Answer
Conversions:
$$
\begin{array}{l}
1 \text { in. }=\frac{1}{12} \mathrm{ft} \\
T_{\mathrm{o}_{\mathrm{R}}}=\left(T_{\mathrm{o} \mathrm{F}}+460\right){ }^{\circ} \mathrm{R}
\end{array}
$$
$1 \mathrm{lb} / \mathrm{in}^{2}=12^{2} \mathrm{lb} / \mathrm{ft}^{2}$
Calculate the volume of basketball:
$$
V=\frac{4}{3} \pi r^{3}
$$
Here, $r$ is the radius of basketball.
Substitute $4.67 \times \frac{1}{12} \mathrm{ft}$ for $r$.
$$
\begin{aligned}
V &=\frac{4}{3} \pi\left(4.67 \times \frac{1}{12}\right)^{3} \\
&=0.24676 \mathrm{ft}^{3}
\end{aligned}
$$
Calculate the mass of the air using ideal gas equation.
$P V=m R T$
Here, $R$ gas constant and $T$ is the temperature.
Substitute $24 \times\left(12^{2}\right) \mathrm{lb} / \mathrm{ft}^{2}$ for $P, 1716 \mathrm{ft} \times \mathrm{lb} / \mathrm{slug} \times{ }^{\circ} \mathrm{R}$ for $R,(70+460)^{\circ} R$ for $T$, and
$0.24676 \mathrm{ft}^{3}$ for $V$
$24\left(12^{2}\right) \times 0.24676=m \times 1716 \times(70+460)$
$m=9.38 \times 10^{-4}$ slugs
Therefore, the mass of the air to inflate the basketball is $9.38 \times 10^{-4}$ slugs
Work Step by Step
Conversions:
$$
\begin{array}{l}
1 \text { in. }=\frac{1}{12} \mathrm{ft} \\
T_{\mathrm{o}_{\mathrm{R}}}=\left(T_{\mathrm{o} \mathrm{F}}+460\right){ }^{\circ} \mathrm{R}
\end{array}
$$
$1 \mathrm{lb} / \mathrm{in}^{2}=12^{2} \mathrm{lb} / \mathrm{ft}^{2}$
Calculate the volume of basketball:
$$
V=\frac{4}{3} \pi r^{3}
$$
Here, $r$ is the radius of basketball.
Substitute $4.67 \times \frac{1}{12} \mathrm{ft}$ for $r$.
$$
\begin{aligned}
V &=\frac{4}{3} \pi\left(4.67 \times \frac{1}{12}\right)^{3} \\
&=0.24676 \mathrm{ft}^{3}
\end{aligned}
$$
Calculate the mass of the air using ideal gas equation.
$P V=m R T$
Here, $R$ gas constant and $T$ is the temperature.
Substitute $24 \times\left(12^{2}\right) \mathrm{lb} / \mathrm{ft}^{2}$ for $P, 1716 \mathrm{ft} \times \mathrm{lb} / \mathrm{slug} \times{ }^{\circ} \mathrm{R}$ for $R,(70+460)^{\circ} R$ for $T$, and
$0.24676 \mathrm{ft}^{3}$ for $V$
$24\left(12^{2}\right) \times 0.24676=m \times 1716 \times(70+460)$
$m=9.38 \times 10^{-4}$ slugs
Therefore, the mass of the air to inflate the basketball is $9.38 \times 10^{-4}$ slugs
