A sample of gas is at 0^{circ}C. To what temp | vedclass

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(B) The $r.m.s.$ speed of gas molecules is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$.This implies that $V_{rms} \propto \sqrt{T}$.Given in

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$819$

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(B) The $r.m.s.$ speed of gas molecules is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$.This implies that $V_{rms} \propto \sqrt{T}$.Given initial temperature $T_{1} = 0^{\circ}C = 273 \; K$.Let the final temperature be $T_{2}$.We want the final $r.m.s.$ speed to be double the initial speed,so $V_{rms_{2}} = 2V_{rms_{1}}$.Using the relation $\frac{V_{rms_{2}}}{V_{rms_{1}}} = \sqrt{\frac{T_{2}}{T_{1}}}$,we get $2 = \sqrt{\frac{T_{2}}{273}}$.Squaring both sides,$4 = \frac{T_{2}}{273}$.Therefore,$T_{2} = 4 	imes 273 = 1092 \; K$.To convert this back to Celsius,$T(^{\circ}C) = T(K) - 273 = 1092 - 273 = 819^{\circ}C$.

$A$ sample of gas is at $0^{\circ}C$. To what temperature must it be raised in order to double the $r.m.s.$ speed of the molecules? (in $^{\circ}C$)

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