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The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K, the root mean square speed of gas molecules is v, then at 480 K it will be (A) 4v (B) 2v (C) 2 v (D) 4 v

The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K, the root mean square speed of gas molecules is v,
then at 480 K it will be
(A) 4v (B) 2v (C) 2
v (D) 4
v

Grade:

2 Answers

Arun
25763 Points
one year ago
Let T1=120K, T2=480K
The root mean square velocity of the gas molecule at 120K= v
The root mean square velocity of the gas molecule at 480K=v2
The ratio of the root mean square velocity at 120K to the ratio of the root mean square velocity at 480K= v/v2
v/v2=
⇒v/v2=
⇒v/v2=
⇒v/v2=1/2
⇒v2=2v
Vrms- root mean square velocity.
Therefore the root mean square speed of gas molecules 480 K, will be 2v
 
Vikas TU
14149 Points
one year ago
Dear student 
Question is not clear 
Please attach an image, 
We will happy to  help you 
Good Luck
Cheers

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