30^ temperature find the root mean square velocity of nitrogen mo… - Vidyadip

Question

$30^{\circ}$ temperature find the root mean square velocity of nitrogen molecules at 76 cm (Mercury) Pressure.

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Answer

Molecular weight of nitrogen $=28$ and $R =8.31$ joule $/ gm$ molecule degree $=8.31 \times$ $10^7$ erg. Kelvin ${ }^{-1} mol^{-1}$
Suppose mole is considered to be one gram molecule of nitrogen at $30^{\circ} C$ temperature
$\begin{aligned}\overline{C^2} & =\frac{3 kT}{m}=\frac{3 kN_{A} T}{m N_{A}}=\frac{3 RT}{M_0} \\\overline{C^2} & =\frac{3 \times 8.3 \times 10^7 \times(273+30)}{28} \\\overline{C^2} & =\frac{3 \times 8.3 \times 303 \times 10^7}{28} \\C_{rms} & =\sqrt{\frac{3 \times 8.3 \times 303 \times 10^7}{28}} \\\therefore \quad C_{\text {rms }} & =5.19 \times 10^4 meter / sec .\end{aligned}$

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