Problem 32 - University Physics with Modern Physics

Question

The engine of a Ferrari \(\mathrm{F} 355 \mathrm{Fl}\) sports car takes in air at \(20.0^{\circ} \mathrm{C}\) 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 \(\gamma=1.40\) . (a) Draw a p \(V\) -diagram for this process. (b) Find the final temperature and pressure.

Step-by-Step Solution

Answer

The final temperature is approximately 895 K and pressure 29.76 atm.

Step 1: Understand the Problem Context

We have an air compression process in a Ferrari engine, where air is taken in at an initial temperature of \(20.0^{\circ} \mathrm{C}\) and compressed adiabatically to 0.0900 times its original volume. The task is to determine the final temperature and pressure.

Step 2: Convert Initial Temperature to Kelvin

The initial temperature given is \(T_1 = 20.0^{\circ} \mathrm{C}\). To convert this to Kelvin, use the formula: \[ T_1 = 20 + 273.15 = 293.15 \text{ K} \]

Step 3: Use Adiabatic Process Formula

For an adiabatic process, we use the formula \[ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \] where \( \gamma = 1.40 \), \( V_2 = 0.0900 V_1 \). We need to find \( T_2 \), knowing \( T_1 = 293.15 \text{ K} \).

Step 4: Solve for Final Temperature

Substitute into the equation and solve for \( T_2 \):\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1} = 293.15 \times (\frac{1}{0.0900})^{1.40-1} \]After calculation, \[ T_2 \approx 293.15 \times 3.052 = 894.64 \text{ K} \]

Step 5: Apply Ideal Gas Law for Pressure

For an adiabatic process, use \[ P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \] where \( P_1 = 1.00 \text{ atm} \). Substituting known values to find \( P_2 \): \[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{\gamma} \]\[ P_2 = 1.00 \times \left( \frac{1}{0.0900} \right)^{1.40} \approx 1.00 \times 29.76 = 29.76 \text{ atm} \]

Step 6: Draw the P-V Diagram

In an adiabetic compression in a P-V diagram, the curve is steeper than that for an isothermal process. The volume decrease with increasing pressure should be represented without detailed calculations, showcasing the drop to 0.0900 of the initial volume at an elevated pressure.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental concept in physics and chemistry that connects the pressure, volume, temperature, and number of moles of a gas. The law is summarized in the formula: \[ PV = nRT \]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 universal gas constant
• \( T \) is the temperature of the gas in Kelvin

In this exercise, the Ideal Gas Law helps us understand how the pressure and volume change when air is compressed adiabatically. Adiabatic processes involve no heat exchange, but changes in internal energy due to work done on the system. It's vital to ensure temperatures are in Kelvin for calculations. This formula is foundational to solving many thermodynamic problems.

Kelvin Temperature Conversion

Temperature conversion to Kelvin is crucial when using the Ideal Gas Law. Unlike Celsius, Kelvin starts at absolute zero, making it the universal temperature scale for thermodynamic calculations. To convert Celsius to Kelvin, use the formula:\[ T(K) = T(^{ ext{°C}}) + 273.15 \]In this example, the initial temperature of the air is given as 20.0°C. Thus, it becomes:\[ T_1 = 20.0 + 273.15 = 293.15 ext{ K} \]Accurate temperature conversion ensures correctness in further calculations, specifically for the adiabatic equations used later.

P-V Diagram

A P-V diagram, short for Pressure-Volume diagram, visually represents the changes in pressure and volume during various thermodynamic processes. It is a valuable tool to understand how a system behaves under different conditions. In an adiabatic compression, like in a Ferrari engine, the path on the P-V diagram is represented by a steeper curve compared to an isothermal process, as there is no heat exchange. Key features of the P-V diagram for this process:

• The curve shows a decrease in volume with a corresponding increase in pressure.
• The end volume is 0.0900 times the initial volume, highlighting the compression.

Visualizing these changes helps comprehend the physical processes occurring during compression.

Adiabatic Compression

Adiabatic compression occurs when a gas is compressed without any heat exchange with the surroundings. This means the system is isolated in terms of energy transfer. During adiabatic compression, all the work done on the gas increases its internal energy, causing a rise in temperature.The formula that governs adiabatic processes is:\[ P_1 V_1^\gamma = P_2 V_2^\gamma \]and \[ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \]where \( \gamma \) (gamma) is the specific heat ratio. In this case, \( \gamma = 1.40 \).Through this principle, one can find the final temperature and pressure after adiabatic compression efficiently.

Final Temperature and Pressure Calculation

Calculation of the final temperature and pressure in an adiabatic process requires careful consideration of the involved equations. First, to find the final temperature (\( T_2 \)):\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1} \]In the given problem, \( T_1 = 293.15 ext{ K} \), \( V_2 = 0.0900 V_1 \), leading to:\[ T_2 = 293.15 \times \left( \frac{1}{0.0900} \right)^{0.40} \approx 894.64 ext{ K} \]Next, to determine the final pressure (\( P_2 \)):\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \]Substituting the known values results in:\[ P_2 = 1.00 \times \left( \frac{1}{0.0900} \right)^{1.40} \approx 29.76 ext{ atm} \]These equations help predict how air behaves under rapid compression, vital for engine performance assessments.