At what temperature will the root mean square | vedclass

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Question

(C) The root mean square speed of gas molecules is given by the formula $
u_{rms} = \sqrt{\frac{3RT}{M}}$.Given: Temperature of hydrogen $T_H = 27 \,

Vedclass · Accepted Answer

$4800 \, K$

Vedclass · Answer

(C) The root mean square speed of gas molecules is given by the formula $
u_{rms} = \sqrt{\frac{3RT}{M}}$.Given: Temperature of hydrogen $T_H = 27 \, ^\circ C = 27 + 273 = 300 \, K$.Molar mass of oxygen $M_{O_2} = 32 \, g \, mol^{-1}$.Molar mass of hydrogen $M_{H_2} = 2 \, g \, mol^{-1}$.Since $
u_{rms(O_2)} = 
u_{rms(H_2)}$,we have $\sqrt{\frac{3RT_O}{M_{O_2}}} = \sqrt{\frac{3RT_H}{M_{H_2}}}$.Squaring both sides and simplifying,we get $\frac{T_O}{M_{O_2}} = \frac{T_H}{M_{H_2}}$.Therefore,$T_O = T_H 	imes \frac{M_{O_2}}{M_{H_2}} = 300 	imes \frac{32}{2} = 300 	imes 16 = 4800 \, K$.

At what temperature will the root mean square speed $(\nu_{rms})$ of oxygen gas molecules be equal to the $\nu_{rms}$ of hydrogen gas molecules at $27 \, ^\circ C$? $(M_{O_2} = 32 \, g \, mol^{-1}, M_{H_2} = 2 \, g \, mol^{-1})$

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