Question

A bicycle tire is filled with air at a pressure of \( 270 \) kPa at \( 27^\circ C \). What is the approximate pressure of the air in the tire when the temperature increases to \( 36^\circ C \)?

• A. \( 270 \) kPa
• B. \( 262 \) kPa
• C. \( 278 \) kPa
• D. \( 360 \) kPa

Hint:

For gases at constant volume, use Gay-Lussac’s Law: \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] where temperatures must be in Kelvin. Pressure changes in direct proportion to temperature.

Answer 1

The Correct Option is C

Step 1: Using Gay-Lussac’s Law (Pressure-Temperature Relationship). \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] where: - \( P_1 = 270 \) kPa (initial pressure), - \( T_1 = 27^\circ C = (27 + 273) = 300 K \) (initial temperature), - \( T_2 = 36^\circ C = (36 + 273) = 309 K \) (final temperature), - \( P_2 \) is the final pressure. 
Step 2: Solve for \( P_2 \). \[ P_2 = P_1 \times \frac{T_2}{T_1} \] \[ P_2 = 270 \times \frac{309}{300} \] \[ P_2 = 270 \times 1.03 \] \[ P_2 = 278.1 { kPa} \approx 278 { kPa} \] 
Final Answer: \[ \boxed{278 { kPa}} \]