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

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Homework Help Overview

The discussion revolves around determining the temperature required for oxygen gas (O2) to achieve a specified root mean square (rms) speed of 699 m/s. The problem involves concepts from kinetic theory and gas behavior, particularly focusing on the rms speed equation.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the rms speed formula, questioning the validity of using different constants for diatomic versus monatomic gases. There is an exploration of the relationship between temperature and rms speed, with attempts to verify calculations.

Discussion Status

Several participants have calculated the temperature to be 627 K, but there is uncertainty as the system marks this answer incorrect. Some participants suggest that the formula may differ for diatomic gases, leading to confusion about the correct approach. The discussion reflects a mix of agreement on calculations and questioning of assumptions regarding the formulas used.

Contextual Notes

Participants note discrepancies in expected results based on different interpretations of the rms speed formula, particularly regarding the degrees of freedom for diatomic gases. There is mention of guidance from a professor that suggests an alternative formula, which adds to the complexity of the discussion.

NasuSama
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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.
 
  • #10


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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