Thermodynamics - Thermal Energy of a gas

AI Thread Summary
The discussion centers on calculating the thermal energy of hydrogen gas using the root mean square (rms) speed and the number of molecules. The initial calculation using the formula provided resulted in an incorrect thermal energy value, prompting confusion regarding the relationship between translational and total internal energy. It was clarified that for diatomic molecules like H2, the total internal energy includes both translational and rotational degrees of freedom, leading to a multiplication factor of 5/3. Participants emphasized the importance of understanding the kinetic theory of gases and the equipartition law to grasp why diatomic molecules have additional degrees of freedom. A recommendation was made to review class materials for a deeper understanding of these concepts.
WolfeSieben
Messages
20
Reaction score
0

Homework Statement



The rms speed of the molecules in 1.2 g of hydrogen gas is 1800 m/s.
What is the thermal energy of the gas?

Homework Equations



m = Mass of 1 H molecule: 1.67 x 10^-27 kg
N = Total # of molecules = 7.19 x 10^23 molecules
c = rms speed

The equation provided by my tutor was:

E(therm) = N x (1/2)(m)(c)^2



The Attempt at a Solution




Using the formula i was provided with above:

E(therm) = (7.19 x 10^23 molecules)(1/2)(1.67x10^-27 kg)(1800^2 m/s)

Which equates to 1,945 J.

This answer is wrong, and I am also confused because this number seems oddly close to the translational energy (e=1/2mv^2) which equates to 1.9 KJ earlier in the assignment)

Can someone tell me if I way out to left field and provide some hints?

Thanks
 
Physics news on Phys.org
"Thermal energy" can refer to the total internal energy of the gas, not just the translational kinetic energy. The H2 molecule has two degrees of rotational freedom in addition to the three degrees of translational freedom. So you have to multiply the translational energy by 5/3 to get the total internal (thermal) energy.

AM
 
Perfect, so I multiplied 1.9 KJ by 5/3 and it gave me the correct answer.

However I am interested in understanding the concept behind the question. Why is it that the degrees of rotational freedom and 3 degrees of translational freedom equate to multiplying by 5/3?

Thanks for your help!
 
WolfeSieben said:
Perfect, so I multiplied 1.9 KJ by 5/3 and it gave me the correct answer.

However I am interested in understanding the concept behind the question. Why is it that the degrees of rotational freedom and 3 degrees of translational freedom equate to multiplying by 5/3?

Thanks for your help!
It is a bit complicated. You should review the kinetic theory of gases and the equipartition law.

A monatomic gas (single atoms) has no moment of inertia (virtually all the mass is located at the tiny centre). So the atoms have translational kinetic energy in three directions and no rotational or vibrational energy. According to the equipartition theory, at thermal equilibrium N atoms having 3 degrees of freedom will have 3NkT/2 energy (each degree of freedom carrying NkT/2 energy).

Now, with an H2 molecule the molecular mass is spread out because there are two nuclei separated by a comparatively large space (ie compared to the size of one nucleus). This gives the diatomic molecule a moment of inertia about two axes (the 2 axes perpendicular to the axis joining the nuclei). It has virtually no moment of inertia about the third axis joining the nuclei because the mass is all on that axis. That gives it 2 additional degrees of freedom for a total of 5, each with NkT/2 energy (according to the equipartition law).

For reasons that I am not quite clear on myself, light nuclei such as H2 lack a vibrational mode at least at lower temperatures so they only have 5 degrees of freedom, 3 translational and 2 rotational each carrying NkT/2 of energy.

AM
 
@WolfeSieben: I am sure your tutor didn't asked you to solve that question without teaching you the theory/concept behind it. I advise you to go back to your class notes/course book and study the chapter dealing with kinetic energy (or internal energy) of non monoatomic (or polyatomic) gases.
Andrew Mason gave a good explanation, but you need study material and also examples.

good luck !
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

Similar threads

Back
Top