The temperature of an ideal gas is increased from 120 K to 1080 K. If at 120 K the root mean square velocity of the gas molecules is ‘V’, then at 1080 K it will be
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a
4V
b
3V
c
V2
d
V4
answer is B.
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Detailed Solution
Vrms =3RTM ⇒V α T ⇒ V1V2=T1T2=1201080=19T is temperature in kelvinV1V2=13substitute given V1=VVV2=13V2=3V=root mean square velocity at 1080K