/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 An automobile tire contains air ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 33

An automobile tire contains air at \(225 \times 10^{3} \mathrm{Pa}\) at \(25.0^{\circ} \mathrm{C} .\) The stem valve is removed and the air is allowed to expand adiabatically against the constant external pressure of one bar until \(P=P_{\text {external}}\). For air, \(C_{V, m}=5 R / 2 .\) Calculate the final temperature. Assume ideal gas behavior

Short Answer

Expert verified
The final temperature of the air inside the tire after adiabatic expansion is approximately \(171.98 \, K\).

Step by step solution

01

Write down the given information and relevant formulas

Given Information: 1. Initial pressure, \(P_1 = 225 \times 10^3 \, Pa\) 2. Initial temperature, \(T_1 = 25.0 + 273.15 = 298.15 \, K\) (converted to Kelvin) 3. Constant external pressure, \(P_2 = 10^5 \, Pa\) (1 bar) 4. Molar heat capacity at constant volume, \(C_{V, m} = \frac{5}{2}R\), where \(R\) is the universal gas constant Adiabatic expansion formula for ideal gases: \[T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{(C_{V,m} - R)/C_{V,m}}\]
02

Rearrange the adiabatic expansion formula to find the final temperature (T2)

We need to find the final temperature (\(T_2\)). Plug the given values in the formula: \[T_2 = 298.15 \, K \times \left(\frac{10^5 \, Pa}{225 \times 10^3 \, Pa}\right)^{(\frac{5}{2}R - R)/(\frac{5}{2}R)}\]
03

Simplify the exponent

Simplify the exponent term in the formula: \(\frac{(\frac{5}{2}R - R)}{(\frac{5}{2}R)} = \frac{\frac{1}{2}R}{\frac{5}{2}R} = \frac{1}{5}\) Now, plug the simplified exponent into the formula: \[T_2 = 298.15 \, K \times \left(\frac{10^5 \, Pa}{225 \times 10^3 \, Pa}\right)^{1/5}\]
04

Calculate the final temperature

Calculate the final temperature by evaluating the expression: \[T_2 = 298.15 \, K \times \left(\frac{10^5 \, Pa}{225 \times 10^3 \, Pa}\right)^{1/5} \approx 171.98 \, K\] The final temperature of the air inside the tire after adiabatic expansion is approximately 171.98 K.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation in understanding the relationship between pressure, volume, temperature, and moles of an ideal gas. It is typically expressed as \( PV = nRT \) where \( P \) is the pressure, \( V \) is the volume, \( n \) is the amount of substance in moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

This law combines several gas laws including Boyle's, Charles's, and Avogadro's laws into one comprehensive equation. In our adiabatic expansion problem, the ideal gas law helps us to understand how gas expands when it is not in thermal equilibrium with its surroundings.
Molar Heat Capacity
Molar heat capacity, denoted as \( C_m \) is the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius. For an ideal gas, the molar heat capacity can be measured at constant volume \( (C_{V,m}) \) or constant pressure \( (C_{P,m}) \). The relationship between the two is given by \( C_{P,m} = C_{V,m} + R \), where \( R \) is the universal gas constant.

Understanding the concept of molar heat capacity is crucial in thermodynamics to predict how a gas will respond to changes in temperature, especially during adiabatic processes where no heat is exchanged with the environment.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It involves studying how energy is transferred within systems and how it affects matter. The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. The second law introduces the concept of entropy, a measure of disorder, which tends to increase over time.

In our context, adiabatic expansion is a thermodynamic process in which a gas expands without exchanging heat with its surroundings. This means that the internal energy change of the gas is used entirely to do work on the surroundings, which manifests as a change in temperature and pressure.
Universal Gas Constant
The universal gas constant \( R \) is a physical constant that is featured in the ideal gas law and appears in many fundamental equations in thermodynamics, including the calculation of molar heat capacities. It links the macroscopic properties of gases with the amount of substance present. The value of \( R \) is \( 8.314 \text{J mol}^{-1} \text{K}^{-1} \). This constant is derived from empirical measurements on ideal gas behavior and has the same value for all ideal gases.

In the solution to our exercise, the universal gas constant was used to find the final temperature of the gas following adiabatic expansion, illustrating its usefulness in practical applications of thermodynamics.

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Most popular questions from this chapter

A 2.25 mole sample of an ideal gas with \(C_{V, m}=3 R / 2\) initially at \(310 . \mathrm{K}\) and \(1.25 \times 10^{5}\) Pa undergoes a reversible adiabatic compression. At the end of the process, the pressure is \(3.10 \times 10^{6}\) Pa. Calculate the final temperature of the gas. Calculate \(q, w, \Delta U,\) and \(\Delta H\) for this process.

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 2.35 mole sample of an ideal gas, for which \(C_{V, m}=\) \(3 R / 2,\) initially at \(27.0^{\circ} \mathrm{C}\) and \(1.75 \times 10^{6} \mathrm{Pa}\), undergoes a twostage transformation. For each of the stages described in the following list, calculate the final pressure, as well as \(q, w, \Delta U,\) and \(\Delta H .\) Also calculate \(q, w, \Delta U,\) and \(\Delta H\) for the complete process. a. The gas is expanded isothermally and reversibly until the volume triples. b. Beginning at the end of the first stage, the temperature is raised to \(105^{\circ} \mathrm{C}\) at constant volume.

A 3.50 mole sample of \(\mathrm{N}_{2}\) in a state defined by \(T_{i}=\) 250\. \(\mathrm{K}\) and \(V_{i}=3.25 \mathrm{L}\) undergoes an isothermal reversible expansion until \(V_{f}=35.5 \mathrm{L}\) Calculate \(w,\) assuming (a) that the gas is described by the ideal gas law, and (b) that the gas is described by the van der Waals equation of state. What is the percent error in using the ideal gas law instead of the van der Waals equation? The van der Waals parameters for \(\mathrm{N}_{2}\) are listed in Table 7.4

A \(2.50 \mathrm{mol}\) sample of an ideal gas for which \(C_{V, m}=3 R / 2\) undergoes the following two-step process: (1) From an initial state of the gas described by \(T=13.1^{\circ} \mathrm{C}\) and \(P=1.75 \times 10^{5} \mathrm{Pa}\), the gas undergoes an isothermal expansion against a constant external pressure of \(3.75 \times 10^{4} \mathrm{Pa}\) until the volume has doubled. (2) Subsequently, the gas is cooled at constant volume. The temperature falls to \(-23.6^{\circ} \mathrm{C}\). Calculate \(q, w, \Delta U,\) and \(\Delta H\) for each step and for the overall process.
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