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Air \((\gamma=1.4)\) is pumped at 2 atm pressure in a motor tyre at \(20^{\circ} \mathrm{C}\). If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre. Neglect any mixing with the atmospheric air.

Short Answer

Expert verified
The air temperature is approximately -16.15°C.

Step by step solution

01

Understand the Process

When the tyre bursts, the process is adiabatic expansion. We must use the adiabatic relation for ideal gases to find the new temperature. The initial conditions are 2 atm and 20°C (293 K), and the final pressure will be 1 atm.
02

Use the Adiabatic Process Formula

The adiabatic process for an ideal gas is given by the formula \( \frac{T_1}{T_2} = \left( \frac{P_1}{P_2} \right)^{\frac{\gamma-1}{\gamma}} \), where \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin, \( P_1 \) and \( P_2 \) are the initial and final pressures, and \( \gamma \) is the heat capacity ratio.
03

Substitute Known Values

Substitute \( T_1 = 293 \, \text{K} \), \( P_1 = 2 \, \text{atm} \), \( P_2 = 1 \, \text{atm} \), and \( \gamma = 1.4 \). The equation becomes: \[ \frac{293}{T_2} = \left( \frac{2}{1} \right)^{\frac{1.4-1}{1.4}} = 2^{0.4/1.4}. \]
04

Solve for Final Temperature \( T_2 \)

Calculate \( 2^{0.4/1.4} \) using a calculator, which is approximately 1.14. Then solve for \( T_2 \): \[ T_2 = \frac{293}{1.14} \approx 257 \, \text{K}. \]
05

Convert Kelvin to Celsius

Convert the final temperature from Kelvin to Celsius using the formula \( T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15 \). Thus, \( T_{\text{Celsius}} = 257 - 273.15 \approx -16.15 \, \text{°C}. \)

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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 principle in thermodynamics that relates the pressure, volume, and temperature of an ideal gas. It is commonly expressed in the equation: \[ PV = nRT \] where:
  • \( P \) is the pressure of the gas.
  • \( V \) is the volume of the gas.
  • \( n \) is the number of moles of the gas.
  • \( R \) is the ideal gas constant.
  • \( T \) is the temperature of the gas in Kelvin.
The Ideal Gas Law is crucial for understanding and predicting the behavior of gases under varying conditions.
In adiabatic processes, like in our exercise, it helps us understand how a change in pressure can affect the temperature, without any heat transfer occurring.
Heat Capacity Ratio
The heat capacity ratio, denoted as \( \gamma \), is a critical parameter in describing the adiabatic processes of gases. It is the ratio of the heat capacity at constant pressure \( C_p \) to the heat capacity at constant volume \( C_v \). Mathematically, it is represented as:\[ \gamma = \frac{C_p}{C_v} \]In the exercise, air has a \( \gamma \) value of 1.4, which is typical for diatomic gases like nitrogen and oxygen, the main components of air.
This ratio determines how much a substance's temperature will change when it expands or is compressed adiabatically.
High \( \gamma \) values often result in more significant temperature changes under adiabatic conditions, demonstrating its importance in calculations of temperature changes when no heat is exchanged with the surroundings.
Temperature Conversion
Converting between temperature units is a necessary skill in thermodynamics. In the context of our exercise, it involves converting Celsius to Kelvin and vice versa. For scientific calculations, the Kelvin scale is often used because it starts at absolute zero, the lowest possible temperature.
  • Convert Celsius to Kelvin: \( T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \)
  • Convert Kelvin to Celsius: \( T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15 \)
These conversions allow scientists to seamlessly incorporate temperatures into formulas consistent with other units in the International System.
Pressure-Temperature Relationship
The Pressure-Temperature Relationship is a key aspect of adiabatic processes. In adiabatic expansion, like when a tire bursts, the temperature of the gas decreases as the pressure decreases, obeying this relationship.
For an adiabatic process of an ideal gas, the relationship is mathematically expressed as:\[ \frac{T_1}{T_2} = \left( \frac{P_1}{P_2} \right)^{\frac{\gamma-1}{\gamma}} \]where \( T_1 \) and \( T_2 \) are initial and final temperatures, \( P_1 \) and \( P_2 \) are initial and final pressures, and \( \gamma \) is the heat capacity ratio.
  • A higher initial pressure compared to the final pressure means the temperature will drop. This principle is evident when the pressure is reduced but adiabatically, keeping the temperature still relevant.
  • In our exercise, this relationship helps in accurately finding the change in temperature of the air escaping the tyre.
Understanding this relationship is vital in processes where temperature control and predictions under changing pressure conditions are crucial.

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Most popular questions from this chapter

Consider a given sample of an ideal gas \(\left(C_{p} / C_{V}=\gamma\right)\) having initial pressure \(p_{0}\) and volume \(V_{c}\). (a) The gas is isothermally taken to a pressure \(p_{0} / 2\) and from there adiabatically to a pressure \(p_{0} / 4\). Find the final volume. (b) The gas is brought back to its initial state. It is adiabatically taken to a pressure \(p_{0} / 2\) and from there isothermally to a pressure \(p_{0} / 4\). Find the final volume.

Figure (27-E5) shows two rigid vessels \(A\) and \(B\), each of volume \(200 \mathrm{~cm}^{3}\) containing an ideal gas \(\left(C_{V}=12 \cdot 5\right.\) \(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) ). The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is \(75 \mathrm{~cm}\) of mercury and the temperature is \(300 \mathrm{~K}\). (a) Find the number of moles of the gas in each vessel. (b) \(5 \cdot 0 \mathrm{~J}\) of heat is supplied to the gas in the vessel \(A\) and \(10 \mathrm{~J}\) to the gas in the vessel \(B\). Assuming no appreciable transfer of heat from \(A\) to \(B\) calculate the difference in the heights of mercury in the two sides of the manometer. Gas constant \(R=8 \cdot 3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

An amount \(Q\) of heat is added to a monatomic ideal gas in a process in which the gas performs a work \(Q / 2\) on its surrounding. Find the molar heat capacity for the process.

\(5 \mathrm{~g}\) of a gas is contained in a rigid container and is heated from \(15^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\). Specific heat capacity of the gas at constant volume is \(0.172 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\) and the mechanical equivalent of heat is \(4 \cdot 2 \mathrm{~J} \mathrm{cal}^{-1} .\) Calculate the change in the internal energy of the gas.

The volume of an ideal gas \((\gamma=1 \cdot 5)\) is changed adiabatically from \(4.00\) litres to \(3.00\) litres. Find the ratio of (a) the final pressure to the initial pressure and (b) the final temperature to the initial temperature.

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