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(D) The root mean square velocity of an ideal gas is given by $v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$.Since $R$ and $M$ are constants,$v_{	ext{rms}}

Vedclass · Accepted Answer

$15.5$

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(D) The root mean square velocity of an ideal gas is given by $v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$.Since $R$ and $M$ are constants,$v_{	ext{rms}} \propto \sqrt{T}$.Initial temperature $T_1 = 27 + 273 = 300 	ext{ K}$.Final temperature $T_2 = 127 + 273 = 400 	ext{ K}$.The ratio of velocities is $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \approx 1.1547$.The percentage increase is given by $\left( \frac{v_2 - v_1}{v_1} ight) 	imes 100 = \left( \frac{v_2}{v_1} - 1 ight) 	imes 100$.Percentage increase $= (1.1547 - 1) 	imes 100 = 0.1547 	imes 100 = 15.47\% \approx 15.5\%$.

When the temperature of an ideal gas is increased from $27^{\circ} C$ to $127^{\circ} C$,calculate the percentage increase in its $v_{\text{rms}}$. (in $\%$)

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