The diameter of an oxygen molecule is 2.94 ti | vedclass

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(D) The Van der Waals constant '$b$' represents the excluded volume per mole of gas.For one mole of gas,$b = N_A 	imes 4 	imes V_{molecule}$,where $

Vedclass · Accepted Answer

$32 	imes 10^{-6}$

Vedclass · Answer

(D) The Van der Waals constant '$b$' represents the excluded volume per mole of gas.For one mole of gas,$b = N_A 	imes 4 	imes V_{molecule}$,where $V_{molecule} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (d/2)^3 = \frac{1}{6} \pi d^3$.Thus,$b = N_A 	imes 4 	imes \frac{1}{6} \pi d^3 = \frac{2}{3} \pi N_A d^3$.Given $N_A = 6.022 	imes 10^{23} \ mol^{-1}$ and $d = 2.94 	imes 10^{-10} \ m$:$b = \frac{2}{3} 	imes 3.14 	imes 6.022 	imes 10^{23} 	imes (2.94 	imes 10^{-10})^3$.$b \approx 2.093 	imes 6.022 	imes 10^{23} 	imes 25.41 	imes 10^{-30} \approx 32 	imes 10^{-6} \ m^3/mol$.

The diameter of an oxygen molecule is $2.94 \times 10^{-10} \ m$. The Van der Waals gas constant '$b$' in $m^3/mol$ will be:

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