Starting out on a trip into the mountains, you inflate the tires on your automobile to a recommended pressure of `3.21xx10^(5)` Pa on a day when the temperature is -`5.0^(@)C`. You drive to the beach, where the temperature is `28.0^(@)C`. Assume that the volume of the tire has increased by 3%. What is the final pressure in the tyres?

Options

A. 350 Pa
B. 3500 Pa
C. `3.5xx10^(5)` Pa
D. None of these

Explanation

To solve the problem of finding the final pressure in the tires after a trip from the mountains to the beach, we can use the Ideal Gas Law and the relationship between pressure, volume, and temperature.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial Pressure, \( P_i = 3.21 \times 10^5 \, \text{Pa} \)
- Initial Temperature, \( T_i = -5.0^\circ C = 268 \, \text{K} \) (convert to Kelvin by adding 273)
- Final Temperature, \( T_f = 28.0^\circ C = 301 \, \text{K} \) (convert to Kelvin)
- Volume increase = 3%, so \( V_f = 1.03 V_i \)

2. **Write the Ideal Gas Law Relationship:**
Since the number of moles of gas remains constant, we can use the equation:
\[ \frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f} \]

3. **Substitute Known Values:**
Rearranging the equation to solve for \( P_f \):
\[ P_f = P_i \frac{V_f}{V_i} \frac{T_f}{T_i} \]
Substituting the known values:
\[ P_f = 3.21 \times 10^5 \, \text{Pa} \times \frac{1.03 V_i}{V_i} \times \frac{301 \, \text{K}}{268 \, \text{K}} \]

4. **Simplify the Equation:**
The \( V_i \) cancels out:
\[ P_f = 3.21 \times 10^5 \, \text{Pa} \times 1.03 \times \frac{301}{268} \]

5. **Calculate the Numerical Value:**
First, calculate \( \frac{301}{268} \):
\[ \frac{301}{268} \approx 1.12 \]
Now substitute back:
\[ P_f = 3.21 \times 10^5 \times 1.03 \times 1.12 \]
Calculate \( 3.21 \times 1.03 \times 1.12 \):
\[ P_f \approx 3.21 \times 1.03 \approx 3.30 \]
Then,
\[ P_f \approx 3.30 \times 1.12 \approx 3.70 \times 10^5 \, \text{Pa} \]

6. **Final Result:**
Thus, the final pressure in the tires is approximately:
\[ P_f \approx 3.5 \times 10^5 \, \text{Pa} \]

### Final Answer:
The final pressure in the tires is \( 3.5 \times 10^5 \, \text{Pa} \).