At which temperature will the velocity of O_2 | vedclass

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(C) The root mean square velocity is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since the velocities are equal, we have $\sqrt{\frac{3RT_{

Vedclass · Accepted Answer

$39$

Vedclass · Answer

(C) The root mean square velocity is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since the velocities are equal, we have $\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{N_2}}{M_{N_2}}}$.Squaring both sides and simplifying, we get $\frac{T_{O_2}}{M_{O_2}} = \frac{T_{N_2}}{M_{N_2}}$, which implies $T_{O_2} = T_{N_2} 	imes \frac{M_{O_2}}{M_{N_2}}$.Given $T_{N_2} = 0 + 273 = 273\,K$, $M_{O_2} = 32\,g/mol$, and $M_{N_2} = 28\,g/mol$.Substituting the values: $T_{O_2} = 273 	imes \frac{32}{28} = 273 	imes \frac{8}{7} = 39 	imes 8 = 312\,K$.Converting back to Celsius: $312 - 273 = 39\,^\circ C$.

At which temperature will the velocity of $O_2$ molecules be equal to the velocity of $N_2$ molecules at $0 ^\circ C$?

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