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(A) The $r.m.s.$ velocity of an ideal gas is given by the formula $\upsilon_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $\upsilon_{rms} \propto \sqrt{

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Doubles

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(A) The $r.m.s.$ velocity of an ideal gas is given by the formula $\upsilon_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $\upsilon_{rms} \propto \sqrt{T}$.Given initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.Given final temperature $T_2 = 927^{\circ}C = 927 + 273 = 1200 \ K$.Using the ratio $\frac{\upsilon_2}{\upsilon_1} = \sqrt{\frac{T_2}{T_1}}$,we get $\frac{\upsilon_2}{\upsilon_1} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.Therefore,$\upsilon_2 = 2\upsilon_1$,meaning the $r.m.s.$ velocity doubles.

If the temperature of an ideal gas is increased from $27^{\circ}C$ to $927^{\circ}C$,what happens to its $r.m.s.$ velocity?

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