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A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Assume the initial pressure and temperature of the air before it is put in the tire to be \(P_{i}=1.00\) bar and \(T_{i}=280 .\) K. The final pressure in the tire is \(P_{f}=3.75\) bar. Calculate the final temperature of the air in the tire. Assume that \(C_{V, \pi}=5 R / 2\)

Short Answer

Expert verified
The final temperature of the air in the tire after the adiabatic compression process is approximately \(312.7\,\text{K}\).

Step by step solution

01

Identify the given values

We are given the initial pressure, initial temperature, and final pressure of the air in the tire: - Initial pressure (\(P_{i}\)): 1.00 bar - Initial temperature (\(T_{i}\)): 280 K - Final pressure (\(P_{f}\)): 3.75 bar
02

Convert the pressure units

We need to convert the pressure from bars to Pascals (Pa) because we'll use the ideal gas law, which requires pressure in Pascals. 1 bar is equal to \(10^{5}\) Pa. - Initial pressure (\(P_{i}\)): \(1.00 \times 10^5\) Pa - Final pressure (\(P_{f}\)): \(3.75 \times 10^5\) Pa
03

Apply the adiabatic compression formula

For a reversible adiabatic process, the temperature and pressure are related by the following equation, where \(\gamma\) is the adiabatic index: \[ \frac{T_{f}}{T_{i}} = \left(\frac{P_{i}}{P_{f}}\right)^{\frac{\gamma-1}{\gamma}} \] We are given that \(C_{V, \pi} = \frac{5R}{2}\). Since \(C_P - C_V = R\) for an ideal gas, we can find the adiabatic index \(\gamma\) by the following relation: \[ \gamma = \frac{C_P}{C_V} = \frac{C_V + R}{C_V} = \frac{\frac{5R}{2} + R}{\frac{5R}{2}} = \frac{7R}{5R} = \frac{7}{5} \] Now, we can use the adiabatic compression formula to find the final temperature: \[ \frac{T_{f}}{280} = \left(\frac{1.00\times10^5}{3.75\times10^5}\right)^{\frac{2}{7}} \]
04

Solve for the final temperature

Now, we can solve the equation for \(T_{f}\): \[ T_{f} = 280 \cdot \left(\frac{1.00\times10^5}{3.75\times10^5}\right)^{\frac{2}{7}} \] Using a calculator, we find that: \[ T_{f} \approx 312.7\,\text{K} \] The final temperature of the air in the tire is approximately 312.7 K.

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Key Concepts

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

Reversible Adiabatic Compression
When you pump up a bicycle tire, you often notice that it gets warmer. This temperature increase is a key feature of a reversible adiabatic compression. In such a process, there is no heat exchange between the system and its surroundings, hence the term "adiabatic." Reversible means that the process could theoretically be reversed without any energy loss.
The temperature increase happens because of the work done on the gas inside the tire. As the tire is inflated, the gas is compressed, making its molecules pack closer together and move faster, thereby increasing the temperature. This is described by the adiabatic relation between pressure and temperature. In this specific scenario, the process is assumed to be adiabatic, allowing us to use neat formulas to predict changes in temperature, assuming all variables other than pressure remain constant.
This concept is crucial in understanding how thermal systems can change internal energy through work while insulated from their environment.
Ideal Gas Law
The ideal gas law forms the foundation of understanding gas behaviors under various conditions. It is generally given by the equation \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature.
In the context of adiabatic processes, the volume of the gas might change due to pressure adjustments, but the ideal gas law helps us convert between different states such as temperature and pressure easily. For an adiabatic process, equations derived from the ideal gas assumptions help compute final states based on known initial conditions.
This law is fundamental, especially when dealing with common gases where intermolecular forces and size effects can be neglected, hence the assumption of being "ideal." While gases in reality can deviate from this behavior, the law provides an excellent approximation for many practical problems.
Adiabatic Index
The adiabatic index, often symbolized by \( \gamma \), is a critical parameter in understanding adiabatic processes. It is defined as the ratio of specific heats at constant pressure (\( C_P \)) and volume (\( C_V \)), given by \( \gamma = \frac{C_P}{C_V} \).
This index determines how the temperature of a gas changes under adiabatic compression or expansion. Calculating \( \gamma \) helps simplify the mathematical treatment of adiabatic processes by relating temperature and pressure ratios over the process, as seen in the equation used for calculating the final temperature of a compressed gas.
In the idealized scenario presented, the specific heats were given or derived based on known properties of gases, allowing us to accurately compute the adiabatic index and thus the behavior of the gas during the compression.
Thermal Physics
Thermal physics examines how energy is transferred in systems and how this affects material properties such as temperature. It blends concepts from thermodynamics, statistical mechanics, and kinetic theory to explain how materials respond to stimuli.
In this exercise with the bicycle tire, thermal physics principles help us understand why inflating a tire increases the internal temperature. Compression increases the kinetic energy of gas molecules, raising the temperature as predicted by thermodynamic principles.
The understanding of these thermal behaviors is crucial for developing efficient thermal management and energy transfer systems in many engineering fields. By delving into thermal physics, students can better grasp why processes like adiabatic compression bring about observable changes in everyday phenomena.

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Most popular questions from this chapter

DNA can be modeled as an elastic rod that can be twisted or bent. Suppose a DNA molecule of length \(L\) is bent such that it lies on the arc of a circle of radius \(R_{c}\). The reversible work involved in bending DNA without twisting is \(w_{b e n d}=\frac{B L}{2 R_{c}^{2}}\) where \(B\) is the bending force constant. The DNA in a nucleosome particle is about \(680 . \AA\) in length. Nucleosomal DNA is bent around a protein complex called the histone octamer into a circle of radius 55 A. Calculate the reversible work involved in bending the DNA around the histone octamer if the force constant \(B=2.00 \times 10^{-28} \mathrm{Jm}\)

Calculate \(q, w, \Delta U,\) and \(\Delta H\) if 2.25 mol of an ideal gas with \(C_{V, m}=3 R / 2\) undergoes a reversible adiabatic expansion from an initial volume \(V_{i}=5.50 \mathrm{m}^{3}\) to a final volume \(V_{f}=25.0 \mathrm{m}^{3} .\) The initial temperature is \(275 \mathrm{K}\)

A 2.50 mole sample of an ideal gas, for which \(C_{V, m}=3 R / 2,\) is subjected to two successive changes in state: (1) From \(25.0^{\circ} \mathrm{C}\) and \(125 \times 10^{3} \mathrm{Pa}\), the gas is expanded isothermally against a constant pressure of \(15.2 \times 10^{3} \mathrm{Pa}\) to twice the initial volume. (2) At the end of the previous process, the gas is cooled at constant volume from \(25.0^{\circ} \mathrm{C}\) to \(-29.0^{\circ} \mathrm{C} .\) Calculate \(q, w, \Delta U,\) and \(\Delta H\) for each of the stages. Also calculate \(q, w, \Delta U,\) and \(\Delta H\) for the complete process.

A 1.50 mole sample of an ideal gas at \(28.5^{\circ} \mathrm{C}\) expands isothermally from an initial volume of \(22.5 \mathrm{dm}^{3}\) to a final volume of \(75.5 \mathrm{dm}^{3} .\) Calculate \(w\) for this process (a) for expansion against a constant external pressure of \(1.00 \times 10^{5} \mathrm{Pa}\) and (b) for a reversible expansion.

A 1.75 mole sample of an ideal gas for which \(C_{V, m}=20.8 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\) is heated from an initial temperature of \(21.2^{\circ} \mathrm{C}\) to a final temperature of \(380 .^{\circ} \mathrm{C}\) at constant volume. Calculate \(q, w, \Delta U,\) and \(\Delta H\) for this process.

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