CLASSICAL MECHANICS & THERMODYNAMICS - Chapter 9, Exercise 2 Solution
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RMS Speed of Oxygen Molecules

PROBLEM
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Oxygen gas (O2, molar mass 0.032 kg/mol) is heated to 400 K. Using
v_rms = sqrt(3RT/M), calculate the RMS speed of oxygen molecules at
this temperature.

SOLUTION
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Using the root-mean-square speed formula:

    v_rms = sqrt(3RT/M)

Substitute the given values:

    R = 8.314 J/(mol.K)
    T = 400 K
    M = 0.032 kg/mol

    v_rms = sqrt(3 x 8.314 x 400 / 0.032)
    v_rms = sqrt(9976.8 / 0.032)
    v_rms = sqrt(311,775)
    v_rms = 558.4 m/s (approximately)

ANSWER: The RMS speed of oxygen molecules at 400 K is approximately
558.4 m/s.

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WHY THIS WORKS AS AN ANSWER

Comparing this result to the chapter's own worked example (nitrogen
at 300 K, giving approximately 517 m/s) shows two real, independent
effects at once: oxygen's own heavier molar mass (0.032 kg/mol vs.
nitrogen's 0.028 kg/mol) would, on its own, tend to REDUCE the RMS
speed (since M sits in the denominator inside the square root), but
the higher temperature (400 K vs. 300 K) more than compensates,
raising the average kinetic energy per molecule and pushing the
overall speed higher despite the heavier mass. This is exactly the
real physical relationship the formula v_rms = sqrt(3RT/M) captures:
faster at higher temperature, slower for heavier molecules, with the
two effects combining rather than acting in isolation.