Question

A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 499 cm\(^3\) of air at atmospheric pressure (1.01 \(\times\) 10\({^5}\) Pa) and a temperature of 27.0\(^\circ\)C. At the end of the stroke, the air has been compressed to a volume of 46.2 cm\(^3\) and the gauge pressure has increased to 2.72 \(\times\) 10\({^6}\) Pa. Compute the final temperature.

Short answer

The final temperature is approximately 751.7 K or 478.55°C.

Step-by-step solution

Convert Initial Conditions

First, convert the initial temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is: \( T(K) = T(\degree C) + 273.15 \). Thus, \( 27.0\degree C = 27.0 + 273.15 = 300.15 \) K.

Determine Initial and Final Pressures

The initial pressure is atmospheric, given as \( P_1 = 1.01 \times 10^5 \) Pa. The problem states the final pressure as gauge pressure, \( P_{gauge} = 2.72 \times 10^6 \) Pa. Add atmospheric pressure to find the final absolute pressure: \( P_2 = P_{gauge} + 1.01 \times 10^5 = 2.82 \times 10^6 \) Pa.

Note Volume Compression

The initial volume is \( V_1 = 499 \, \text{cm}^3 \) and the final volume is \( V_2 = 46.2 \, \text{cm}^3 \). We will use these values in subsequent calculations.

Apply the Ideal Gas Law

The ideal gas law states \( PV = nRT \). According to transformations where mass and gas constant are constant, relation can be expressed as \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) for initial and final states. Substitute known values: \[ \frac{1.01 \times 10^5 \times 499}{300.15} = \frac{2.82 \times 10^6 \times 46.2}{T_2} \]

Solve for Final Temperature

Rearrange the equation to solve for \( T_2 \): \[ T_2 = \frac{2.82 \times 10^6 \times 46.2 \times 300.15}{1.01 \times 10^5 \times 499} \] Calculate \( T_2 \) to find the final temperature. \( T_2 \approx 751.7 \) K.

Convert to Celsius if Needed

If the question requires the final answer in Celsius, convert back using \( T(\degree C) = T(K) - 273.15 \): \( 751.7 - 273.15 = 478.55 \degree C \).

Key concepts

Compression Stroke

The compression stroke is a crucial part of an engine's operation. In the Jaguar XK8 convertible with an eight-cylinder engine, the compression stroke plays a vital role in preparing the air-fuel mixture for combustion. During this stroke, a piston moves up within a cylinder, compressing the air inside it. This process increases the air's pressure and temperature significantly.
A smaller, compressed volume leads to these changes as the same amount of air is squeezed into a tighter space. The compression ratio, which is the comparison of the initial volume of the air to the final compressed volume, is a key factor that affects engine performance. In this exercise, the initial volume was 499 cm\(^3\) and compressed to 46.2 cm\(^3\).

• The compression stroke enables the engine to generate more power by igniting a denser air-fuel mixture.
• This stroke helps increase the thermal efficiency of the engine.
• It's a fundamental step in the four-stroke cycle, which also includes intake, power, and exhaust strokes.

Pressure Conversion

When dealing with engines, converting between different types of pressure is often necessary. In the exercise, the initial condition was atmospheric pressure, denoted as 1.01 \(\times\) 10\(^5\) Pa. Pressure in this problem is measured in Pascals (Pa), which is the SI unit for pressure.
The "gauge pressure" provided during the exercise signifies the pressure relative to the atmospheric pressure. Therefore, to obtain the actual or "absolute pressure," the atmospheric pressure must be added to the gauge pressure. In this case:
\[ P_2 = P_{gauge} + P_{atmospheric} = 2.72 \times 10^6 + 1.01 \times 10^5 = 2.82 \times 10^6\] Pa.

• Gauge pressure excludes the atmospheric pressure, measuring only the pressure above atmospheric conditions.
• Absolute pressure is critical for accurate calculations in gas law problems, especially when determining temperature changes.
• Not converting correctly can lead to incorrect results in physics problems dealing with gas laws.

Temperature Conversion

Temperature conversion is necessary when solving problems using the Ideal Gas Law, as it requires temperature in Kelvin. The Kelvin scale is an absolute temperature scale, starting at absolute zero, and is widely used in scientific calculations.
In this exercise, the initial temperature was given in Celsius (27.0\(^\circ\)C). To convert it to Kelvin:
\[ T(K) = T(\degree C) + 273.15 \T(K) = 27.0 + 273.15 = 300.15 \] K.
The Kelvin scale ensures no negative temperatures, providing a consistent base for gas law calculations.

• Kelvin is necessary for mathematical consistency when using physical laws like the Ideal Gas Law.
• For final results in Celsius, conversion back from Kelvin is used: \( T(\degree C) = T(K) - 273.15 \).
• Neglecting temperature conversion can result in substantial errors in calculations.

Gas Laws

The Ideal Gas Law is a fundamental equation used to relate the pressure, volume, and temperature of an ideal gas. It is expressed as \( PV = nRT \). For ideal gases, this equation allows us to make predictions about one property when the others are known. Here, with a constant amount of gas and using the assumption that the gas behaves ideally, the relationship can simplify during state transformation:
\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]
In this problem, this condensed form was specifically used to compute the unknown final temperature \( T_2 \) of the compressed air in the engine cylinder:

• This form considers the conservation of gas properties before and after compression.
• It helps calculate missing information by keeping the mass and the specific gas constant \( R \) unchanged.
• Assuming ideal behavior means ignoring factors like gas interactions which tend to be negligible in many engineering situations at high temperature and low pressure.