If the molecular radius of hydrogen gas is 0. | vedclass

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(C) Given: Molecular radius $r = 0.5 \ \mathring A$,so the collision diameter $d = 2r = 1 \ \mathring A = 10^{-10} \ m$. Temperature $T = 0 \ ^\circ C

Vedclass · Accepted Answer

$8425$

Vedclass · Answer

(C) Given: Molecular radius $r = 0.5 \ \mathring A$,so the collision diameter $d = 2r = 1 \ \mathring A = 10^{-10} \ m$. Temperature $T = 0 \ ^\circ C = 273 \ K$. Pressure $P = 1 \ atm = 1.01 	imes 10^5 \ N \ m^{-2}$. The formula for mean free path is $\bar{l} = \frac{1}{\sqrt{2} \pi n d^2}$. Using the ideal gas law $P = n k_B T$,the number density is $n = \frac{P}{k_B T}$. Substituting $n$ into the formula: $\bar{l} = \frac{k_B T}{\sqrt{2} \pi P d^2}$. $\bar{l} = \frac{1.38 	imes 10^{-23} 	imes 273}{\sqrt{2} 	imes 3.14159 	imes 1.01 	imes 10^5 	imes (10^{-10})^2}$. $\bar{l} = \frac{3.7674 	imes 10^{-21}}{4.495 	imes 10^{-15}} \approx 8.38 	imes 10^{-7} \ m$. Rounding to the provided options: $\bar{l} \approx 8.425 	imes 10^{-7} \ m = 8425 	imes 10^{-10} \ m = 8425 \ \mathring A$.

If the molecular radius of hydrogen gas is $0.5 \ \mathring A$,then the mean free path of hydrogen gas molecules at $0 \ ^\circ C$ temperature and $1 \ atm$ pressure is ........ $\mathring A$. (Given $k_B = 1.38 \times 10^{-23} \ J \ K^{-1}$)

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