Question 993: Thermodynamics & KTG

Question

A flask contains argon and chlorine in the ratio of $2: 1$ by mass, the temperature of mixture is $27^{\circ} \mathrm{C}$, the ratio of root mean square ( $U_{\mathrm{rms}}$ ) of the molecule of two gases (Given, Atomic mass of argon $=39.9 v$, Molecular mass of chlorine $=70.9 u$ )

Accepted Answer

If $m$ is the mass of single molecule and $M$ the molecular mass, then $\frac{1}{2} m v_{\mathrm{rms}}^2=\frac{3}{2} k_B T=$ at given temperature it is constant
$$ \begin{aligned} \because \quad \frac{\left(V_{\mathrm{ms}}^2ight) \mathrm{Ar}}{\left(\mathrm{~N}_{\mathrm{ms}}^2ight) \mathrm{Cl}_2} & =\frac{(m) \mathrm{Cl}_2}{(m) \mathrm{Ar}} \ & =\frac{M \mathrm{Cl}_2}{M_{\mathrm{Ar}}}=\frac{70.9}{39.9}=1.777 \ 	ext { or } \quad \frac{\left(\mathrm{V}_{\mathrm{ms}}ight) \mathrm{Ar}}{\left(\mathrm{~N}_{\mathrm{rms}}ight) \mathrm{Cl}_2} & =\sqrt{1.777}=1.33 \end{aligned} $$

Answer

$\frac{2}{3}$

Answer

0.1

Answer

$\frac{10}{3}$

Answer

1.33

Thermodynamics & KTG

A flask contains argon and chlorine in the ratio of 2: 1 by mass, the temperature of mixture is 27^{\circ} \mathrm{C}, the ratio of root mean square ( U_{\mathrm{rms}} ) of the molecule of two gases (Given, Atomic mass of argon =39.9 v, Molecular mass of chlorine =70.9 u )

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