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the average velocity of CO2 at T1 Kelvin and maximum most probable velocity at T2 Kelvin is 9×10^4.

aabid ahmed , 7 Years ago
Grade 11
anser 2 Answers
Ravleen Kaur

Last Activity: 5 Years ago

The question is incomplete. Please post one more time.

Vikas TU

Last Activity: 5 Years ago

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/πM
and Most probable velocity = √2RT/M
Given –For CO2
Average 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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