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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 \(\mathrm{cm}^{3}\) 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 \(2.72 \times 10^{5}\) Pa. Compute the final temperature.

Short Answer

Expert verified
The final temperature is approximately 906.1 K.

Step by step solution

01

Convert Temperature to Kelvin

The initial temperature is given as 27.0掳C. We first convert this temperature to Kelvin using the formula: \[ T(K) = T(掳C) + 273.15 \]Thus, the initial temperature is \( T_i = 27.0 + 273.15 = 300.15 \) K.
02

Identify Known Values

List down all known values:- Initial volume \( V_i = 499 \) cm鲁- Final volume \( V_f = 46.2 \) cm鲁- Initial pressure \( P_i = 1.01 \times 10^5 \) Pa- Gauge pressure \( P_g = 2.72 \times 10^5 \) Pa- Atmospheric pressure contributes to the total, making final pressure \( P_f = P_g + 1.01 \times 10^5 = 3.73 \times 10^5 \) Pa.
03

Use Ideal Gas Law (Relation between states)

According to the ideal gas law for constant number of moles:\[ \frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f} \]Substitute all known values to find \( T_f \):\[ T_f = \frac{P_f V_f T_i}{P_i V_i} \]
04

Calculate Final Temperature

Plug the values into the formula:\[ T_f = \frac{3.73 \times 10^5 \, \mathrm{Pa} \times 46.2 \, \mathrm{cm}^3 \times 300.15 \, \mathrm{K}}{1.01 \times 10^5 \, \mathrm{Pa} \times 499 \, \mathrm{cm}^3} \]Calculate to find \( T_f \approx 906.1 \, \mathrm{K} \).

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Key Concepts

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

Compression Stroke
When an engine operates, it goes through several cycles, and one of these crucial phases is the compression stroke. This step involves the compression of the air-fuel mixture in an engine's cylinder. In the case of the Jaguar XK8 convertible, during the compression stroke, the volume of air in the cylinder reduces significantly, going from 499 cm鲁 to just 46.2 cm鲁.
  • The primary purpose of this stroke is to increase the pressure and temperature of the air-fuel mixture, making it more efficient for subsequent combustion.
  • As the piston moves upward, it compresses the air, decreasing its volume and increasing its pressure.
  • The compression of this air ensures that when the fuel is ignited, it combusts more explosively, maximizing the power output.
This initial compression is critical for an engine's efficiency, effectiveness, and overall performance, as it dictates how much power can be extracted from the combustion process.
Final Temperature Calculation
To find the final temperature in the cylinder after compression, we apply the ideal gas law, which provides a relation between pressure, volume, and temperature. This relationship is crucial as it helps us understand how these variables interact within a closed system like a car engine's cylinder.

Converting Temperature

Initially, the temperature needs to be in Kelvin, which is suitable for thermodynamic calculations. The given initial temperature in Celsius is converted to Kelvin by adding 273.15, resulting in 300.15 K.

Using the Ideal Gas Law

To compute the final temperature, we use the formula that relates initial and final conditions: \[\frac{P_i V_i}{T_i} = \frac{P_f V_f}{T_f}\]
By arranging the formula to solve for the final temperature \( T_f \), and substituting the known values into this equation, we get:\[T_f = \frac{3.73 \times 10^5 \, \mathrm{Pa} \times 46.2 \, \mathrm{cm}^3 \times 300.15 \, \mathrm{K}}{1.01 \times 10^5 \, \mathrm{Pa} \times 499 \, \mathrm{cm}^3}\]
Completing the calculation provides the final temperature of approximately 906.1 K.
Pressure and Volume Relationship
In an engine like the Jaguar XK8's, the relationship between pressure and volume is defined by physical principles stated in the ideal gas law. This relationship helps understand how variations in volume and pressure directly influence temperature during the compression stroke.
  • According to the law, when the volume of a gas is decreased, as happens during the compression stroke, its pressure increases if the temperature and the amount of gas remain constant.
  • This is expressed by Boyle鈥檚 Law, a part of the ideal gas equation, stating \( P_1 V_1 = P_2 V_2 \) for a constant temperature.
  • However, in a practical scenario inside an engine, both pressure and temperature increase, following the combined form of the ideal gas law: \( PV = nRT \).
Understanding this relationship allows automotive engineers and physicists to design engines that effectively harness the energy produced during the explosion of fuel, combining mechanical elegance with thermodynamic efficiency.

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Most popular questions from this chapter

(a) A deuteron, \(^{2}_{1} \mathrm{H},\) is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a nuclear fusion reactor must be heated to about 300 million \(\mathrm{K}\). What is the ms speed of the deuterons? Is this a significant fraction of the speed of light \(\left(c=3.0 \times 10^{8} \mathrm{m} / \mathrm{s}\right) ?\) (b) What would the temperature of the plasma be if the deuterons had an ms speed equal to 0.10\(c ?\)

The speed of propagation of a sound wave in air at \(27^{\circ} \mathrm{C}\) is about 350 \(\mathrm{m} / \mathrm{s}\) . Calculate, for comparison, (a) \(v_{\mathrm{ms}}\) for nitrogen molecules and (b) the rms value of \(v_{x}\) at this temperature. The molar mass of nitrogen \(\left(\mathrm{N}_{2}\right)\) is \(28.0 \mathrm{g} / \mathrm{mol} .\)

During a test dive in \(1939,\) prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 \(\mathrm{m}\) . The temperature at the surface was \(27.0^{\circ} \mathrm{C},\) and at the bottom it was \(7.0^{\circ} \mathrm{C}\) . The density of seawater is 1030 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder 2.30 \(\mathrm{m}\) high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: You may ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell. \((b)\) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?

A \(3.00-\mathrm{L}\) tank contains air at 3.00 \(\mathrm{atm}\) and \(20.0^{\circ} \mathrm{C} .\) The tank is sealed and cooled until the pressure is 1.00 atm. (a) What is the temperature then in degrees Celsius? Assume the volume of the tank is constant. (b) If the temperature is kept at the value found in part (a) and the gas is compressed, what is the vohme when the pressure again becomes 3.00 \(\mathrm{atm} ?\)

A welder using a tank of volume 0.0750 \(\mathrm{m}^{3}\) fills it with oxygen \((\text { molar mass } 32.0 \mathrm{g} / \mathrm{mol})\) at a gauge pressure of \(3.00 \times 10^{5} \mathrm{Pa}\) and tenperature of \(37.0^{\circ} \mathrm{C}\) . The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is \(22.0^{\circ} \mathrm{C}\) , the gauge pressure of the oxygen in the tank is \(1.80 \times 10^{5}\) Pa. Find (a) the initial mass of oxygen and (b) the mass of oxygen that has leaked out.
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