Answer
(a) $0.032m^3$
(b) $1.92\times 10^{-4}$
(c) We know that the volume of a molecule is $0.02\%$ of the total volume of the gas; therefore, the assumption is valid.
Work Step by Step
(a) We know that
$V=\frac{nRT}{P}$
We plug in the known values to obtain:
$V=\frac{(1.25)(8.31J/mol.K)(310K)}{101\times 10^3Pa}=0.032m^3$
(b) We know that
$V_{mol}=\frac{\pi}{6}(2.5\times 10^{-10}m)^3=8.177\times 10^{-30}m^3$
Now, the fractional volume of $1.25$ moles of a gas can be determined as
$V^{\prime}=\frac{nV_{mol}}{V}$
We plug in the known values to obtain:
$V^{\prime}=\frac{(1.25mol)(6.022\times 10^{23}molecules/mol)(8.177\times 10^{-30}m)}{0.032m^3}$
$V^{\prime}=1.92\times 10^{-4}$
(c) We know that the volume of a molecule is $0.02\%$ of the total volume of the gas; therefore, the assumption is valid.
