Answer:
a. 9947 m
b. 99476 times
c. 2*10^11 molecules
Explanation:
a) To find the mean free path of the air molecules you use the following formula:
[tex]\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}[/tex]
R: ideal gas constant = 8.3144 Pam^3/mol K
P: pressure = 1.5*10^{-6} Pa
T: temperature = 300K
N_A: Avogadros' constant = 2.022*10^{23}molecules/mol
d: diameter of the particle = 0.25nm=0.25*10^-9m
By replacing all these values you obtain:
[tex]\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m[/tex]
b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:
[tex]n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times[/tex]
c) By using the equation of the ideal gases you obtain:
[tex]PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules[/tex]
