In a sample of an ideal gas, the root-mean-square speed of the molecules is 400 m/s at a temperature of 300 K. If the temperature is increased to 1200 K, what is the new root-mean-square speed?
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Correct answer
A. 800 m/s
Principle or equation
For an ideal gas, the root-mean-square speed is proportional to the square root of the absolute temperature: v_rms ∝ √T.
Why this answer is correct
Since v_rms ∝ √T, if T increases by a factor of 4 (from 300 K to 1200 K), v_rms increases by a factor of √4 = 2. Therefore, the new v_rms = 400 m/s × 2 = 800 m/s.
Example
If v_rms = 300 m/s at 300 K, at 1200 K (4 times the temperature) v_rms becomes 300 × 2 = 600 m/s.
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