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the average velocity of CO2 at T1 Kelvin and maximum most probable velocity at T2 Kelvin is 9×10^4.
The question is incomplete. Please post one more time.
Dear student Th question is incomplete , but if you want to find the alue of T1 and T2 . here is the solution Average velocity = √8RT/πMand Most probable velocity = √2RT/MGiven –For CO2Average velocity at T1 Most probable velocity at T2= 9 * 104cm/sec = 9 * 104/100 m/sec.= 9 * 102= m/sec.∴ 9 * 102= √8 * 8.314 9 T1/3.14 *44 * 10-3 …..(A)[Average velocity at T1K]and 9 * 102= √2 * 8.314 * T2/44 * 10-3 …..(B)[Most probable velocity at T2]On solving, T1 = 1682.5 K, T2 = 2143.4 K
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