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

[laser]

Homework Statement

See description

Relevant Equations

Kavg = 3/2kt

Source: Shankar Yale OCW physics

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!

 

[kuruman]

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.

 

[laser]

Also note that the answer T = 244000 K is 10 times larger than the correct answer.

I think it is correct

 

[kuruman]

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.

 

[laser]

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

 

[kuruman]

Ah, yes. That's what I missed.