Question

The temperature of an ideal gas is increased from 120K to 480K. If the root mean square velocity
(Vrms) of the gas molecule at 120K is v , then at 480K it becomes
A.40
B.v/2
C. 2v.
D.v/4
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Answer 1

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

$Vrms=\sqrt{3RT/M$

v/v2=$\sqrt{T1/T2}$

⇒v/v2=$\sqrt{120/480}$

⇒v/v2=$\sqrt{1/4}$

⇒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

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Answer 2

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

option c is correct