it seems fine as long as your gas is ideal.

5. May 23, 2015

### Sipko

So I have looked into the Avogadros constant and the boltzmann constant and I came up with this formula:
Ek = 1/2mv2 = 3/2kT
where k = boltzmann constant = 1.3806x10-23, or R/NA. Where NA= Avogadros constant, and R = 8.3145 (Gas Constant)
from there I go:

Ek = 3/2x(1.3806x10-23)x273 = 5.65x10-21 (for 0.0 C or 273K)
Ek = 3/2x(13806x10-23)x373 = 7.7244x10-21 (for 100 C or 373K)

Right?

6. May 23, 2015

### Orodruin

Staff Emeritus
Yes. Apart from one very important fact:
Boltzmann's constant is ca 1.38x10-23 J/K, just as the gas constant is ca 8.31 J/(mol K).

It is good practice to always state the units in physics, even in intermediate computations. Apart from being more correct, it will also help you spot errors and dimensional inconsistencies. Also, your result should have units of Joule, if you only state a number, it has no meaning as an energy - you must to specify the units here!

7. May 23, 2015

### Sipko

Yes I always forget that.
So the answers would be :
0.0C = 5.65x10-21 J/K
100C = 7.7244x10-21 J/K
Now it should be correct.

8. May 23, 2015

### Orodruin

Staff Emeritus
Yes. With the (somewhat picky) comment that C is a unit of charge while °C is a unit of temperature.

9. May 23, 2015

### Sipko

Ok ok just my keyboard layout does not allow me to place that sign quickly. But as long as its understood what is meant then it shouldn't be a problem. Thanks for the help. This task is done.

10. May 23, 2015

### Orodruin

Staff Emeritus
If you click the ∑ symbol in the PF editor, you will get a selection of useful symbols which you can simply click to insert.