# Kinetic Energy of gas Molecules

1. Dec 7, 2008

### Dulcis21

Hi, the question, I'm having problems with is this:

A 1.0 mol sample of hydrogen gas has a temperature of 30 C. What is the total kinetic energy of all the gas molecules in the sample? How fast would a 75 kg person have to run to have the same kinetic energy?

Relevant equations:
K = (3/2)kT
or K = (3/2)nRT

Ok, so I tried using both formulas:

(3/2)(1.38*10^-23)(30+273) = 6.2721 * 10^-21

Then, since its one mole and the question asks for the total KE, I multiplied this answer by Avogadro's number (6.02*10^23) and got

= 3775.8042 Joules.

For the second part, I would just set this number equal to 0.5mv^2 and solve for v but since I got the first one wrong, it doesn't quite work :(

I really need help; I thought I had the right idea but it's not working out. Thanks

Last edited: Dec 7, 2008
2. Dec 7, 2008

### Andrew Mason

Your method is right. Using $U = \frac{3}{2}nRT$, U = 1.5*8.314*303 = 3778.7 Joules.

Think of one mole or 1 gram of hydrogen possessing 3778.7 J of kinetic energy. The effective average (rms) velocity is:

$$v = \sqrt{2E/m}$$

AM

3. Dec 7, 2008

### Dulcis21

I thought my method was right but my homework web-assign isn't accepting that answer :(
It asked for it in 2 sig figs so i put it in as 3.8 *10^3 but it still said no...

4. Dec 7, 2008

### Dulcis21

Oh I figured it out lol, it needed (5/2)kT to work and then multiply by Avogadro's number.

5. Dec 8, 2008

### Andrew Mason

Of course. H2 is diatomic.

AM