A) \[{{H}_{2}}\]
B) \[{{F}_{2}}\]
C) \[C{{l}_{2}}\]
D) \[{{O}_{2}}\]
Correct Answer: A
Solution :
The velocity of gas at temperature T is given by \[v=\frac{\sqrt{3RT}}{M}\] where, R is gas constant and M the molecular weight. Given, \[R=8.3\text{ }J/mol-K,\] \[T=27{}^\circ C+2736=300K\] \[v=1933m/s\] \[\therefore \] \[M=\frac{3RT}{{{v}^{2}}}=\frac{3\times 8.3\times 300}{{{(1933)}^{2}}}\] \[\therefore \] \[=\frac{7470}{3736489}\] \[\approx 0.001999\] \[\therefore \] \[M\approx 2\times {{10}^{-3}}kg\] which is molecular weight of\[{{H}_{2}}\].
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