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(C) The pressure $P$ is given by $P = h ho g$,where $h = 1.2 	imes 10^{-7} 	ext{ mm} = 1.2 	imes 10^{-10} 	ext{ m}$,$ho = 13600 	ext{ kg/m}^3

Vedclass · Accepted Answer

$3.86 	imes 10^{11}$

Vedclass · Answer

(C) The pressure $P$ is given by $P = h ho g$,where $h = 1.2 	imes 10^{-7} 	ext{ mm} = 1.2 	imes 10^{-10} 	ext{ m}$,$ho = 13600 	ext{ kg/m}^3$,and $g = 9.8 	ext{ m/s}^2$.Using the ideal gas law $PV = NkT$,the number of molecules $N$ is given by $N = \frac{PV}{kT}$.Here,$V = 100 	ext{ cm}^3 = 100 	imes 10^{-6} 	ext{ m}^3 = 10^{-4} 	ext{ m}^3$,$k = 1.38 	imes 10^{-23} 	ext{ J/K}$,and $T = 27 + 273 = 300 	ext{ K}$.Substituting the values:$N = \frac{(1.2 	imes 10^{-10} 	imes 13600 	imes 9.8) 	imes 10^{-4}}{1.38 	imes 10^{-23} 	imes 300}$$N = \frac{1.59936 	imes 10^{-8} 	imes 10^{-4}}{4.14 	imes 10^{-21}}$$N = \frac{1.59936 	imes 10^{-12}}{4.14 	imes 10^{-21}} \approx 0.386 	imes 10^9 	imes 10^3 = 3.86 	imes 10^{11}$ molecules.

An electron tube was sealed off during manufacture at a pressure of $1.2 \times 10^{-7} \text{ mm}$ of mercury at $27^{\circ}C$. Its volume is $100 \text{ cm}^3$. The number of molecules that remain in the tube is

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