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Q.

The temperature of an ideal is increased from 120 K to 480 K. If at 120 K the root mean square velocity of the gas molecules is ‘V’ , at 480 K it becomes

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a

4V

b

2V

c

V/2

d

V/4

answer is B.

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Detailed Solution

Vrms=3RTM  ⇒ Vrms∝T⇒  V1V2=T1T2=120480   obtained after substituting given valueshere R is universal gas constantV=V; T is absolute temperature;M is molecular weight;V1V2=14given V1=V substitute in above equation,VV2=12⇒V2=2V
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