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At what temperature will the r.m.s. velocity of oxygen be one and half times of its value at N.T.P.? Solution: 1/2 mc 2 = 3/2 KT Suppose the temperature required is T¢ when the velocity will be 3/2 C 3/2C / C = √T'/T or, T' = 9/4 × 273 = 614.25°K Plzzź explain solution or use any other method i dont understand the way they solved it


3 years ago

Arun
25768 Points
							Dear Azhar, As we knowE = 1/2 m* c²Also E = 3/2 *K* TNow1/2 * m * c² = 3/2* K*TNow C² is proportional to TC is proportional to $\sqrt$THenceC1/C2 = (T1/T2)1/2But C1 = 3/2 * CC2 = CHence3/2 = (T1/273)1/2T1 = 273 *9/4 = 614.25 K Hope it helps Regards,Arun ( AskIITians forum Expert)

3 years ago
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