At what temperature will the rms speed of air molecules be double that of NTP?

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At what temperature will the rms speed of air molecules be double that of NTP?

Options

(a) 519⁰C
(b) 619⁰C
(c) 719⁰C
(d) 819⁰C

Correct Answer:

819⁰C

Explanation:

v(rms) = √(3RT / M)
At T = T₀, v(rms) = √(3RT / M)
But at temperature T, v(rms) = 2 × √(3RT₀ / M)
⇒ √(3RT / M) = 2 √(3RT₀ / M) ⇒ √T = √(4T₀) ⇒ T = 4T₀
T = 4 × 273 K = 1092 K
Therefore, T = 819°C

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Topics: Behavior of Perfect Gas and Kinetic Theory (34)
Subject: Physics (2479)

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