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Chapter 5: Problem 54

A bicycle tire is filled with air to a pressure of 75 . psi at a temperature of \(19^{\circ} \mathrm{C}\). Riding the bike on asphalt on a hot day increases the temperature of the tire to \(58^{\circ} \mathrm{C}\). The volume of the tire increases by \(4.0 \%\). What is the new pressure in the bicycle tire?

Short Answer

Expert verified
The new pressure in the bicycle tire at a temperature of \(58^{\circ} \mathrm{C}\) and a 4.0% increase in volume is approximately 66.15 psi.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given Celsius temperatures to Kelvin, as the ideal gas law requires temperatures to be in Kelvin. \(T_1 = 19 + 273.15 = 292.15\, K\) \(T_2 = 58 + 273.15 = 331.15\, K\)
02

Calculate the ratio of the volume

Since the volume increases by 4.0%, we can express the final volume as 104% of the initial volume: \(V_2 = 1.04 V_1\)
03

Use the ideal gas law to find the final pressure

Rewrite the ideal gas law equation as: \(P_1V_1 / T_1 = P_2V_2 / T_2\) Substitute the given values and expressions: \((75\, psi)V_1 / (292.15\, K) = P_2(1.04 V_1) / (331.15\, K)\) Observe that the initial volume \(V_1\) can be canceled out from both sides of the equation. Now, solve for the final pressure \(P_2\): \(P_2 = 75\, psi \times \frac{292.15\, K}{331.15\, K} \times \frac{1.04}{1}\) \(P_2 = 75\, psi \times 0.882\) \(P_2 = 66.15\, psi\)
04

Conclusion

The new pressure in the bicycle tire at a temperature of \(58^{\circ} \mathrm{C}\) and a 4.0% increase in volume is approximately 66.15 psi.

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