# At a given temperature the ratio of root-mean-square velocities of two different gases of molecular weights M1 and M2 respectively is:

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At a given temperature the ratio of root-mean-square velocities of two different gases of molecular weights M1 and M2 respectively is:

1. $\sqrt{\dfrac{M_1}{M_2}}$
2. $\sqrt{\dfrac{M_2}{M_1}}$
3. $\dfrac{M_1}{M_2}$
4. $\dfrac{M_2}{M_1}$

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Correct Answer - Option 2 : $\sqrt{\dfrac{M_2}{M_1}}$

Concept:

• Root Mean Square Speed: It is defined as the square root of the mean of squares of the speed of different molecules.
• The root-mean-square speed takes into account both molecular weight and temperature, two factors that directly affect the kinetic energy of a material.

From the expansion of pressure,

$P = \frac{1}{3}{\rm{\rho v}}_{rms}^2$

${v_{rms}} = \sqrt {\frac{{3P}}{\rho }}$

$V_{rms}= \sqrt {\frac{{3PV}}{{Mass\;of\;gas}}} = \sqrt {\frac{{3RT}}{M}}$

$[\because \rho = \frac{M}{V}]$

⇒ vrms ∝ T

Where, R = universal gas constant, M = molar mass, P = pressure due to density,ρ = density.

Explanation:

As, the two gases molecular weights M1 and M2, given,

we have the formula,

$V_{rms}= \sqrt {\frac{{3PV}}{{Mass\;of\;gas}}} = \sqrt {\frac{{3RT}}{M}}$

${V_{rms}} = \sqrt {\frac{{3RT}}{M}}$

$\therefore {V_{rms}}\alpha \frac{1}{M}$

$\therefore \frac{{{V_1}}}{{{V_2}}} = \sqrt {\frac{{{M_2}}}{{{M_1}}}}$