To solve this problem, we can use the ideal gas law in the form of the combined gas law, which relates pressure and temperature for a fixed volume:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

where:

• $P_1$ is the initial pressure,
• $T_1$ is the initial temperature (in Kelvin),
• $P_2$ is the final pressure,
• $T_2$ is the final temperature (in Kelvin).

Step 1: Convert Temperatures to Kelvin

The initial and final temperatures in Celsius need to be converted to Kelvin:

$T_1 = 35.0^\circ \text{C} + 273.15 = 308.15 \text{ K}$ $T_2 = -37.0^\circ \text{C} + 273.15 = 236.15 \text{ K}$

Step 2: Set Up the Combined Gas Law

We know the initial gauge pressure $P_1 = 2.30 \times 10^5 \, \text{N/m}^2$.

Now, we apply the combined gas law to solve for $P_2$:

$\frac{P_2}{P_1} = \frac{T_2}{T_1}$ $P_2 = P_1 \cdot \frac{T_2}{T_1}$

Substituting the values:

$P_2 = (2.30 \times 10^5) \cdot \frac{236.15}{308.15}$ $P_2 \approx 1.76 \times 10^5 \, \text{N/m}^2$

Step 3: Convert Pressure to Atmospheres

To convert $P_2$ from pascals (N/m²) to atmospheres, we use the conversion $1 \, \text{atm} = 1.013 \times 10^5 \, \text{N/m}^2$.

$P_2 \text{ (in atm)} = \frac{1.76 \times 10^5}{1.013 \times 10^5}$ $P_2 \approx 1.74 \, \text{atm}$

Final Answer:

The gauge pressure in the tires at $-37.0^\circ \text{C}$ is approximately 1.74 atm.

Would you like more details on any specific step?

Here are some related questions:

1. How does a change in temperature affect tire pressure?
2. Why do we need to convert temperatures to Kelvin in gas law calculations?
3. What happens to the pressure if the volume of the tire changes as well?
4. How would the answer change if we used absolute pressure instead of gauge pressure?
5. What is the significance of gauge pressure versus absolute pressure in real-world applications?

Tip: Always convert temperatures to Kelvin when working with gas laws, as they rely on absolute temperature measurements.