A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixture is 27°C. The ratio of root mean square speed of the molecules of the two gases (v_rms(Ar)/v_rms(Cl₂)) is: (Atomic mass of argon = 40.0 u and molecular mass of chlorine = 70.0 u) ____.

First, write the expression for the root‑mean‑square speed of a gas species:

$v\_\{\backslash mathrm\{rms\}\}=\backslash sqrt\{\backslash dfrac\{3RT\}\{M\}\}\backslash$

Because both gases are at the same temperature, the factor $3RT$ cancels when we take a ratio. Thus the ratio of the rms speeds of argon (Ar) and chlorine (Cl₂) becomes

$\backslash dfrac\{v\_\{\backslash mathrm\{rms\}\}(\backslash mathrm\{Ar\})\}\{v\_\{\backslash mathrm\{rms\}\}(\backslash mathrm\{Cl\_\{2\}\})\}=\backslash dfrac\{\backslash sqrt\{\backslash dfrac\{3RT\}\{M\_\{\backslash mathrm\{Ar\}\}\}\}\}\{\backslash sqrt\{\backslash dfrac\{3RT\}\{M\_\{\backslash mathrm\{Cl\_\{2\}\}\}\}\}\}=\backslash sqrt\{\backslash dfrac\{M\_\{\backslash mathrm\{Cl\_\{2\}\}\}\}\{M\_\{\backslash mathrm\{Ar\}\}\}\}\backslash$

Insert the given molar masses (atomic mass of Ar = 40 u, molecular mass of Cl₂ = 70���u):

$\backslash sqrt\{\backslash dfrac\{70\}\{40\}\}=\backslash sqrt\{\backslash dfrac\{7\}\{4\}\}\backslash$

Now, using the mass ratio of the mixture (argon : chlorine = 2 : 1 by mass), multiply the above result by the mass‑ratio factor 2:

$2\backslash ;\backslash times\backslash ;\backslash sqrt\{\backslash dfrac\{7\}\{4\}\}=2\backslash ;\backslash times\backslash ;\backslash dfrac\{7\}\{4\}=\backslash dfrac\{7\}\{2\}\backslash$

Therefore, the required ratio of the rms speeds is

$\backslash dfrac\{v\_\{\backslash mathrm\{rms\}\}(\backslash mathrm\{Ar\})\}\{v\_\{\backslash mathrm\{rms\}\}(\backslash mathrm\{Cl\_\{2\}\})\}=\backslash dfrac\{7\}\{2\}\backslash$