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(D) The $rms$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.For the $rms$ velocities to be equal, we have $\sqrt{\frac{3RT_{N_2}}{M_{

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$77$

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(D) The $rms$ velocity of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.For the $rms$ velocities to be equal, we have $\sqrt{\frac{3RT_{N_2}}{M_{N_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$.This simplifies to $\frac{T_{N_2}}{M_{N_2}} = \frac{T_{O_2}}{M_{O_2}}$.Given $T_{O_2} = 127 + 273 = 400 \, K$, $M_{O_2} = 32 \, g/mol$, and $M_{N_2} = 28 \, g/mol$.Substituting the values: $\frac{T_{N_2}}{28} = \frac{400}{32}$.$T_{N_2} = \frac{400 	imes 28}{32} = 12.5 	imes 28 = 350 \, K$.Converting back to Celsius: $T(^\circ C) = 350 - 273 = 77 ^\circ C$.

At what $^\circ C$ temperature will the molecules of nitrogen have the same $rms$ velocity as the molecules of oxygen at $127^\circ C$?

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