Problem 34 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

(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 1: 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.

Step 1: 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 \( p_2 \).

Step 2: 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 \( \left( \dfrac{V_1}{V_2} \right) = \dfrac{1}{0.0900} \).

Step 3: 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 \( T_2 \approx 774.11 \, K \).

Step 4: 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 \( p_2 \approx 1810.44 \, \text{kPa} \).

Step 5: 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.

Key Concepts

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.