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Chapter 5: Problem 56

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 is approximately 79.88 psi.

Step by step solution

01

Convert Celsius to Kelvin

It's essential to use Kelvin as the unit for temperature in gas law equations. Convert the given temperatures from Celsius to Kelvin using the formula: \[ T(K) = T(^\circ \mathrm{C}) + 273.15 \] Initial temperature: \[ T_1 = 19^\circ \mathrm{C} + 273.15 = 292.15 \mathrm{K} \] Final temperature: \[ T_2 = 58^\circ \mathrm{C} + 273.15 = 331.15 \mathrm{K} \]
02

Find the initial and final volumes' ratio

Given that the volume of the tire increases by 4%, we can represent the initial volume as \(V_1\) and the final volume as \(V_2 = 1.04 V_1\).
03

Write down the combined gas law and rearrange for the new pressure

The combined gas law is given by: \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \] We want to find \(P_2\), so we can rearrange the equation: \( P_2 = \frac{P_1V_1T_2}{V_2T_1} \)
04

Plug in the known values and solve for the new pressure

Now, we can plug in the initial pressure (\(P_1\) = 75 psi), initial temperature (\(T_1\) = 292.15 K), final temperature (\(T_2\) = 331.15 K), and the final volume (\(V_2\) = 1.04\(V_1\)): \[ P_2 = \frac{75 \mathrm{psi} \cdot V_1 \cdot 331.15 \mathrm{K}}{1.04 V_1 \cdot 292.15 \mathrm{K}} \] The \(V_1\) terms cancel out, and we can solve the equation: \[ P_2 = \frac{75 \mathrm{psi} \cdot 331.15 \mathrm{K}}{1.04 \cdot 292.15 \mathrm{K}} = 79.88 \mathrm{psi} \] The new pressure in the bicycle tire is approximately 79.88 psi.

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Most popular questions from this chapter

A particular balloon is designed by its manufacturer to be inflated to a volume of no more than \(2.5 \mathrm{~L}\). If the balloon is filled with \(2.0 \mathrm{~L}\) helium at sea level, is released, and rises to an altitude at which the atmospheric pressure is only \(500 . \mathrm{mm} \mathrm{Hg}\), will the balloon burst? (Assume temperature is constant.)

A compressed gas cylinder contains \(1.00 \times 10^{3} \mathrm{~g}\) argon gas. The pressure inside the cylinder is \(2050 . \mathrm{psi}\) (pounds per square inch) at a temperature of \(18^{\circ} \mathrm{C}\). How much gas remains in the cylinder if the pressure is decreased to 650 . psi at a temperature of \(26^{\circ} \mathrm{C}\) ?

Consider two different containers, each filled with 2 moles of \(\mathrm{Ne}(\mathrm{g})\). One of the containers is rigid and has constant volume. The other container is flexible (like a balloon) and is capable of changing its volume to keep the external pressure and internal pressure equal to each other. If you raise the temperature in both containers, what happens to the pressure and density of the gas inside each container? Assume a constant external pressure.

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a 1.0000-L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results.

A \(2.747-\mathrm{g}\) sample of manganese metal is reacted with excess HCl gas to produce \(3.22 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})\) at \(373 \mathrm{~K}\) and \(0.951\) atm and a manganese chloride compound \(\left(\mathrm{MnCl}_{x}\right)\). What is the formula of the manganese chloride compound produced in the reaction?
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