Rms speed

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Question

When the temperature of a gas is raised from 30° C to 90° C, the percentage increase in the r.m.s. velocity of the molecules will be :

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Solution

$\begin{array}{l} V_{r m s} \propto \sqrt{T} T_{1} = 30 + 273 = 303 K \\ T_{2} = 90 + 273 = 363 K \\ \frac{V_{1}}{V_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{303}{363}} \\ \frac{V_{2} - V_{1}}{V_{1}} \times 100 = (\sqrt{\frac{363}{303}} - 1) \times 100 = 9.5 % \end{array}$

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