The rms speed of an oxygen molecule at a cert | vedclass

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(C) The formula for the rms speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Initially,for oxygen molecules $(O_2)$,the molar mass is $M_1 = 32 \,g/mol$ and

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$1200$

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(C) The formula for the rms speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Initially,for oxygen molecules $(O_2)$,the molar mass is $M_1 = 32 \,g/mol$ and temperature is $T_1$. Given $v_1 = 600 \,ms^{-1}$.So,$600 = \sqrt{\frac{3RT_1}{32}} \quad ... (i)$After dissociation,the oxygen molecule $(O_2)$ becomes atomic oxygen $(O)$. The new molar mass is $M_2 = 16 \,g/mol$ and the new temperature is $T_2 = 2T_1$.The new rms speed $v_2$ is given by $v_2 = \sqrt{\frac{3RT_2}{M_2}} = \sqrt{\frac{3R(2T_1)}{16}} \quad ... (ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{v_2}{600} = \frac{\sqrt{\frac{6RT_1}{16}}}{\sqrt{\frac{3RT_1}{32}}} = \sqrt{\frac{6RT_1}{16} 	imes \frac{32}{3RT_1}} = \sqrt{2 	imes 2} = \sqrt{4} = 2$.Therefore,$v_2 = 600 	imes 2 = 1200 \,ms^{-1}$.

The rms speed of an oxygen molecule at a certain temperature is $600 \,ms^{-1}$. If the temperature is doubled and the oxygen molecule dissociates into atomic oxygen atoms,the new rms speed is (in $\,ms^{-1}$)

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