The temperature at which the rms speed of oxy | vedclass

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(A) The $RMS$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$, which implies $v_{rms} \propto \sqrt{\frac{T}{M}}$.Given that $v_{rms, O} = 0.75 	i

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$87$

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(A) The $RMS$ speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$, which implies $v_{rms} \propto \sqrt{\frac{T}{M}}$.Given that $v_{rms, O} = 0.75 	imes v_{rms, N} = \frac{3}{4} v_{rms, N}$.Using the ratio: $\frac{v_{rms, O}}{v_{rms, N}} = \sqrt{\frac{T_O}{T_N} 	imes \frac{M_N}{M_O}}$.Here, $T_N = 287 + 273 = 560 \ K$, $M_N = 28 \ g/mol$, and $M_O = 32 \ g/mol$.Substituting the values: $\frac{3}{4} = \sqrt{\frac{T_O}{560} 	imes \frac{28}{32}}$.Squaring both sides: $\frac{9}{16} = \frac{T_O}{560} 	imes \frac{7}{8}$.$T_O = \frac{9}{16} 	imes \frac{8}{7} 	imes 560 = \frac{9}{2} 	imes \frac{560}{7} = 9 	imes 40 = 360 \ K$.Converting to Celsius: $T_O = 360 - 273 = 87^{\circ} C$.

The temperature at which the rms speed of oxygen molecules is $75 \%$ of the rms speed of nitrogen molecules at a temperature of $287^{\circ} C$ is: (in $^{\circ} C$)

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