Chapter 17 - Phases and Phase Changes - Problems and Conceptual Exercises - Page 608: 83

$1.1\times 10^{24}~$molecules

We can find the number of air molecules as follows: $N=\frac{P(A(2\pi r))}{kT}$ We plug in the known values to obtain: $N=\frac{42lb/in^2(\frac{6894.76Pa}{1lb/in^2})(0.0028m^2)(2\pi(0.68m))}{(1.38\times 10^{-23}J/K)(297.15K)}$ $N=1.1\times 10^{24}~$molecules