The engine of a Ferrari F355 F1 sports car takes in air at 20.020.0°C and 1.001.00 atm and compresses it adiabatically to 0.09000.0900 times the original volume. The air may be treated as an ideal gas with g=1.40g = 1.40. Find the final temperature and pressure.

Ch 19: The First Law of Thermodynamics

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 19: The First Law of Thermodynamics

Problem 28b

Chapter 19, Problem 28b

The engine of a Ferrari F355 F1 sports car takes in air at $20.0$ °C and $1.00$ atm and compresses it adiabatically to $0.0900$ times the original volume. The air may be treated as an ideal gas with $g equals 1.40$ . Find the final temperature and pressure.

Verified step by step guidance

Understand the concept of adiabatic processes: In an adiabatic process, no heat is exchanged with the surroundings. For an ideal gas undergoing an adiabatic process, the relationship between pressure, volume, and temperature is governed by the adiabatic condition: $ PV^\gamma = \text{constant} $, where $ \gamma $ is the adiabatic index.

Use the adiabatic relation for temperature: The relation between initial and final temperatures in an adiabatic process is given by $ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} $. Here, $ T_1 $ is the initial temperature, $ V_1 $ is the initial volume, $ V_2 $ is the final volume, and $ \gamma = 1.40 $. Convert the initial temperature from Celsius to Kelvin by adding 273.15.

Calculate the final temperature: Substitute the given values into the temperature relation. $ T_1 = 20.0 + 273.15 $ K, $ V_2 = 0.0900 \times V_1 $, and $ \gamma = 1.40 $. Compute $ T_2 $ using the formula $ T_2 = T_1 \left( \frac{1}{0.0900} \right)^{0.40} $.

Use the adiabatic relation for pressure: The relation between initial and final pressures in an adiabatic process is given by $ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{\gamma} $. Here, $ P_1 = 1.00 $ atm is the initial pressure.

Calculate the final pressure: Substitute the given values into the pressure relation. Compute $ P_2 $ using the formula $ P_2 = 1.00 \left( \frac{1}{0.0900} \right)^{1.40} $. This will give you the final pressure in atm.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Adiabatic Process

An adiabatic process is a thermodynamic process in which no heat is exchanged with the surroundings. For an ideal gas undergoing an adiabatic process, the relationship between pressure, volume, and temperature is governed by the adiabatic condition: PV^γ = constant, where γ (gamma) is the heat capacity ratio (Cp/Cv). This concept is crucial for determining the final temperature and pressure of the gas after compression.

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics, expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. This law helps relate the macroscopic properties of gases and is essential for calculating changes in state variables when the gas is treated as ideal, as in this problem.

Heat Capacity Ratio (γ)

The heat capacity ratio, γ (gamma), is the ratio of the heat capacity at constant pressure (Cp) to the heat capacity at constant volume (Cv). It is a critical parameter in adiabatic processes, affecting how pressure and temperature change with volume. For diatomic gases like air, γ is typically around 1.40, which influences the calculations of final temperature and pressure in adiabatic compression.

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Related Practice

Textbook Question

A monatomic ideal gas that is initially at $1.50 \times 10^{5}$ Pa and has a volume of $0.0800$ m³ is compressed adiabatically to a volume of $0.0400$ m³. What is the final pressure?

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Textbook Question

A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at $1.00 \times 10^{5}$ Pa and occupies a volume of $2.50 \times 10^{- 3}$ m³. If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.

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Textbook Question

On a warm summer day, a large mass of air (atmospheric pressure $1.01 \times 10^{5}$ Pa) is heated by the ground to $26.0$ °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 $0.850 \times 10^{5}$ Pa. Assume that air is an ideal gas, with $g equals 1.40$ . (This rate of cooling for dry, rising air, corresponding to roughly $1$ C° per $100$ m of altitude, is called the dry adiabatic lapse rate.)

Textbook Question

A monatomic ideal gas that is initially at $1.50$\times$10^5$ Pa and has a volume of $0.0800$ m³ is compressed adiabatically to a volume of $0.0400$ m³. What is the ratio of the final temperature of the gas to its initial temperature? Is the gas heated or cooled by this compression?

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Textbook Question

Five moles of monatomic ideal gas have initial pressure $2.50 times 10 cubed$ Pa and initial volume $2.10$ m³. While undergoing an adiabatic expansion, the gas does $1480$ J of work. What is the final pressure of the gas after the expansion?

Textbook Question

A player bounces a basketball on the floor, compressing it to $80.0 percent sign$ of its original volume. The air (assume it is essentially N₂ gas) inside the ball is originally at $20.0$ °C and $2.00$ atm. The ball's inside diameter is $23.9$ cm. What temperature does the air in the ball reach at its maximum compression? Assume the compression is adiabatic and treat the gas as ideal.