What Is the Correct Temperature for O2 Gas to Achieve an RMS Speed of 699 m/s?

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The discussion revolves around calculating the temperature required for O2 gas to achieve a root mean square (RMS) speed of 699 m/s using the formula v_{rms} = √(3RT/M). The initial calculations yield a temperature of 627 K, which is questioned for accuracy. Participants clarify that the formula applies to both monatomic and diatomic gases, despite the latter having additional degrees of freedom. A participant mentions their professor suggested using v_{rms} = √(5RT/M), which leads to confusion, as the consensus is that the correct formula is indeed the one with 3 in the numerator. The conversation emphasizes the importance of using the correct formula for accurate results in gas speed calculations.
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V_RMS problem... Need help with it!

Homework Statement



Find T, the temperature at which O2 gas molecules have an rms speed of v = 699 m/s.

Homework Equations



It's the rms form I have used.

v_{rms} = √(3RT/M)

The Attempt at a Solution



Let

M = 32 g/mol ≈ 32 * 10^(-3) kg/mol
v_rms = 699 m/s
R = 8.314 J/(mol * K)

Attempted to solve for T. Here is what I have:

v_{rms}^{2} = 3RT/M
T = v_{rms}^{2}M/(3R)

Plug and chug in the values, and I got 627 K, which is not correct.
 
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Your work looks good to me. Are you sure the answer is incorrect?
 


I seem to recall that the available energy modes for a diatomic gas are different than for a monatomic gas. 5/2 vs 3/2 as I recall. Some googling may be in order. Try "kinetic energy diatomic" or something along those lines.
 


TSny said:
##v_{rms} = \sqrt{3RT/M}## is valid for both monatomic and diatomic gases even though diatomic gases have additional degrees of freedom of motion.
http://en.wikipedia.org/wiki/Root-mean-square_speed

Well, so much for my memory then. Carry on, nothing to see here... :smile:
 


Check your calculation. What do you get? I have the answer 627 K, but the system marks it incorrect.
 


I also get 627 K. I assume the answer into be in Kelvins.
 


TSny said:
I also get 627 K. I assume the answer into be in Kelvins.

Increasingly strange.

My professor told me to use this form which is right. v_rms = √(5RT/M).

I entered in the value of T for the HW assignment, and I got the right answer. How strange... :\
 


NasuSama said:
Increasingly strange.

My professor told me to use this form which is right. v_rms = √(5RT/M).

I entered in the value of T for the HW assignment, and I got the right answer. How strange... :\

Yes, strange. Don't think he/she is correct. The 5 should be a 3. Just web-search "rms speed gas" for many examples.
 
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Actually, he said that because the velocity is diatomic, but as you said; no matter if the particles are atomic or not, v_rms = √(3RT/M). I believe he is thinking of the kinetic energy of the diatomic particles.
 
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