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# At what temperature the molecule of nitrogen will have the same rms velocity as the molecule of oxygen at $\;127^{0}C\;$ ?

$(a)\;457^{0}C\qquad(b)\;273^{0}C\qquad(c)\;350^{0}C\qquad(d)\;77^{0}C$

Answer : $\;457^{0}C$
Explanation :
$V_{m}=\sqrt{\large\frac{3RT}{M}}$
$\large\frac{V_{N_{2}}}{V_{O_{2}}} = 1$
$\sqrt{\large\frac{3RT_{1}}{M_{N_{2}}}} = \sqrt{\large\frac{3RT_{2}}{M_{O_{2}}}}$
$T_{1} = T_{2} \times \large\frac{M_{N_{2}}}{M_{O_{2}}}$
$= \large\frac{400 \times 28}{32} = 350 K$
Therefore , $\;t=350 - 273 =77^{0}C$