Question

the total pressure in a 12.1 l automobile tire is 43 psi at 13 °c. how much does the pressure in the tire rise if its temperature increases to 38 °c and the volume remains at 12.1 l? express your answer in pounds per square inch to one significant figure.

Explanation:

Step1: Convert temperatures to Kelvin

To use the ideal gas law (since volume is constant, we can use Gay - Lussac's law, which is a special case of the ideal gas law \( \frac{P_1}{T_1}=\frac{P_2}{T_2} \) where \( V \) and \( n \) are constant), we first convert the Celsius temperatures to Kelvin. The conversion formula is \( T(K)=T(^{\circ}C) + 273.15 \).

For \( T_1 = 13^{\circ}C \), \( T_1=13 + 273.15=286.15\ K \)

For \( T_2 = 38^{\circ}C \), \( T_2=38 + 273.15 = 311.15\ K \)

Step2: Apply Gay - Lussac's law

Gay - Lussac's law states that \( \frac{P_1}{T_1}=\frac{P_2}{T_2} \) (at constant \( V \) and \( n \)). We know \( P_1 = 43\ psi \), \( T_1 = 286.15\ K \), and \( T_2=311.15\ K \). We can solve for \( P_2 \):

\( P_2=\frac{P_1\times T_2}{T_1} \)

Substitute the values: \( P_2=\frac{43\ psi\times311.15\ K}{286.15\ K} \)

First, calculate the ratio of temperatures: \( \frac{311.15}{286.15}\approx1.087 \)

Then, \( P_2 = 43\times1.087\approx46.741\ psi \)

Step3: Calculate the pressure change \( \Delta P \)

The pressure change \( \Delta P=P_2 - P_1 \)

\( \Delta P=46.741\ psi- 43\ psi = 3.741\ psi \)

Step4: Round to one significant figure

When rounding to one significant figure, we look at the first non - zero digit. The number \( 3.741 \) rounded to one significant figure is \( 4\ psi \) (since the first significant figure is 3, and the next digit is 7 which is greater than 5, so we round up the 3 to 4).

Answer:

\( \boxed{4} \)