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Chapter 19: Problem 34

The engine of a Ferrari \(\mathrm{F} 355 \mathrm{FI}\) sports car takes in air at \(20.0^{\circ} \mathrm{C}\) and 1.00 \(\mathrm{atm}\) and compresses it adiabatically to 0.0900 times the original volume. The air may be treated as an ideal gas with \(\gamma=1.40 .\) (a) Draw a \(p V\) -diagram for this process. (b) Find the final temperature and pressure.

Short Answer

Expert verified
(a) Draw a steeper curve from \( P_1, V_1 \) to \( P_2, V_2 \). (b) \( T_2 \approx 774.11 \, K \), \( p_2 \approx 1810.44 \, \text{kPa} \).

Step by step solution

01

Understanding Adiabatic Process

An adiabatic process is one where heat is not transferred into or out of the system, so any change in internal energy results only from work done on or by the system. We will use the relation for adiabatic processes: \( pV^\gamma = \text{constant} \). The problem states that \( \gamma = 1.40 \), indicating the ratio of specific heats for air.
02

Initial Setup and Variables

For the problem, we know the initial temperature \( T_1 = 20.0^\circ C = 293.15 \, K \) and initial pressure \( p_1 = 1.00 \, \text{atm} = 101.3 \, \text{kPa} \). The volume is compressed to 0.0900 times the initial volume, meaning \( V_2 = 0.0900 V_1 \). The goal is to find the final temperature \( T_2 \) and final pressure \( p_2 \).
03

Using the Ideal Gas Law

According to the ideal gas law, we can express the relationship between pressure, volume, and temperature as: \( TV^{\gamma-1} = \text{constant} \). We can use this equation to find the final temperature: \( T_2 = T_1 \left( \dfrac{V_1}{V_2} \right)^{\gamma-1} \), where \( \left( \dfrac{V_1}{V_2} \right) = \dfrac{1}{0.0900} \).
04

Calculating the Final Temperature

Substitute the known values into the equation: \( T_2 = 293.15 \left( \dfrac{1}{0.0900} \right)^{1.40-1} \). Simplifying inside the parenthesis, \( T_2 = 293.15 \left( 11.111 \right)^{0.40} \). Calculating gives \( T_2 \approx 774.11 \, K \).
05

Calculating the Final Pressure

Using the adiabatic condition for pressure and volume \( p_1V_1^\gamma = p_2V_2^\gamma \). Rearranging this gives \( p_2 = p_1 \left( \dfrac{V_1}{V_2} \right)^\gamma \). Substituting in, \( p_2 = 101.3 \times (11.111)^{1.40} \). This gives \( p_2 \approx 1810.44 \, \text{kPa} \).
06

Drawing the PV Diagram

On a PV diagram, the initial point \( P_1 \) represents the initial pressure and volume \((V_1)\). As the volume decreases to \(0.0900V_1\), the pressure increases to \(p_2\). The curve between these points should be steeper than an isothermal because it is adiabatic.

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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 crucial concept in understanding gas behavior under varying conditions. It is expressed as \( PV = nRT \), where:
  • \( P \) represents pressure
  • \( V \) is volume
  • \( n \) denotes the number of moles
  • \( R \) is the ideal gas constant
  • \( T \) is the absolute temperature in Kelvin
This equation shows the direct relationship between pressure, volume, and temperature. In adiabatic processes, although the total energy remains constant, changes in these state variables occur due to compression or expansion. The ideal gas law ensures that we can relate these changes accurately, essential for calculating outcomes like final pressure and temperature.
Specific Heat Ratio
The specific heat ratio \( \gamma \) is a key factor in adiabatic processes. Defined as \( \gamma = \frac{C_p}{C_v} \), it signifies the ratio between specific heat at constant pressure \( C_p \) and specific heat at constant volume \( C_v \).
  • For air, \( \gamma = 1.40 \), often used in these calculations.
This ratio is vital in equations governing adiabatic processes, such as \( pV^\gamma = \text{constant} \) and \( TV^{\gamma-1} = \text{constant} \). Understanding \( \gamma \) helps in predicting how gases behave during compression or expansion without heat exchange.
PV Diagram
A PV diagram is a graphical representation of the relationship between pressure (P) and volume (V) of a gas. It visually illustrates how these variables change throughout a process.
  • In an adiabatic process, like engine compression, the curve is steep compared to an isothermal process.
  • Initially, volume decreases while pressure increases significantly due to the absence of heat transfer.
The diagram of an adiabatic process reflects these dynamics with a steep curve, indicating rapid pressure increase with decreasing volume, as observed from the initial to final states in the given Ferrari engine problem.
Final Temperature and Pressure
Determining the final temperature and pressure in adiabatic compression involves using specific equations adapted from the ideal gas law.For temperature, the relation is \( T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1} \). In the solution:
  • The initial temperature \( T_1 \) is converted from Celsius to Kelvin, providing a base for calculation.
  • A volume ratio \( \frac{V_1}{V_2} \) dictates the magnitude of temperature increase.
Regarding pressure, use \( p_2 = p_1 \left( \frac{V_1}{V_2} \right)^\gamma \), emphasizing how compressing volume elevates pressure.From the problem solution, the calculations reveal a final temperature of approximately 774.11 K and pressure at 1810.44 kPa, highlighting significant changes due to adiabatic compression.

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Most popular questions from this chapter

A cylinder with a movable piston contains 3.00 \(\mathrm{mol}\) of \(\mathrm{N}_{2}\) gas (assumed to behave like an ideal gas). (a) The \(\mathrm{N}_{2}\) is heated at constant volume until 1557 \(\mathrm{J}\) of heat have been added. Calculate the change in temperature. (b) Suppose the same amount of heat is added to the \(\mathrm{N}_{2}\) , but this time the gas is allowed to expand while remaining at constant pressure. Calculate the temperature change. (c) In which case, (a) or \((b),\) is the final internal energy of the \(\mathbf{N}_{2}\) higher? How do you know? What accounts for the difference between the two cases?

On a warm summer day, a large mass of air (armospheric pressure \(1.01 \times 10^{5} \mathrm{Pa}\) ) is heated by the ground to a temperature of \(26.0^{\circ} \mathrm{C}\) and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why? Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only \(0.850 \times 10^{5} \mathrm{Pa}\) . Assume that air is an ideal gas, with \(\gamma=1.40\) . (This rate of cool- ing for dry, rising air, corresponding to roughly \(1^{\circ} \mathrm{C}\) per 100 \(\mathrm{m}\) of attitude, is called the dry adiabatic lapse rate.)

In a cylinder sealed with a piston, you rapidly compress 3.00 \(\mathrm{L}\) of \(\mathrm{N}_{2}\) gas initially at 1.00 atm pressure and \(0.00^{\circ} \mathrm{C}\) to half its original volume. Assume the \(\mathrm{N}_{2}\) behaves like an ideal gas. (a) Calculate the final temperature and pressure of the gas. (b) If you now cool the gas back to \(0.00^{\circ} \mathrm{C}\) without changing the pressure, what is its final volume?

A cylinder contains 0.0100 mol of helium at \(T=27.0^{\circ} \mathrm{C}\) (a) How much heat is needed to raise the temperature to \(67.0^{\circ} \mathrm{C}\) while keeping the volume constant? Draw a \(p V\) -diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from \(27.0^{\circ} \mathrm{C}\) to \(67.0^{\circ} \mathrm{C}\) ? Draw a pV-diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? d) If the gas is ideal, what is the change in its internal energy in part (a)? In part \((b) ?\) How do the two answers compare? Why?

A Thermodynamic \(\mathbf{P} \mathbf{r} \mathbf{o}-\) cess In an Insect. The Africanbombardier beetle Stenaptinus insignis can emit a jet of defensive spray from the movable tip of its abdomen (Fig, 19.32\()\) . The beetle's body has reservoirs of two different chemicals; when the beetle is disturbed, these chemicals are combined in a reaction chamber, producing compound that is warmed from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) by the heat of reaction. The high pressure produced allows the compound to be sprayed out at speeds up to 19 \(\mathrm{m} / \mathrm{s}(68 \mathrm{km} / \mathrm{h})\) , scaring away predators of all kinds. (The beetle shown in the figure is 2 \(\mathrm{cm}\) long.) Calculate the heat of reaction of the two chemicals (in J/kg). Assume that the specific heat capacity of the two chemicall and the spray is the same as that of water, \(4.19 \times 10^{3} \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , and that the initial temperature of the chemicals is \(20^{\circ} \mathrm{C}\) .
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