Question
An automobile tire having a temperature of 0.55◦C (a cold tire on a cold day) is filled to a gauge pressure of 27 lb/in^2. What would be the gauge pressure in the tire when its temperature rises to 33◦C? For simplicity, assume that the volume of the tire remains constant, that the air does not leak out and the atmospheric pressure remains constant at 14.7 lb/in^2. Answer in units of lb/in^2.
Answers
GPT-4o mini
To determine the gauge pressure of the tire when its temperature rises, we can use the ideal gas law, which in a simplified form for constant volume is given by:
\[
\frac{P_1}{T_1} = \frac{P_2}{T_2}
\]
where:
- \( P_1 \) is the initial absolute pressure,
- \( T_1 \) is the initial temperature in Kelvin,
- \( P_2 \) is the final absolute pressure, and
- \( T_2 \) is the final temperature in Kelvin.
First, we need to convert the temperatures from Celsius to Kelvin:
\[
T_1 = 0.55^\circ C + 273.15 = 273.70 \, K
\]
\[
T_2 = 33^\circ C + 273.15 = 306.15 \, K
\]
Next, we need to convert the gauge pressure to absolute pressure. The gauge pressure is given as 27 lb/in², and the atmospheric pressure is 14.7 lb/in², so the initial absolute pressure \( P_1 \) is:
\[
P_1 = 27 \, lb/in^2 + 14.7 \, lb/in^2 = 41.7 \, lb/in^2
\]
Now we can rearrange the ideal gas law to solve for \( P_2 \):
\[
P_2 = P_1 \cdot \frac{T_2}{T_1}
\]
Substituting in the values we have:
\[
P_2 = 41.7 \, lb/in^2 \cdot \frac{306.15 \, K}{273.70 \, K}
\]
Calculating the fraction:
\[
\frac{306.15}{273.70} \approx 1.118
\]
Now substituting this value back into the equation for \( P_2 \):
\[
P_2 \approx 41.7 \, lb/in^2 \cdot 1.118 \approx 46.6 \, lb/in^2
\]
Finally, to find the new gauge pressure \( P_{gauge} \), we subtract the atmospheric pressure from the absolute pressure:
\[
P_{gauge} = P_2 - 14.7 \, lb/in^2
\]
Calculating the gauge pressure:
\[
P_{gauge} \approx 46.6 \, lb/in^2 - 14.7 \, lb/in^2 \approx 31.9 \, lb/in^2
\]
Thus, the gauge pressure in the tire when its temperature rises to 33°C will be approximately:
\[
\boxed{31.9} \, lb/in^2
\]
\[
\frac{P_1}{T_1} = \frac{P_2}{T_2}
\]
where:
- \( P_1 \) is the initial absolute pressure,
- \( T_1 \) is the initial temperature in Kelvin,
- \( P_2 \) is the final absolute pressure, and
- \( T_2 \) is the final temperature in Kelvin.
First, we need to convert the temperatures from Celsius to Kelvin:
\[
T_1 = 0.55^\circ C + 273.15 = 273.70 \, K
\]
\[
T_2 = 33^\circ C + 273.15 = 306.15 \, K
\]
Next, we need to convert the gauge pressure to absolute pressure. The gauge pressure is given as 27 lb/in², and the atmospheric pressure is 14.7 lb/in², so the initial absolute pressure \( P_1 \) is:
\[
P_1 = 27 \, lb/in^2 + 14.7 \, lb/in^2 = 41.7 \, lb/in^2
\]
Now we can rearrange the ideal gas law to solve for \( P_2 \):
\[
P_2 = P_1 \cdot \frac{T_2}{T_1}
\]
Substituting in the values we have:
\[
P_2 = 41.7 \, lb/in^2 \cdot \frac{306.15 \, K}{273.70 \, K}
\]
Calculating the fraction:
\[
\frac{306.15}{273.70} \approx 1.118
\]
Now substituting this value back into the equation for \( P_2 \):
\[
P_2 \approx 41.7 \, lb/in^2 \cdot 1.118 \approx 46.6 \, lb/in^2
\]
Finally, to find the new gauge pressure \( P_{gauge} \), we subtract the atmospheric pressure from the absolute pressure:
\[
P_{gauge} = P_2 - 14.7 \, lb/in^2
\]
Calculating the gauge pressure:
\[
P_{gauge} \approx 46.6 \, lb/in^2 - 14.7 \, lb/in^2 \approx 31.9 \, lb/in^2
\]
Thus, the gauge pressure in the tire when its temperature rises to 33°C will be approximately:
\[
\boxed{31.9} \, lb/in^2
\]