At what temperature is the root mean square s | vedclass

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(B) The root mean square (rms) speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given temperature of helium,$T_{He} = -20^{\circ}C

Vedclass · Accepted Answer

$2.52 	imes 10^{3} \; K$

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(B) The root mean square (rms) speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given temperature of helium,$T_{He} = -20^{\circ}C = 253 \; K$.Atomic mass of argon,$M_{Ar} = 39.9 \; u$.Atomic mass of helium,$M_{He} = 4.0 \; u$.We are given that $(v_{rms})_{Ar} = (v_{rms})_{He}$.Therefore,$\sqrt{\frac{3RT_{Ar}}{M_{Ar}}} = \sqrt{\frac{3RT_{He}}{M_{He}}}$.Squaring both sides and simplifying,we get $\frac{T_{Ar}}{M_{Ar}} = \frac{T_{He}}{M_{He}}$.Rearranging for $T_{Ar}$,we get $T_{Ar} = T_{He} 	imes \frac{M_{Ar}}{M_{He}}$.Substituting the values: $T_{Ar} = 253 	imes \frac{39.9}{4.0}$.$T_{Ar} = 253 	imes 9.975 = 2523.675 \; K$.Rounding to significant figures,$T_{Ar} \approx 2.52 	imes 10^{3} \; K$.

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at $-20\,^{\circ} C$? (Atomic mass of $Ar = 39.9 \; u$,of $He = 4.0 \; u$)

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