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(C) The root mean square speed of a gas is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the te

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$1212$

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(C) The root mean square speed of a gas is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the temperature in Kelvin,and $M$ is the molar mass.Given,$(V_{rms})_{Ne} = (V_{rms})_{He}$.Substituting the formula: $\sqrt{\frac{3RT_{Ne}}{M_{Ne}}} = \sqrt{\frac{3RT_{He}}{M_{He}}}$.Squaring both sides and canceling common terms: $\frac{T_{Ne}}{M_{Ne}} = \frac{T_{He}}{M_{He}}$.Given $T_{He} = -33^{\circ} C = (-33 + 273) \ K = 240 \ K$,$M_{Ne} = 20.2 \ u$,and $M_{He} = 4.0 \ u$.Substituting the values: $\frac{T_{Ne}}{20.2} = \frac{240}{4.0}$.$\frac{T_{Ne}}{20.2} = 60$.$T_{Ne} = 60 	imes 20.2 = 1212 \ K$.

At what temperature is the root mean square (rms) speed of Neon gas atoms equal to the rms speed of Helium gas atoms at $-33^{\circ} C$ (in $K$)? (Atomic mass of $Ne = 20.2 \ u$ and that of $He = 4.0 \ u$)

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