/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 \(\bullet\) A Jaguar XK8 convert... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 11

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

Short Answer

Expert verified
The final temperature is approximately 1733 K.

Step by step solution

01

Convert initial temperature to Kelvin

The initial temperature is given as \(27.0^{\circ} \mathrm{C}\). We convert this to Kelvin by adding 273.15.\[ T_1 = 27.0 + 273.15 = 300.15 \, \mathrm{K} \]
02

Use ideal gas law in initial state

The equation for the ideal gas law is \( PV = nRT \). We rearrange to solve for \( nRT_1 = P_1V_1 \), where\( P_1 = 1.01 \times 10^5 \, \mathrm{Pa} \), \( V_1 = 499 \, \mathrm{cm}^3 = 4.99 \times 10^{-4} \, \mathrm{m}^3 \), and \( T_1 = 300.15 \, \mathrm{K} \).We convert \( V_1 \) to cubic meters for consistency.Calculate \( nRT_1 = P_1V_1 \).\[ nRT_1 = (1.01 \times 10^5) \times (4.99 \times 10^{-4}) \]
03

Find final pressure in absolute terms

Gauge pressure is the pressure above atmospheric pressure. To find the absolute pressure, add atmospheric pressure to the gauge pressure.\[ P_2 = 2.72 \times 10^6 + 1.01 \times 10^5 = 2.821 \times 10^6 \, \mathrm{Pa} \]
04

Use ideal gas law in final state

Use the equation \( nRT_2 = P_2V_2 \), where \( P_2 = 2.821 \times 10^6 \, \mathrm{Pa} \) and \( V_2 = 46.2 \, \mathrm{cm}^3 = 4.62 \times 10^{-5} \, \mathrm{m}^3 \).Calculate:\[ nRT_2 = (2.821 \times 10^6) \times (4.62 \times 10^{-5}) \]
05

Equate and solve for final temperature

Since \( nR \) is constant throughout the process, equate the two expressions from the ideal gas law.\[ nRT_1 = nRT_2 \]\[ P_1V_1 = P_2V_2 \]Solve for \( T_2 \):\[ T_2 = \frac{P_2V_2}{P_1V_1} \times T_1 \]Substitute the values:\[ T_2 = \frac{2.821 \times 10^6 \times 4.62 \times 10^{-5}}{1.01 \times 10^5 \times 4.99 \times 10^{-4}} \times 300.15 \approx 1732.59 \text{ K} \]
06

Conclusion

The final temperature of the air after compression is approximately 1732.59 K.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Compression Stroke
In an internal combustion engine, the compression stroke is one of the critical steps where the piston moves in the cylinder to compress the air-fuel mixture. This compression increases the air pressure significantly, which is essential for efficient combustion when the fuel is ignited.
  • During the compression stroke, the volume of air inside the cylinder is drastically reduced.
  • This process results in a temperature rise due to the work done on the gas, following the principles of thermodynamics.
Understanding the compression stroke is fundamental for appreciating how engines work and why the temperature and pressure of the air change so dramatically.
Temperature Conversion
Temperature conversion often involves changing temperature values between Celsius, Fahrenheit, and Kelvin. In scientific calculations, it's crucial to use the Kelvin scale because it starts at absolute zero, where all thermal motion ceases.
  • The conversion from Celsius to Kelvin is simple: add 273.15 to the Celsius temperature.
  • For example, an initial temperature of 27.0掳C converts to 300.15 K (i.e., 27.0 + 273.15).
Using Kelvin in calculations ensures precision since the Kelvin scale is directly proportional to the kinetic energy of particles.
Gauge Pressure
Gauge pressure measures the pressure in a system over and above the atmospheric pressure. It's distinct from absolute pressure, which includes atmospheric pressure in its measurement.
  • Gauge pressure is often provided in tasks where the pressure into the system needs to be determined independently of atmospheric conditions.
  • This is why, to find the absolute pressure, we need to add the gauge pressure to the atmospheric pressure (approximately 101,325 Pa at sea level).
Understanding gauge pressure is crucial when assessing mechanical systems' pressures, like in car engines.
Cylinder Volume
Cylinder volume refers to the space within an engine cylinder available for the air-fuel mixture. The volume changes as the piston moves, influencing the pressure and temperature within the cylinder.
  • The initial volume is larger as the piston is at the bottom of its stroke, and it compresses the air to a much smaller final volume.
  • This change in volume is significant because, combined with the properties of gases, it leads to changes in pressure and temperature according to the ideal gas law.
In practical applications, knowing the volume and its changes helps us calculate the other gas properties when using the ideal gas law.

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Most popular questions from this chapter

\(\bullet$$\bullet\) You blow up a spherical balloon to a diameter of 50.0 \(\mathrm{cm}\) until the absolute pressure inside is 1.25 atm and the temperature is \(22.0^{\circ} \mathrm{C}\) . Assume that all the gas is \(\mathrm{N}_{2},\) of molar mass 28.0 \(\mathrm{g} / \mathrm{mol} .\) (a) Find the mass of a single \(\mathrm{N}_{2}\) molecule. (b) How much translational kinetic energy does an average \(\mathrm{N}_{2}\) molecule have? (c) How many \(\mathrm{N}_{2}\) molecules are in this balloon? (d) What is the total translational kinetic energy of all the molecules in the balloon?

\(\bullet\) Five moles of an ideal monatomic gas with an initial temperature of \(127^{\circ} \mathrm{C}\) expand and, in the process, absorb 1200 \(\mathrm{J}\) of heat and do 2100 \(\mathrm{J}\) of work. What is the final temperature of the gas?

\(\bullet$$\bullet\) At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at \(20.0^{\circ} \mathrm{C} ?\) (Hint: The periodic table in Appendix \(C\) shows the molar mass (in \(\mathrm{g} / \mathrm{mol} )\) of each element under the chemical symbol for that element. The molar mass of \(\mathrm{H}_{2}\) is twice the molar mass of hydrogen atoms, and similarly for \(\mathrm{N}_{2}.)\)

\(\bullet\) Perfectly rigid containers each hold \(n\) moles of ideal gas, one being hydrogen \(\left(\mathrm{H}_{2}\right)\) and other being neon \((\mathrm{Ne}) .\) If it takes 100 \(\mathrm{J}\) of heat to increase the temperature of the hydrogen by \(2.50 \mathrm{C}^{\circ},\) by how many degrees will the same amount of heat raise the temperature of the neon?

\(\bullet$$\bullet\) In a cylinder, 4.00 \(\mathrm{mol}\) of helium initially at \(1.00 \times 10^{6} \mathrm{Pa}\) and 300 \(\mathrm{K}\) expands until its volume doubles. Compute the work done by the gas if the expansion is (a) isobaric, (b) adiabatic. (c) Show each process on a \(p V\) diagram. In which case is the magnitude of the work done by the gas the greatest? (d) In which case is the magnitude of the heat transfer greatest? (e) In which case is the magnitude of the change in internal energy greatest?
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