Calculate the mean free path of air molecules at a pressure of 7.00×10^−13 atm and a temperature of 303 k . (this pressure is readily attainable in the laboratory.) model the air molecules as spheres with a radius of 2.00×10^−10 m .

82.8986 km

Explanation:

Given:

Pressure = 7.00×10⁻¹³ atm

Since , 1 atm = 101325 Pa

So, Pressure = 7.00×10⁻¹³×101325 Pa = 7.09275×10⁻⁸ Pa

Radius = 2.00×10⁻¹⁰ m

Diameter = 4.00×10⁻¹⁰ m (2× Radius)

Temperature = 303 K

The expression for mean free path is:

Boltzmann Constant = 1.38×10⁻²³ J/K

So,

Mean free path = 82.8986×10³ m = 82.8986 km

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2