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Before beginning a long trip on a hot day, a driver inflates an automobile tire to a gauge pressure of \(1.80\) atm at \(300 \mathrm{~K}\). At the end of the trip, the gauge pressure has increased to \(2.20\) atm. (a) Assuming the volume has remained constant, what is the temperature of the air inside the tire? (b) What percentage of the original mass of air in the tire should be released so the pressure returns to its original value? Assume the temperature remains at the value found in part (a) and the volume of the tire remains constant as air is released.

Short Answer

Expert verified
The final temperature of the air inside the tire is approximately 366.67K. The driver should release approximately 18.18% of the original mass of air in the tire to return the pressure to its original value.

Step by step solution

01

Apply the Ideal Gas Law

Use the ideal gas law in the form \(P_1/T_1 = P_2/T_2\), because V and n are constant, and simplify to find \(T_2\). Here \(P_1 = 1.80\) atm, \(T_1 = 300K\), and \(P_2 = 2.20\) atm, but \(T_2\) is unknown.
02

Calculate Final Temperature

Solve for \(T_2\), \(T_2 = P_2*T_1 / P_1 = 2.20*300 / 1.80 \approx 366.67K\). The temperature of the air inside the tire after the trip is approximately 366.67K.
03

Calculate Final Moles of Gas

For the second part of the question, once again use the ideal gas law, but this time we are considering change in number of moles of the gas to bring back the pressure to its original value. So, assuming \(T = T_2\), \(P = P_1\), we set up a ratio \(n_1/n_2 = P_1/P_2 = 1.80/2.20\). As n1 is taken as 100%, solve for n2 = n1 * (1.80/2.20) = 81.82% of n1.
04

Calculate Percentage of Air to be Release

The percentage of the original mass of air in the tire to be released is given by: \(percentage = 100 - 81.82 = 18.18\%\).

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships and conversions between heat and other forms of energy. It plays a fundamental role in understanding how systems operate, including gases in an automobile tire. In this context, understanding how temperature and pressure of a gas change when conditions vary is essential.

Thermodynamics provides the underlying principles that explain the behavior of gases, particularly through various laws like the Ideal Gas Law. This law simplifies the complexities of thermodynamics by describing the state of an ideal gas and providing a foundation for solving real-world problems. It mainly concerns energy transfer and conversion, laying the groundwork for the study of pressure, volume, and temperature relationships in gases.
Pressure-Temperature Relationship
The pressure-temperature relationship of gases is crucial in thermodynamics. It outlines how changes in temperature can lead to changes in pressure, assuming constant volume. This relationship is part of Gay-Lussac's law, stating that the pressure of a fixed mass of gas is directly proportional to its absolute temperature, provided the volume remains constant.
  • When temperature increases, the gas molecules move more vigorously, increasing the pressure inside a fixed volume, like a tire.
  • Conversely, if temperature decreases, the pressure goes down as well.
This concept is mathematically represented through the Ideal Gas Law when it states that \[\frac{P_1}{T_1} = \frac{P_2}{T_2}\]In our problem, knowing the initial pressure and temperature allows us to calculate the final temperature, given a change in pressure. This understanding is fundamental in predicting how temperature impacts gas pressure, which was necessary to solve for the new temperature of the tire.
Gas Laws
The Ideal Gas Law is one of the fundamental gas laws in thermodynamics, combining several individual laws into a single equation. It is expressed as \(PV = nRT\), where:
  • \(P\) is the pressure of the gas.
  • \(V\) is the volume.
  • \(n\) is the number of moles.
  • \(R\) is the universal gas constant.
  • \(T\) is the temperature in Kelvin.
While the full formula accounts for multiple variables, in many exercises, such as the one with the automobile tire, simplification occurs when certain quantities like volume and number of moles are constant.

In such cases, the simplified version used can be \[\frac{P_1}{T_1} = \frac{P_2}{T_2}\] This version shows the intrinsic link between the pressure and temperature of the gas when other factors don't change. Understanding this allows students not only to solve academic problems but also to see real-world applications, like why tire pressure increases on a hot day.
Temperature Calculation
Temperature calculation in gases often revolves around changes when using the Ideal Gas Law. In this exercise, finding the final temperature after a change in pressure is crucial.

The formula above is rearranged to solve for the unknown temperature:\[T_2 = \frac{P_2 \times T_1}{P_1} \]Plugging the values from our problem:
  • Initial pressure \( P_1 = 1.80 \text{ atm} \)
  • Final pressure \( P_2 = 2.20 \text{ atm} \)
  • Initial temperature \( T_1 = 300 \text{ K} \)
The calculation gives:\[T_2 = \frac{2.20 \times 300}{1.80} \approx 366.67 \text{ K}\]This result shows how the temperature inside the tire increased to approximately 366.67 K. It explains step by step, ensuring clarity, and connects academic understanding with practical scenarios like adjusting tire pressure on a hot day. Understanding each step and how numbers fit into the equation aids in mastering these thermodynamic calculations.

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Most popular questions from this chapter

A grandfather clock is controlled by a swinging brass pendulum that is \(1.3 \mathrm{~m}\) long at a temperature of \(20^{\circ} \mathrm{C}\). (a) What is the length of the pendulum rod when the temperature drops to \(0.0^{\circ} \mathrm{C}_{7}^{\text {(b) If a pendulum's period is given }}\) by \(T=2 \pi \sqrt{L / g}\), where \(L\) is its length, does the change in length of the rod cause the clock to rum fast or slow?

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Two small containers, each with a volume of \(100 \mathrm{~cm}^{3}\), contain helium gas at \(0^{-} \mathrm{C}\) and \(1.00\) atm pressure. The two containers are joined by a small open tube of negligible volume, allowing gas to flow from one container to the other. What common pressure will exist in the two containers if the temperature of one container is raised to \(100^{\circ} \mathrm{C}\) while the other container is kept at \(0^{\circ} \mathrm{C}\) ?

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