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A bicycle tire has an internal volume of \(1.52 \mathrm{L}\) and contains 0.406 mol of air. The tire will burst if its internal pressure reaches 7.25 atm. To what temperature, in degrees Celsius, does the air in the tire need to be heated to cause a blowout?

Short Answer

Expert verified
The tire needs to be heated to approximately 53.7 °C.

Step by step solution

01

Understand the problem

We need to find the temperature at which the pressure inside the bicycle tire reaches 7.25 atm, causing it to burst. We are given the internal volume of the tire, the amount of air in moles, and we need to find the temperature in degrees Celsius.
02

Recall the Ideal Gas Law

The Ideal Gas Law is given by the formula \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (0.0821 L⋅atm/mol⋅K), and \( T \) is the temperature in Kelvin.
03

Rearrange the Ideal Gas Law to solve for temperature

We can rearrange the Ideal Gas Law to solve for \( T \): \[ T = \frac{PV}{nR} \].
04

Plug in the known values

Substitute the given values into the equation: pressure \( P = 7.25 \) atm, volume \( V = 1.52 \) L, moles \( n = 0.406 \) mol, and the ideal gas constant \( R = 0.0821 \) L⋅atm/mol⋅K. This gives us:\[ T = \frac{7.25 \, \text{atm} \times 1.52 \, \text{L}}{0.406 \, \text{mol} \times 0.0821 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}} \].
05

Calculate the temperature in Kelvin

Calculate the value to find the temperature in Kelvin: \[ T = \frac{7.25 \times 1.52}{0.406 \times 0.0821} \approx 326.85 \, \text{K} \].
06

Convert Kelvin to Celsius

The temperature in Kelvin is approximately 326.85 K. To convert Kelvin to Celsius, subtract 273.15 from the Kelvin temperature: \[ T_{\text{Celsius}} = 326.85 - 273.15 = 53.7 \, ^\circ\text{C} \].
07

Conclusion

The air in the tire needs to be heated to approximately 53.7 °C to reach the blowout pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure and Volume Relationship
When dealing with gases, particularly in enclosed spaces, pressure and volume have a specific relationship governed by Boyle's Law. According to Boyle's Law, if the temperature of a gas remains constant, the pressure of a gas is inversely proportional to its volume. This means that as you decrease the volume of a gas, its pressure increases, and vice versa. However, this is under the condition that the temperature and the amount of gas moles remain constant.

In the case of our bicycle tire exercise, we observe a scenario influenced by the Ideal Gas Law, a more comprehensive model that includes temperature as a variable. Here, the tire maintains a constant volume, but as the temperature of the gas inside the tire increases, so does the pressure, eventually reaching a point where the tire may burst. When volume is fixed and moles of gas are constant, according to the Ideal Gas Law, increased temperature results in increased pressure. This integrated approach involving all these variables helps us understand what conditions lead to the tire reaching its maximum pressure.
Temperature Conversion
In scientific calculations and exercises involving gases, it is crucial to convert temperatures into a consistent unit for accurate results, typically Kelvin. The Kelvin scale is an absolute metric unit of measurement that begins at absolute zero (0 Kelvin). It is widely used in gas law calculations due to its direct proportionality to the kinetic energy and pressure in gases.

To convert temperatures from Celsius to Kelvin, you merely need to add 273.15 to the Celsius temperature. Conversely, to find the temperature in Celsius from Kelvin, you subtract 273.15. This straightforward conversion ensures that our calculations align with the parameters of the Ideal Gas Law, allowing for the consistent application of the formula without errors stemming from incorrect units.

For example, if you calculate a gas temperature under pressure conditions of 326.85 K, subtract 273.15 to convert to Celsius, resulting in 53.7 °C, the temperature at which the bicycle tire bursts.
Bicycle Tire Pressure
Understanding how pressure builds within a bicycle tire is essential for maintaining safety and performance. Tire pressure is influenced by a variety of factors, including the number of air moles, volume of the tire, and especially the temperature, as is demonstrated by the Ideal Gas Law. A key concern is reaching the tire's maximum capacity.

Bicycle tires are designed to withstand specific pressures; however, increasing the temperature can lead to a proportional increase in pressure. In warmer climates or during high-intensity cycling, the air inside tires can warm up, potentially increasing the risk of exceeding the tire's pressure limit. Keeping an eye on ambient temperature changes and adjusting tire pressure accordingly can prevent blowouts and ensure smooth riding.

Regularly checking and regulating tire pressure by mechanical means allows for the maintenance of an optimal pressure that balances performance and safety, without overstepping the structural limits of the tire. Being aware of the relationship outlined in the Ideal Gas Law is essential for all cyclists.

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Most popular questions from this chapter

A sample of gas occupies \(135 \mathrm{mL}\) at \(22.5^{\circ} \mathrm{C} ;\) the pressure is \(165 \mathrm{mm}\) Hg. What is the pressure of the gas sample when it is placed in a 252 -mL. flask at a temperature of \(0.0^{\circ} \mathrm{C} ?\)

A xenon fluoride can be prepared by heating a mixture of \(\mathrm{Xe}\) and \(\mathrm{F}_{2}\) gases to a high temperature in a pressureproof container. Assume that xenon gas was added to a \(0.25-\mathrm{L}\) container until its pressure reached \(0.12 \mathrm{atm}\) at \(0.0^{\circ} \mathrm{C} .\) Fluorine gas was then added until the total pressure reached 0.72 atm at \(0.0^{\circ} \mathrm{C}\). After the reaction was complete, the xenon was consumed completely, and the pressure of the \(\mathrm{F}_{2}\) remaining in the container was 0.36 atm at \(0.0^{\circ} \mathrm{C} .\) What is the empirical formula of the xenon fluoride?

You have \(3.6 \mathrm{L}\) of \(\mathrm{H}_{2}\) gas at \(380 \mathrm{mm} \mathrm{Hg}\) and \(25^{\circ} \mathrm{C}\) What is the pressure of this gas if it is transferred to a \(5.0-1 .\) flask at \(0.0^{\circ} \mathrm{C} 2\)

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A balloon for long-distance flying contains \(1.2 \times 10^{7} \mathrm{L}\) of helium. If the helium pressure is \(737 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C},\) what mass of helium (in grams) does the balloon contain? (See Study Question \(14 .\) )

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