Molecular velocities of two gases at the same | vedclass

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(D) The root mean square velocity $(v_{rms})$ of a gas is given by the formula: $v = \sqrt{\frac{3RT}{M}}$,where $R$ is the gas constant,$T$ is the te

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$m_1 v_1^2 = m_2 v_2^2$

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(D) The root mean square velocity $(v_{rms})$ of a gas is given by the formula: $v = \sqrt{\frac{3RT}{M}}$,where $R$ is the gas constant,$T$ is the temperature,and $M$ is the molar mass.Since the temperature $T$ is the same for both gases,$v \propto \frac{1}{\sqrt{M}}$.Therefore,$\frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}}$.Squaring both sides,we get $\frac{v_1^2}{v_2^2} = \frac{M_2}{M_1}$.Rearranging the terms,we obtain $M_1 v_1^2 = M_2 v_2^2$. Since mass $m$ is proportional to molar mass $M$,the expression holds as $m_1 v_1^2 = m_2 v_2^2$.

Molecular velocities of two gases at the same temperature are $v_1$ and $v_2$. Their masses are $m_1$ and $m_2$ respectively. Which of the following expressions is correct?

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