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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

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 mc2 = 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

Grade:12th pass

1 Answers

Arun
25750 Points
6 years ago
Dear Azhar,
 
As we know
E = 1/2 m* c²
Also E = 3/2 *K* T
Now
1/2 * m * c² = 3/2* K*T
Now 
C² is proportional to T
C is proportional to \sqrtT
Hence
C1/C2 = (T1/T2)1/2
But C1 = 3/2 * C
C2 = C
Hence
3/2 = (T1/273)1/2
T1 = 273 *9/4 = 614.25 K
 
Hope it helps
 
Regards,
Arun ( AskIITians forum Expert)
 

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