# Kinetic energy for a mole of a gas discrepancy

1. Jul 22, 2011

### gkangelexa

Exam krackers says that the average kinetic energy for a mole of a gas is K.E.avg = 3/2 RT

but my chemistry book says that the total kinetic energy for a mole of gas is = 3/2 RT
meaning that
NA (average K.E. of one molecule) = 3/2 RT

These 2 statements are different, for the bottom one says that the total K.E. for a mole of gas is = 3/2 RT, yet the top one says that the average K.E. for a mole is =3/2RT

Which is correct?

2. Jul 22, 2011

### Ygggdrasil

What units does the expression 3/2 RT have?

3. Jul 22, 2011

### gkangelexa

R = 8.314 J/(K x mol)

so 3/2 RT units are J/ mol

Average kinetic energy for one molecule is: 1/2 m (mean square speed)...
so the units for that are: (kg m2) / (s2 molecule)

so this means my chemistry book is correct because

(average kinetic energy for one molecule) (NA) works because the units work:

since NA = molecules/ mol.... the multiplication equals 1 J/mol....

right??

So then why does my Exam Krackers book say that the average kinetic energy for a mole of gas = 3/2 RT...?

4. Jul 22, 2011

### Ygggdrasil

You are correct that (3/2)RT gives the total energy per mole of a monatomic ideal gas. The average kinetic energy of a molecule of that gas is given by a very similar formula, (3/2)kBT, where kB is the Boltzman constant. As you can probably guess, kB = R/NA.

I'm not sure why your other book says that (3/2)RT is the average kinetic energy for a mole of gas.