Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
the average velocity of CO2 at T1 Kelvin and maximum most probable velocity at T2 Kelvin is 9×10^4. 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
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -