A gas mixture consists of molecules of type 1 | vedclass

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(A) The $r.m.s.$ speed of a gas molecule is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$.Since the temperature $T$ is the same for all gases

Vedclass · Accepted Answer

$(V_{rms})_1 < (V_{rms})_2 < (V_{rms})_3$ and $(\overline{K})_1 = (\overline{K})_2 = (\overline{K})_3$

Vedclass · Answer

(A) The $r.m.s.$ speed of a gas molecule is given by the formula $V_{rms} = \sqrt{\frac{3RT}{M}}$.Since the temperature $T$ is the same for all gases in a mixture at thermal equilibrium,$V_{rms} \propto \frac{1}{\sqrt{M}}$.Given $m_1 > m_2 > m_3$,it follows that $(V_{rms})_1 The average kinetic energy per molecule is given by $\overline{K} = \frac{3}{2}kT$.Since the temperature $T$ is the same for all molecules in the mixture,the average kinetic energy $\overline{K}$ is the same for all types of molecules,i.e.,$(\overline{K})_1 = (\overline{K})_2 = (\overline{K})_3$.

$A$ gas mixture consists of molecules of type $1, 2$ and $3$,with molar masses $m_1 > m_2 > m_3$. Let $(V_{rms})$ and $\overline{K}$ be the $r.m.s.$ speed and average kinetic energy of the gases,respectively. Which of the following is true?

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