Internal energy of a gas and kinetic energy, "typical velocity"

AI Thread Summary
The discussion addresses the internal energy of a gas, specifically questioning the kinetic energy formula K_avg = 3/2kT and its application to one particle versus a mole, highlighting the need to multiply by Avogadro's number. It also critiques the use of root mean squared velocity in deriving "typical velocity," suggesting a misunderstanding of the term's definition in the context of probability distributions. Additionally, there is confusion regarding the internal energy of diatomic gases, with participants noting that the expected factor should be 5/2 instead of 3/2. Errors in calculations leading to an incorrect temperature of 244,000 K are pointed out, emphasizing the importance of considering volume in these calculations. Overall, the conversation reveals significant discrepancies in understanding fundamental concepts in gas physics.
laser
Messages
104
Reaction score
17
Homework Statement
See description
Relevant Equations
Kavg = 3/2kt
Source: Shankar Yale OCW physics
Screenshot_1.png

I have three questions here:

1. K_avg is 3/2kT, sure. But isn't this the kinetic energy of one particle only? So why isn't the answer multiplied by avogadro's number (because one mole).

2. When doing the "typical velocity" derivation, I noticed that they used the root mean squared formula to derive the expression there. But I would think "typical velocity" means the probable velocity, i.e. the velocity when dP/dv = 0 in the probability vs velocity curve.

3. Just wondering, as nitrogen is a diatomic gas, why doesn't the internal energy have a factor of 5/2 as opposed to 3/2?

Thanks!
 
Physics news on Phys.org
laser said:
Homework Statement: See description
Relevant Equations: Kavg = 3/2kt

Source: Shankar Yale OCW physics
View attachment 345547
I have three questions here:

1. K_avg is 3/2kT, sure. But isn't this the kinetic energy of one particle only? So why isn't the answer multiplied by avogadro's number (because one mole).

2. When doing the "typical velocity" derivation, I noticed that they used the root mean squared formula to derive the expression there. But I would think "typical velocity" means the probable velocity, i.e. the velocity when dP/dv = 0 in the probability vs velocity curve.

3. Just wondering, as nitrogen is a diatomic gas, why doesn't the internal energy have a factor of 5/2 as opposed to 3/2?

Thanks!
  1. You are right and the answer is wrong. The average energy needs to be multiplied by Avogadro's number.
  2. "Typical" velocity is not typical use. It could mean r.m.s. velocity in the problem author's mind.
  3. Because the author of the solution apparently doesn't know the difference between the two or has calculated the internal energy of a diatomic molecule in the approximation ##2 \approx 1.##
Also note that the answer T = 244000 K is 10 times larger than the correct answer. The author of the solution has noted that it "is clearly a very high temperature" but has not considered that there might be a calculational error here.

I suggest that you reconsider this site as a source of learning.

Edited to fix problem with not considering the volume. See posts #5 and #6.
 
Last edited:
kuruman said:
Also note that the answer T = 244000 K is 10 times larger than the correct answer.
I think it is correct
 
laser said:
I think it is correct
I don't think so.
##2\times 1.013\times 10^5=202,600##. If you divide by 8.314, you don't get ##244,000## which is a larger number. when multiplied by the volume gives the right answer.

Edited to fix problem with not considering the volume. See posts #5 and #6.
 
Last edited:
kuruman said:
I don't think so.
##2\times 1.013\times 10^5=202,600##. If you divide by 8.314, you don't get ##244,000## which is a larger number.
volume is 10 m^3
 
Ah, yes. That's what I missed.
 
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...
Back
Top