A) 110.2 K
B) 90.2 K
C) 190.2 K
D) 100.2 K
Correct Answer: C
Solution :
[c] If T be the required temperature, then, \[{{v}_{rms}}-{{v}_{p}}=200m/s\] \[\text{or }\sqrt{\frac{3RT}{{{M}_{He}}}}-\sqrt{\frac{2RT}{{{M}_{He}}}}=200m/s\] \[\text{or }\sqrt{T}\text{ }\left[ \sqrt{3}-\sqrt{2} \right]=\sqrt{\frac{{{M}_{He}}}{R}}\left( 200m/s \right)\] Substituting \[{{M}_{He}}=4\times {{10}^{-3}}kg,\,R=8.134\,J/mol-K,\] \[T=190.2K.\]
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