Calculating Translational Energy in a Nitrogen Gas Vessel

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Homework Help Overview

The discussion revolves around calculating the translational kinetic energy of nitrogen gas contained in a 5L vessel at a temperature of 27 degrees Celsius and a pressure of 3 atm. Participants explore the relevant equations and concepts related to kinetic energy in gases.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss how to determine the number of moles of nitrogen gas and subsequently the total number of molecules. There are questions about the application of Avogadro's number and the distinction between total translational energy and average kinetic energy per molecule.

Discussion Status

Some participants have provided clarifications on the calculations needed to find the total translational energy and the average kinetic energy per molecule. There is an ongoing exploration of the relationships between the variables involved, with multiple interpretations of the equations being discussed.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for solving the problem. There is also a focus on ensuring the correct application of physical constants and definitions related to gas properties.

trah22
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Homework Statement


a 5L verssel contains notrogen gas at 27 degrees celsius and 3 atm. Find a) total translational kinetic energy of gas molecules and b)average kinetic energy per molecule


Homework Equations


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

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 don't 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 I am not sure how to apply it...could someone help me out
 
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trah22 said:
2. Homework Equations [/b]
Etrans=3/2nRT and etrans=3/2NkT, N=6.022x10^23 n=mass/molarmass,
R=.0821

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" (Boltzmann'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
 
the density is 1.251 g/l so multiplying that by 5 gives 6.255 but i don't understand one thing, after u divide this mass by 14(molarmass of Nitrogen) don't u get the number of moles because n=m/M so wouldn't 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 don't know what to apply to calculate the average KE per molecule
 
Last edited:
trah22 said:
the density is 1.251 g/l so multiplying that by 5 gives 6.255 but i don't understand one thing, after u divide this mass by 14(molarmass of Nitrogen) don't u get the number of moles because n=m/M so wouldn't u then multiply that by avogadros number to get the total number of molecules?

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
[tex] N=\frac{6.255}{m_{\textrm{mole}}}N_{\textrm{avogadro}}=\frac{6.255}{14}N_{\textrm{avogadro}}=<br /> \frac{6.255}{14/N_{\textrm{avogadro}}}=\frac{6.255}{m_{\textrm{molecule}}}[/tex]

Regardless how you choose to break up the calculation. The next step is to take the above value of [tex]N[/tex] and multiply it by [tex]3kT/2[/tex]. Cheers.
 
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 doesn't quite make sense...
 
trah22 said:
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 doesn't quite make sense...

yeah, it does make sense. The total translational energy is
[tex] \frac{3}{2}Nk_bT[/tex]
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 [tex]\frac{3}{2}Nk_bT[/tex] divided by the total number of molecules [tex]N[/tex]. In other words:
[tex] \left(\frac{3}{2}Nk_bT\right)/N[/tex]

But that *is* just [tex]\frac{3}{2}k_bT[/tex]
 
thanks for that great explanation man, u cleared it up completely for me:cool:
 
I'm glad that I helped. Cheers.
 

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