Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A tire is inflated to a gauge pressure of \(28.0\) psi at \(71^{\circ} \mathrm{F}\). Gauge pressure is the pressure above atmospheric pressure, which is \(14.7\) psi. After several hours of driving, the air in the tire has a temperature of \(115^{\circ} \mathrm{F}\). What is the gauge pressure of the air in the tire? What is the actual pressure of the air in the tire? Assume that the tire volume changes are negligible.

Short Answer

Expert verified
Answer: The new gauge pressure is 30.62 psi, and the actual pressure of the air in the tire is 45.32 psi.

Step by step solution

01

Convert Temperature to Kelvin

To use the Gay-Lussac's Law, we need to convert the temperatures given in degrees Fahrenheit to Kelvin. The conversion is as follows: \(K = \frac{5}{9} (F - 32) + 273.15\) For \(T_1 = 71^{\circ}\mathrm{F}\): \(T_1K = \frac{5}{9}(71-32) + 273.15 = 294.26\mathrm{K}\) For \(T_2 = 115^{\circ}\mathrm{F}\): \(T_2K = \frac{5}{9}(115-32) + 273.15 = 318.71\mathrm{K}\)
02

Use Gay-Lussac's Law to find the new gauge pressure

Using the given initial gauge pressure, \(P_1 = 28.0\,\text{psi}\), and the temperatures in Kelvin, we can find the new gauge pressure, \(P_2\), using Gay-Lussac's Law: \(\frac{P_1}{T_1K}=\frac{P_2}{T_2K}\) Rearranging for \(P_2\): \(P_2 = \frac{P_1 \cdot T_2K}{T_1K} = \frac{28.0\,\text{psi} \cdot 318.71\,\text{K}}{294.26\,\text{K}} = 30.62\,\text{psi}\) So, the new gauge pressure is \(30.62\,\text{psi}\).
03

Convert gauge pressure to actual pressure

To find the actual pressure of the air in the tire, we need to add the atmospheric pressure, \(P_{atm}=14.7\,\text{psi}\), to the gauge pressure: \(P_\text{actual}=P_\text{gauge}+P_\text{atm}=30.62\,\text{psi} +14.7\,\text{psi}=45.32\,\text{psi}\) The actual pressure of the air in the tire is \(45.32\,\text{psi}\). The new gauge pressure is \(30.62\,\text{psi}\), and the actual pressure of the air in the tire is \(45.32\,\text{psi}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Gauge Pressure
Gauge pressure is a term you'll often encounter in physics, especially when dealing with fluids and gases. It refers to the pressure of a system above the atmospheric pressure. Essentially, it's the difference between the absolute pressure within a container (like a tire or a boiler) and the pressure of the air outside. This distinction is crucial because the pressure we experience everyday, atmospheric pressure, acts on everything.

In the exercise with the tire, gauge pressure is what a tire pressure gauge measures. The value doesn't consider the omnipresent atmospheric pressure. That's why, when calculating the actual pressure inside the tire, the atmospheric pressure of about 14.7 psi (pounds per square inch) at sea level is added to the gauge pressure to get the full picture of forces acting on the tire walls.

Understanding gauge pressure is essential for various applications - from inflating tires to industrial processes. It helps ensure safety and efficiency in systems where pressure is a critical factor.
Temperature Conversion
Temperature conversion is an indispensable tool in science for communicating findings and performing calculations when temperatures are expressed in different scales. There are three primary temperature scales used globally: Fahrenheit (°F), Celsius (°C), and Kelvin (K). Fahrenheit is often used for everyday temperature readings in the United States, while Celsius is more common internationally. Kelvin, on the other hand, is the temperature scale of choice for scientific endeavors.

In our exercise's context, converting Fahrenheit to Kelvin is necessary to apply Gay-Lussac's Law correctly, since the law requires absolute temperature for accurate calculations. The Kelvin scale starts at absolute zero, the theoretical point where particles have minimal thermal motion. Understanding how to convert between these scales assures you can apply scientific laws accurately and bridge the communication gap between different systems of measurement.

As demonstrated in the solution, creating a reliable method to convert temperatures is key to performing accurate physics calculations. Whether it's monitoring weather patterns, designing HVAC systems, or working out the expansion of gases, mastering temperature conversion unlocks the ability to navigate between different measurement systems.
Kelvin Temperature Scale
The Kelvin temperature scale is fundamental to the physical sciences because it is based on absolute zero, the coldest possible temperature where molecular energy is at a minimum. Unlike Celsius and Fahrenheit, the Kelvin scale does not use degrees. Instead, it uses Kelvin (K) as the unit of measurement. One of the most important characteristics of the Kelvin scale for scientists and engineers is the direct proportionality to energy.

When working with gas laws, the Kelvin scale provides a way to compare temperatures on an absolute scale – that's why temperatures for our tire pressure problem are converted to Kelvin. This scale ensures that calculations take into account the absolute energy present in the system. Using Kelvin allows for a clearer understanding of thermal dynamics and aids in accurate predictions of gas behavior under varying temperature conditions.

To students and professionals alike, getting familiar with the Kelvin scale and how it relates to other temperature measurement scales is pivotal for a thorough grasp of thermodynamics, physics, and chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), is burned in oxygen, carbon dioxide and steam are formed. A sample of acetylene with a volume of \(7.50 \mathrm{~L}\) and a pressure of \(1.00\) atm is burned in excess oxygen at \(225^{\circ} \mathrm{C}\). The products are transferred without loss to a 10.0-L flask at the same temperature. (a) Write a balanced equation for the reaction. (b) What is the total pressure of the products in the \(10.0\) - \(\mathrm{L}\) flask? (c) What is the partial pressure of each of the products in the flask?

Consider a vessel with a movable piston. A reaction takes place in the vessel at constant pressure and a temperature of \(200 \mathrm{~K}\). When reaction is complete, the pressure remains the same and the volume and temperature double. Which of the following balanced equations best describes the reaction? (a) \(\mathrm{A}+\mathrm{B}_{2} \longrightarrow \mathrm{AB}_{2}\) (b) \(\mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}\) (c) \(2 \mathrm{AB}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}_{2}\) (d) \(2 \mathrm{AB}_{2} \longrightarrow \mathrm{A}_{2}+2 \mathrm{~B}_{2}\)

Dinitrogen oxide, commonly called nitrous oxide, is used as a propellant gas for whipped-cream dispensers. It is prepared by heating ammonium nitrate to \(250^{\circ} \mathrm{C}\). Water vapor is also formed. (a) Write a balanced equation for the decomposition of ammonium nitrate into nitrous oxide and steam. (b) What volume of dinitrogen oxide gas is formed at \(250^{\circ} \mathrm{C}\) and 1,0 atm when \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is heated? Assume \(100 \%\) yield.

Calculate the densities (in grams per liter) of the following gases at \(75^{\circ} \mathrm{F}\) and \(1.33\) bar. (a) argon (b) ammonia (c) acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\)

A balloon filled with helium has a volume of \(1.28 \times 10^{3} \mathrm{~L}\) at sea level where the pressure is \(0.998\) atm and the temperature is \(31^{\circ} \mathrm{C}\). The balloon is taken to the top of a mountain where the pressure is \(0.753 \mathrm{~atm}\) and the temperature is \(-25^{\circ} \mathrm{C}\). What is the volume of the balloon at the top of the mountain?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free