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Chapter 3: Q34E (page 612)

Calculate the mean free path of air molecules at 3.50 x 10-13 atm and 300 K. (This pressure is readily attainable in the laboratory; see Exercise 18.23.) As in Example 18.8, model the air molecules as spheres of radius 2.0 x 10-10 m.

Short Answer

Expert verified

The mean free path of the air molecule is1.648×105m

Step by step solution

01

Data and Formulas

Given data;

The pressure of the air molecule in the laboratory isP=3.5×10-3atm

The temperature of the gas is T = 300K

The radius of the spheres of the air molecule is r=2×10-10m

ConstantK=1.38×10-23J/K

Formula;

Mean free path

λ=KT4π2r2P ........... (1)

02

Find mean free path

From equation (1), put the all values and get the mean free path of the air molecule

λ=1.38×10-23J/K×300K4π×2×2×10-10m2×3.5×10-13atm×1.01×105Pa/atm=1.648×105m

Hence, the mean free path of the air molecule is 1.648×105m

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