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Engineering Entrance Exams

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

Profile image of aniket anand
12 Years agoGrade
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2 Answers

Profile image of Arun
6 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
 
Profile image of Vikas TU
6 Years ago
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