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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
25750 Points
3 years 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
3 years ago
Dear student 
Question is not clear 
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We will happy to  help you 
Good Luck
Cheers

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