At temperature T, the r.m.s. speed of helium | vedclass

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(A) The $r.m.s.$ speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since the $r.m.s.$ speeds are equal, we have $\sqrt{\

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$273$

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(A) The $r.m.s.$ speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since the $r.m.s.$ speeds are equal, we have $\sqrt{\frac{3RT_{He}}{M_{He}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}$.This implies $\frac{T_{He}}{M_{He}} = \frac{T_{H_2}}{M_{H_2}}$, or $T_{He} = T_{H_2} 	imes \frac{M_{He}}{M_{H_2}}$.Given that $T_{H_2} = 273 \, K$ (at $NTP$, $0^\circ C = 273 \, K$), $M_{He} = 4 \, g/mol$, and $M_{H_2} = 2 \, g/mol$.Substituting the values: $T_{He} = 273 	imes \frac{4}{2} = 273 	imes 2 = 546 \, K$.To convert to Celsius: $T(^\circ C) = 546 - 273 = 273^\circ C$.

At temperature $T,$ the $r.m.s.$ speed of helium molecules is the same as the $r.m.s.$ speed of hydrogen molecules at normal temperature and pressure $(NTP)$. The value of $T$ is ....... $^\circ C$.

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