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# Helium gas is filled vessel ( having negligible thermal expansion coefficeint ) when heated from 300 K to 600 K . Then the average kinetic energy of atom will be

$(a)\;\sqrt{2} \;times\qquad(b)\;2 \;times\qquad(c)\;unchanged\qquad(d)\;half$

Answer : Twice the original value
Explanation :
$KE_{average} = \large\frac{3}{2} $$KT\; for a single atom KE_{1} = \large\frac{3}{2}$$K (300)$
$KE_{2} = \large\frac{3}{2}$$K (600)$
Hence it doubles its original value.
edited Aug 9, 2014