The rms velocity of molecules of a given mass | vedclass

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Question

(A) The $rms$ velocity of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.From this formula,it is clear that $v_{rms} \propto \sqrt{T}$

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$\frac{400}{\sqrt{3}} \, m \, s^{-1}$

Vedclass · Answer

(A) The $rms$ velocity of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.From this formula,it is clear that $v_{rms} \propto \sqrt{T}$.The $rms$ velocity is independent of the pressure of the gas.Given: $T_1 = 27^{\circ}C = 300 \, K$,$v_1 = 200 \, m \, s^{-1}$.New temperature: $T_2 = 127^{\circ}C = 400 \, K$.Using the proportionality $v_2 = v_1 \sqrt{\frac{T_2}{T_1}}$:$v_2 = 200 \sqrt{\frac{400}{300}} = 200 \sqrt{\frac{4}{3}} = 200 	imes \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}} \, m \, s^{-1}$.

The $rms$ velocity of molecules of a given mass of gas at $27^{\circ}C$ and $1.0 \times 10^5 \, Nm^{-2}$ pressure is $200 \, m \, s^{-1}$. What will be the $rms$ velocity if the temperature and pressure are $127^{\circ}C$ and $0.5 \times 10^5 \, Nm^{-2}$ respectively?

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