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In a diesel engine, the piston compresses air at \(305 \mathrm{~K}\) to a volume that is onesixteenth of the original volume and a pressure that is 48.5 times the original pressure. What is the temperature of the air after the compression?

Short Answer

Expert verified
The final temperature of the air is approximately 924.69 K.

Step by step solution

01

Understand the Problem

We know that in a diesel engine, the piston compresses air, increasing its temperature and pressure according to the Ideal Gas Law. Before compression, air is at 305 K, and after compression, the volume is reduced to \(\frac{1}{16}\)th of the original volume and the pressure increases to 48.5 times the original pressure. We need to find the final temperature.
02

Write the Ideal Gas Law Equations

The Ideal Gas Law can be expressed as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the amount of substance, \(R\) is the gas constant, and \(T\) is temperature. For the initial state (1) and final state (2), we have \(P_1V_1 = nRT_1\) and \(P_2V_2 = nRT_2\).
03

Set Up the Relationship Between the Two States

Since \(nR\) is constant and doesn't change, we can use the equation \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\). Substitute \(P_2 = 48.5P_1\) and \(V_2 = \frac{1}{16}V_1\). The equation then becomes \(\frac{P_1V_1}{T_1} = \frac{48.5P_1 \times \frac{1}{16}V_1}{T_2}\).
04

Simplify the Equation

Cancel out the common terms. After substituting \(T_1 = 305 \, \mathrm{K}\), we have \(\frac{1}{305} = \frac{48.5}{16T_2}\), as \(P_1V_1\) cancels on both sides.
05

Solve for the Final Temperature \(T_2\)

Rearrange \(\frac{1}{305} = \frac{48.5}{16T_2}\) to find \(T_2\). Multiply both sides by \(16T_2\) to get \(16T_2 = 305 \times 48.5\). Simplify and solve: \(T_2 = \frac{305 \times 48.5}{16}\). Calculating it gives \(T_2 \approx 924.69\, \mathrm{K}\).
06

Conclusion

Based on the calculations, the final temperature of the air after compression in the diesel engine is approximately 924.69 K.

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Key Concepts

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

Pressure and Volume Relationship
In the realm of thermodynamics, the relationship between pressure and volume in a gas is crucial for understanding how systems like diesel engines work. According to the Ideal Gas Law, pressure
  • is directly proportional to the product of volume and temperature
  • when the amount of gas remains constant
This is expressed by the equation: \[ PV = nRT \].
Here, \(P\) represents pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature.

When volume decreases, pressure increases if the temperature remains constant. This behavior aligns with Boyle's Law, which states that at a constant temperature, the pressure of a gas varies inversely with its volume. In simpler terms, when you compress a gas (decrease its volume), the pressure goes up, provided that the temperature doesn't change.

In the context of the diesel engine exercise, the pressure increase is quite dramatic—48.5 times the initial pressure, caused by the significant drop in volume when the piston compresses the air to just one-sixteenth of its original size. Understanding this concept helps to comprehend how the mechanical action of a Diesel engine leads to drastic changes in the properties of the air inside the cylinder.
Diesel Engine Thermodynamics
Diesel engines operate based on several key thermodynamic processes that revolve around the behavior of gases. The principle utilized in this scenario is the compression of air within the cylinder. In a diesel engine:
  • Air is compressed until it reaches a high pressure and temperature.
  • The fuel is then injected and ignites due to the high temperature of the compressed air, combusting spontaneously.
The compression stroke is the first major act, where the volume inside the cylinder is significantly reduced. During this compression:
  • No heat exchange with the surroundings occurs, making it an adiabatic process.
  • The intense compression results in an internal temperature rise of the gas.

After the compression, as depicted in our original problem, the compressed air's extreme pressure and concurrent temperature boost set the perfect conditions for fuel ignition. Diesel engines excel here, because the air-fuel mixture reaches a higher temperature without requiring any spark, making them more efficient overall.

Therefore, understanding the thermodynamics of diesel engines not only helps with solving temperature problems but also highlights the efficiency and robustness of these engines compared to typical gasoline engines.
Temperature Calculation
Calculating the temperature change during the compression process in a diesel engine is key to understanding engine efficiency. Using the Ideal Gas Law and the relationships between the thermodynamic variables:
  • The equation \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \) is essential.
  • Given values are substituted into this equation to find the unknown variables.

In the example provided, we know:
  • Initial temperature, \( T_1 = 305 \text{ K} \).
  • Final pressure is 48.5 times the initial pressure.
  • Final volume is one-sixteenth of the initial volume.
By substituting these into the relationship and simplifying, we calculate the final temperature of the compressed air using:\[ T_2 = \frac{P_2V_2}{P_1V_1} \cdot T_1 \].
This leads to determining: \[ T_2 = \frac{305 \times 48.5}{16} \approx 924.69 \text{ K} \].
This calculation helps illustrate the dramatic increase in temperature as air gets compressed in a diesel engine, providing insight into the key factors that make diesel engines efficient in converting fuel into mechanical energy.

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Most popular questions from this chapter

A spherical balloon is made from a material whose mass is \(3.00 \mathrm{~kg} .\) The thickness of the material is negligible compared to the 1.50 -m radius of the balloon. The balloon is filled with helium (He) at a temperature of \(305 \mathrm{~K}\) and just floats in air, neither rising nor falling. The density of the surrounding air is \(1.19 \mathrm{~kg} / \mathrm{m}^{3}\). Find the absolute pressure of the helium gas.

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Insects do not have lungs as we do, nor do they breathe through their mouths. Instead, they have a system of tiny tubes, called tracheae, through which oxygen diffuses into their bodies. The tracheae begin at the surface of the insect's body and penetrate into the interior. Suppose that a trachea is \(1.9 \mathrm{~mm}\) long with a cross-sectional area of \(2.1 \times 10^{-9} \mathrm{~m}^{2}\). The concentration of oxygen in the air outside the insect is \(0.28 \mathrm{~kg} / \mathrm{m}^{3}\) and the diffusion constant is \(1.1 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). If the mass per second of oxygen diffusing through a trachea is \(1.7 \times 10^{-12} \mathrm{~kg} / \mathrm{s}\), find the oxygen concentration at the interior end of the tube.

A frictionless gas-filled cylinder is fitted with a movable piston, as the drawing shows. The block resting on the top of the piston determines the constant pressure that the gas has. The height \(h\) is \(0.120 \mathrm{~m}\) when the temperature is \(273 \mathrm{~K}\) and increases as the temperature increases. What is the value of \(h\) when the temperature reaches \(318 \mathrm{~K}\) ?

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