# Homework Help: Rms speed

1. Feb 17, 2016

### KUphysstudent

1. The problem statement, all variables and given/known data
Calculate the rms speed of one nitrogen molecule at 27 Celcius

2. Relevant equations
sqrt(v^2) = sqrt((3kT)/(m))

3. The attempt at a solution
sqrt(v^2) = sqrt((3kT)/(m))
k = 8.62*10^-5 eV/K
T = 300 K
3kT = 0.078 eV
m = 14 g/mol
m = 0.014 kg/mol

3kT/m = 0.078 eV/0.014 kg/mol = 5.571
if 5.571 was 557.1 m/s I would assume it was correct but I doing something completely wrong here, any help will be appreciated. edit nevermind that statement since I have to take the sqaure root also

sqrt(5.571)

Last edited: Feb 17, 2016
2. Feb 17, 2016

### Bystander

3. Feb 17, 2016

### KUphysstudent

Yea sorry it is 27 Celcius which is 300 K :)

4. Feb 17, 2016

### Bystander

Diatomic molecule.

5. Feb 17, 2016

### KUphysstudent

increasing the weight of the molecule does not solve the problem when you would expect it going about 500 m/s on average.

6. Feb 17, 2016

### Bystander

300 m/s: eV conversion, or something equally fatal.

7. Feb 17, 2016

### Staff: Mentor

Units?

8. Feb 17, 2016

### KUphysstudent

3kT/m = 0.078 eV/0.014 kg/mol = 5.571 eV*kg/mol

9. Feb 17, 2016

### Staff: Mentor

And how do you go from there to (m/s)2?

10. Feb 17, 2016

### KUphysstudent

1 eV = 1.6 * 10^-19 J, so adding this to the 3kT/m = (0.078 eV *(1.6 *10*-19 J))/0.014 kg/mol = 8.9264^-19 kg*J/mol
J = N*m = (kg*m^2)/s^2
8.9264^-19 kg^2*m^2/s^2*mol
hmm something fishy is going on here, no idea how kg^2/mol relate if they do at all

Oh wait.
it is 8.9264^-19 m^2/s^2*mol
Still have the 1/mol problem though

11. Feb 17, 2016

### Staff: Mentor

Yes, because you are taking m as the mass of one mole of atomic N, not as the mass of one molecule of N2.

12. Feb 17, 2016

### KUphysstudent

According to my book Boltzmann's constant is 8.62*10^-5 eV/K
but every place online I look at problems similar they use 8.3145 J/mol*K
and if I do this problem method I have seen online, I get 3*(8.3145 J/mol*K)(300 K) / 0.014 kg/mol = 5.345^5 J/kg which makes the units a lot easier to process for me, kg *m^2/ kg*s^2
which gives sqrt(5.345^5 m^2/s^2) = 66.049 m/s , which is still quite slow but better.

13. Feb 17, 2016

### Staff: Mentor

Your value for the Boltzmann constant is correct. The other you cite is the gas constant, defined as $R = N_A k_B$, where $N_A$ is the Avogadro constant. It is used mostly by chemists.

You are still not taking into account that you have diatomic molecules of N2.

That should be 5.345×105, not 5.3455.

14. Feb 17, 2016

### KUphysstudent

Oh yea im missing Avogadro's constant and I also forgot 1 atom is not 1 molecule.

(28.0134 g/mol) / (6.02*10^23 mol^-1) = 4.65*10^-23 g = 4.65*10^-26 kg

3*(8.62*10^-5 eV/K)*(300 K) / (4.65*10^-26 kg) = 1.6092^29 m^2/s^2

v = sqrt(1.6092^29 m^2/s^2) = 990.479 m/s
that is more like it, now it is super fast. thank you very much for all your help :)

15. Feb 17, 2016

### KUphysstudent

took me some time to write the above, but thank you for clarifying :)

16. Feb 17, 2016

### Staff: Mentor

The number is not correct for units of m2/s2

That speed is not the correct one, it is too big. Are you converting from eV to J correctly?

17. Feb 17, 2016

### KUphysstudent

No it looks like I made a mistake with the power.
28.0134/6.02*10^23 = 4.6533^-23
(3(8.62*10^-5)(300))/(4.65*10^-26) = 1.66838*10^24
sqrt(1.66838^24) = 465.09 m/s