When the temperature of a gas is raised from 27`c to 90`c the percentage increase in the ram velocity of molecular

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Arun · Answer

V rms = &radic;(3RT / M o ) i.e. V rms &prop; &radic;T T 1 = 27&deg;C = 27 + 273 = 300K T 2 = 90&deg;C = 90 + 273 = 363K &there4; {V rms2 / V rms1 } = &radic;(T 2 / T 1 ) = &radic;(363 / 300) = 1.1 % increase in rms velocity = {(V rms2 &ndash; V rms1 ) / (V rms1 )} &times; 100 = {(1.1 &ndash; 1) / 1} &times; 100 = 0.1 &times; 100 % increase in rms velocity = 10%

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When the temperature of a gas is raised from 27`c to 90`c the percentage increase in the ram velocity of molecular

When the temperature of a gas is raised from 27`c to 90`c the percentage increase in the ram velocity of molecular

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