Chapter 18: Problem 7
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
Step by step solution
Convert Initial Conditions
Determine Initial and Final Pressures
Note Volume Compression
Apply the Ideal Gas Law
Solve for Final Temperature
Convert to Celsius if Needed
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Compression Stroke
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
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
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
\[\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\]
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.