Answer
Calculate the volume of air in the automobile tire.
$$
\begin{aligned}
V &=V_{1}-V_{2} \\
&=\frac{\pi}{4} d_{1}^{2} h-\frac{\pi}{4} d_{2}^{2} h \\
&=\frac{\pi}{4} h\left(d_{1}^{2}-d_{2}^{2}\right)
\end{aligned}
$$
Here, $V$ is the volume of the air, $V_{1}$ is the volume of the outer cylinder, $V_{2}$ is the volume of inner cylinder, $d_{1}$ is the diameter of the outer cylinder, $d_{2}$ is the diameter of the inner cylinder, and $h$ is the height of the cylinder.
Substitute $52 \mathrm{~cm}$ for $d_{1}, 33 \mathrm{~cm}$ for $d_{2}$, and $13 \mathrm{~cm}$ for $h$.
$$
\begin{aligned}
V &=\frac{\pi}{4}(13)\left(52^{2}-33^{2}\right) \\
&=16489.434 \mathrm{~cm}^{3} \times \frac{1 \mathrm{~m}^{3}}{10^{6} \mathrm{~cm}^{3}} \\
&=0.016489 \mathrm{~m}^{3}
\end{aligned}
$$
Calculate the mass of air present in the tire initially at $25^{\circ} \mathrm{C}$ and $202 \mathrm{kPa}$ using ideal gas law.
$$
\begin{array}{l}
p_{1}=\rho_{1} R_{1} T_{1} \\
p_{1}=\frac{m_{1}}{V_{1}} R_{1} T_{1} \\
m_{1}=\frac{p_{1} V_{1}}{R_{1} T_{1}}
\end{array}
$$
Here, $p$ is the pressure in the tire, $R$ is the gas constant for air, and $T$ is the temperature of air in
the tire.
Substitute $202 \mathrm{kPa}$ for $p_{1}, 25^{\circ} \mathrm{C}$ for $T_{1}, 0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ for $R$, and $0.016489 \mathrm{~m}^{3}$ for $V_{1}$.
$$
\begin{aligned}
m_{1} &=\frac{(202)(0.016489)}{(0.287)(25+273)} \\
&=0.038944 \mathrm{~kg}
\end{aligned}
$$
Calculate the final mass of air present in the tire after the air is pumped in to it.
$$
m_{2}=\frac{p_{2} V_{2}}{R_{2} T_{2}}
$$
Substitute $303 \mathrm{kPa}$ for $p_{2}, 30^{\circ} \mathrm{C}$ for $T_{2}, 0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ for $R$, and $0.016489 \mathrm{~m}^{3}$ for $V_{2}$.
$$
\begin{aligned}
m_{2} &=\frac{(303)(0.016489)}{(0.287)(30+273)} \\
&=0.05745 \mathrm{~kg}
\end{aligned}
$$
Calculate the mass of air added to the tire.
$$
m=m_{2}-m_{1}
$$
Substitute $0.05745 \mathrm{~kg}$ for $m_{2}$ and $0.038944 \mathrm{~kg}$ for $m_{1}$.
$$
\begin{aligned}
m &=0.05745-0.038944 \\
&=0.0185 \mathrm{~kg}
\end{aligned}
$$
Therefore, the mass of air added to the tire is $0.0185 \mathrm{~kg}$.
Calculate the air pressure in the tire if the air is cooled to $0^{\circ} \mathrm{C}$ using ideal gas law.
$$
\begin{array}{l}
p_{f}=\rho_{f} R_{f} T_{f} \\
p_{f}=\frac{m_{f}}{V_{f}} R_{f} T_{f}
\end{array}
$$
Here, $m_{f}$ is the final mass in the tire that is equal to $m_{2}$.
Substitute $0^{\circ} \mathrm{C}$ for $T, 0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ for $R$, and $0.016489 \mathrm{~m}^{3}$ for $V$, and $0.05745 \mathrm{~kg}$ for
$m_{2} .$
$$
\begin{aligned}
p &=\frac{0.05745}{0.016489}(0.287)(0+273) \\
&=272.985 \\
& \approx 273 \mathrm{kPa}
\end{aligned}
$$
Therefore, the air pressure when the temperature in the tire is cooled to $0^{\circ} \mathrm{C}$ is $273 \mathrm{kPa}$.
Work Step by Step
Calculate the volume of air in the automobile tire.
$$
\begin{aligned}
V &=V_{1}-V_{2} \\
&=\frac{\pi}{4} d_{1}^{2} h-\frac{\pi}{4} d_{2}^{2} h \\
&=\frac{\pi}{4} h\left(d_{1}^{2}-d_{2}^{2}\right)
\end{aligned}
$$
Here, $V$ is the volume of the air, $V_{1}$ is the volume of the outer cylinder, $V_{2}$ is the volume of inner cylinder, $d_{1}$ is the diameter of the outer cylinder, $d_{2}$ is the diameter of the inner cylinder, and $h$ is the height of the cylinder.
Substitute $52 \mathrm{~cm}$ for $d_{1}, 33 \mathrm{~cm}$ for $d_{2}$, and $13 \mathrm{~cm}$ for $h$.
$$
\begin{aligned}
V &=\frac{\pi}{4}(13)\left(52^{2}-33^{2}\right) \\
&=16489.434 \mathrm{~cm}^{3} \times \frac{1 \mathrm{~m}^{3}}{10^{6} \mathrm{~cm}^{3}} \\
&=0.016489 \mathrm{~m}^{3}
\end{aligned}
$$
Calculate the mass of air present in the tire initially at $25^{\circ} \mathrm{C}$ and $202 \mathrm{kPa}$ using ideal gas law.
$$
\begin{array}{l}
p_{1}=\rho_{1} R_{1} T_{1} \\
p_{1}=\frac{m_{1}}{V_{1}} R_{1} T_{1} \\
m_{1}=\frac{p_{1} V_{1}}{R_{1} T_{1}}
\end{array}
$$
Here, $p$ is the pressure in the tire, $R$ is the gas constant for air, and $T$ is the temperature of air in
the tire.
Substitute $202 \mathrm{kPa}$ for $p_{1}, 25^{\circ} \mathrm{C}$ for $T_{1}, 0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ for $R$, and $0.016489 \mathrm{~m}^{3}$ for $V_{1}$.
$$
\begin{aligned}
m_{1} &=\frac{(202)(0.016489)}{(0.287)(25+273)} \\
&=0.038944 \mathrm{~kg}
\end{aligned}
$$
Calculate the final mass of air present in the tire after the air is pumped in to it.
$$
m_{2}=\frac{p_{2} V_{2}}{R_{2} T_{2}}
$$
Substitute $303 \mathrm{kPa}$ for $p_{2}, 30^{\circ} \mathrm{C}$ for $T_{2}, 0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ for $R$, and $0.016489 \mathrm{~m}^{3}$ for $V_{2}$.
$$
\begin{aligned}
m_{2} &=\frac{(303)(0.016489)}{(0.287)(30+273)} \\
&=0.05745 \mathrm{~kg}
\end{aligned}
$$
Calculate the mass of air added to the tire.
$$
m=m_{2}-m_{1}
$$
Substitute $0.05745 \mathrm{~kg}$ for $m_{2}$ and $0.038944 \mathrm{~kg}$ for $m_{1}$.
$$
\begin{aligned}
m &=0.05745-0.038944 \\
&=0.0185 \mathrm{~kg}
\end{aligned}
$$
Therefore, the mass of air added to the tire is $0.0185 \mathrm{~kg}$.
Calculate the air pressure in the tire if the air is cooled to $0^{\circ} \mathrm{C}$ using ideal gas law.
$$
\begin{array}{l}
p_{f}=\rho_{f} R_{f} T_{f} \\
p_{f}=\frac{m_{f}}{V_{f}} R_{f} T_{f}
\end{array}
$$
Here, $m_{f}$ is the final mass in the tire that is equal to $m_{2}$.
Substitute $0^{\circ} \mathrm{C}$ for $T, 0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ for $R$, and $0.016489 \mathrm{~m}^{3}$ for $V$, and $0.05745 \mathrm{~kg}$ for
$m_{2} .$
$$
\begin{aligned}
p &=\frac{0.05745}{0.016489}(0.287)(0+273) \\
&=272.985 \\
& \approx 273 \mathrm{kPa}
\end{aligned}
$$
Therefore, the air pressure when the temperature in the tire is cooled to $0^{\circ} \mathrm{C}$ is $273 \mathrm{kPa}$.
