Answer
(a) $65.2mol$
Work Step by Step
We know that
$n=\frac{PV}{RT}$......eq(1)
$r=\frac{d}{2}=\frac{4.75ft}{2}=0.7239m$
and $V=\frac{4}{3}\pi r^3$
$V=\frac{4}{3}\pi(0.7239m)^3=1.59m^3$
We plug in the known values in eq(1) to obtain:
$n=\frac{(1.01\times 10^5Pa)(1.59m^3)}{(8.31 J/mol.K)(273K)}$
$n=65.2mol$
