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# The rms speed of oxygen molecules in a gas is V . If the temperature is doubled and the $\;O_{2}\;$ molecule dissociates into oxygen atoms , the rms speed will become

$(a)\;V\qquad(b)\;\sqrt{2} V\qquad(c)\;2V\qquad(d)\;4V$

Answer : $\;2V$
Explanation :
$V_{rms} = \sqrt{\large\frac{3RT}{M}}$
$V_{rms}^{'} = \sqrt{\large\frac{3RT^{1}}{M^{'}}}$
$T^{'} = 2T$
$M^{'} = \large\frac{M}{2}$
$V_{rms}^{'} = 2 V_{rms}\;.$