# Homework Help: Total translational energy

1. Jul 14, 2007

### trah22

1. The problem statement, all variables and given/known data
a 5L verssel contains notrogen gas at 27 degrees celcius and 3 atm. Find a) total translational kinetic energy of gas molecules and b)average kinetic energy per molecule

2. Relevant equations
Etrans=3/2nRT and etrans=3/2NkT, N=6.022x10^23 n=mass/molarmass,
R=.0821

3. The attempt at a solution

for part a)3/2(not sure how to get n)(3(.0821)(300)

i multiplied .0821 by 3 because the pressure in the question is 3 atm and one atm is .0821 not sure if this is right... i dont quite understand what to do with the volume for part b, i know i have to use avogadro's number since its asking per molecule but im not sure how to apply it...could someone help me out

2. Jul 14, 2007

### olgranpappy

In your equation N is *not* Avagadro's number. N is the number of Nitrogen molecules. You need to look up the density of Nitrogen gas and multiply that by the given volume to find the total mass of the gas. Then lookup the mass of a single molecule of Nitrogen and divide the total mass of the gas by the mass of a single molecule to find N.

Once you have N, multiply it by 3/2 then multiply it by "k" (Boltzman's Constant) then multiply it by the temperature in Kelvin. This will give you the total translational energy of the 5 liters of Nitrogen. Cheers

3. Jul 15, 2007

### trah22

the density is 1.251 g/l so multiplying that by 5 gives 6.255 but i dont understand one thing, after u divide this mass by 14(molarmass of Nitrogen) dont u get the number of moles because n=m/M so wouldnt u then multiply that by avogadros number to get the total number of molecules?

could u also tell me the difference between find the total trslational kinetic energy of the gas molecules vs the average KE per molecule, i understand that Etrans=nKt gives total translational but i dont know what to apply to calculate the average KE per molecule

Last edited: Jul 15, 2007
4. Jul 15, 2007

### olgranpappy

The mass of one mole of Nitrogen is 14 grams. That is not the same as the mass of one molecule of Nitrogen. The mass of one molecule of Nitrogen is 14/(6.02*10^23) grams.

So, yes, if you dived 6.255 grams by 14 grams you do get the number of *moles* and then multiplying by Avogadro's number gives the number of molecules.

That works, but that's not what I told you to do. If you do what I told you to do, which was to divide 6.255 grams by the mass of one nitrogen *molecule*, you will also get the same answer since
$$N=\frac{6.255}{m_{\textrm{mole}}}N_{\textrm{avogadro}}=\frac{6.255}{14}N_{\textrm{avogadro}}= \frac{6.255}{14/N_{\textrm{avogadro}}}=\frac{6.255}{m_{\textrm{molecule}}}$$

Regardless how you choose to break up the calculation. The next step is to take the above value of $$N$$ and multiply it by $$3kT/2$$. Cheers.

5. Jul 15, 2007

### trah22

alright thanks for clearing that up,

kb=1.38x10^-23
N=2.689x10^23

Etrans=3NkT/2
=3(2.689x10^23)(1.38x10^-23)(300K)/2
=1669.9J/K

this is the amount of total translational energy of the gas molecules, for part b of the question (finding the average kinetic energy per molecule) is correct equation (im looking at my notes) just 3kbT/2 without any N, but that doesnt quite make sense.....

6. Jul 15, 2007

### olgranpappy

yeah, it does make sense. The total translational energy is
$$\frac{3}{2}Nk_bT$$
which you just found. This is the *total* translational energy. I.e. the sum of the translational kinetic energies of all of the molecules.

The "average kinetic energy per molecule" is defined to mean "that quantity which, when multiplied by the total number of molecules, gives the total energy." It's just the total energy $$\frac{3}{2}Nk_bT$$ divided by the total number of molecules $$N$$. In other words:
$$\left(\frac{3}{2}Nk_bT\right)/N$$

But that *is* just $$\frac{3}{2}k_bT$$

7. Jul 15, 2007

### trah22

thanks for that great explanation man, u cleared it up completely for me

8. Jul 15, 2007

### olgranpappy

I'm glad that I helped. Cheers.