At what temperature will the molecules of nit | vedclass

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(A) The formula for $r.m.s.$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms}$ and $R$ are constant for both gases,we have $T \pro

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$77$

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(A) The formula for $r.m.s.$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms}$ and $R$ are constant for both gases,we have $T \propto M$.Therefore,$\frac{T_{N_2}}{T_{O_2}} = \frac{M_{N_2}}{M_{O_2}}$.Given $T_{O_2} = 127^{\circ}C = 127 + 273 = 400 \ K$.The molar mass of $N_2$ is $M_{N_2} = 28 \ g/mol$ and for $O_2$ is $M_{O_2} = 32 \ g/mol$.Substituting the values: $\frac{T_{N_2}}{400} = \frac{28}{32}$.$T_{N_2} = 400 	imes \frac{28}{32} = 400 	imes 0.875 = 350 \ K$.Converting back to Celsius: $T_{N_2} = 350 - 273 = 77^{\circ}C$.

At what temperature will the molecules of nitrogen have the same $r.m.s.$ velocity as the molecules of oxygen at $127^{\circ}C$ (in $^{\circ}C$)?

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