A diesel engine works at a high compression ratio to compress air until it reaches a temperature high enough to ignite the diesel fuel. Suppose the compression ratio (ratio of volumes) of a specific diesel engine is 20 to \(1 .\) If air enters a cylinder at 1.00 atm and is compressed adiabatically, the compressed air reaches a pressure of 66.0 atm. Assuming that the air enters the engine at room temperature \(\left(25.0^{\circ} \mathrm{C}\right)\) and that the air can be treated as an ideal gas, find the temperature of the compressed air.

Short Answer

Expert verified
Answer: The final temperature of the compressed air is approximately 980.15 K.

Step by step solution

01

Write down the given information

We are given the following information: Initial pressure, \(P_1 = 1.00 \, \text{atm}\) Final pressure, \(P_2 = 66.0 \, \text{atm}\) Initial temperature, \(T_1 = 25.0 \, \text{°C} = 298.15 \, \text{K}\) Compression ratio, \(V_1 / V_2 = 20\) Since the compression is adiabatic, we can use the adiabatic equation for ideal gases to relate pressures, volumes, and temperatures.
02

Use the adiabatic equation

The adiabatic equation for an ideal gas can be expressed as: \(P_1^{\gamma - 1} V_1^{\gamma} = P_2^{\gamma - 1} V_2^{\gamma}\) where \(\gamma\) is the heat capacity ratio, which for diatomic gases like air (approximately) is \(1.4\). Since we know the compression ratio \(V_1 / V_2 = 20\), we can rewrite the equation above as: \(P_1^{\gamma - 1} (20 V_2)^{\gamma} = P_2^{\gamma - 1} V_2^{\gamma}\) Now, divide both sides by \((20 V_2)^{\gamma}\): \(\left(\frac{P_1}{P_2}\right)^{\gamma - 1} = \frac{1}{20^{\gamma}}\)
03

Solve for the final temperature

Next, we will use the ideal gas relation to relate temperatures to pressures and volumes: \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) We can rewrite the above equation using the compression ratio: \(\frac{P_1 (20 V_2)}{T_1} = \frac{P_2 V_2}{T_2}\) Cancelling out \(V_2\) and solving for \(T_2\) gives: \(T_2 = \frac{P_2 T_1}{20 P_1}\) Substituting the values gives: \(T_2 = \frac{66.0 \, \text{atm} \cdot 298.15 \, \text{K}}{20 \cdot 1.00 \, \text{atm}} = 980.15 \, \text{K}\)
04

Express the final answer

The temperature of the compressed air inside the diesel engine is approximately \(980.15 \, \text{K}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Adiabatic Equation
In thermodynamics, the adiabatic process is characterized by the absence of heat transfer between the system and its surroundings. When considering such a process in the context of diesel engines, understanding the adiabatic equation becomes crucial.

This equation, which applies to ideal gases under adiabatic compression or expansion, can be expressed as:
\[P_1 V_1^\gamma = P_2 V_2^\gamma\]
where:
  • \(P_1\) and \(P_2\) are the initial and final pressures, respectively,
  • \(V_1\) and \(V_2\) are the initial and final volumes,
  • and \(\gamma\) is the heat capacity ratio (discussed in a later section).
In a diesel engine, the air undergoes adiabatic compression, increasing its pressure and temperature until it's hot enough to ignite diesel fuel. The adiabatic equation allows us to relate the change in volume and pressure to the change in temperature, which is vital for determining the conditions necessary for fuel combustion.
Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry and physics that relates the pressure, volume, and temperature of an ideal gas. It's typically stated as:
\[PV = nRT\]
where:
  • \(P\) is the pressure,
  • \(V\) is the volume,
  • \(n\) is the number of moles,
  • \(R\) is the universal gas constant,
  • and \(T\) is the temperature in Kelvin.
In the context of a diesel engine, the ideal gas law is used after applying the adiabatic equation to relate the pressures and volumes at two different states. As the ideal gas law assumes no interaction between gas molecules, it greatly simplifies calculations for the temperature and pressure conditions within the engine's cylinders.
Heat Capacity Ratio
The heat capacity ratio, often denoted as \(\gamma\) (gamma), is a dimensionless quantity and an essential parameter in adiabatic processes. It's defined as the ratio of the specific heat capacity at constant pressure \(C_p\) to the specific heat capacity at constant volume \(C_v\):
\[\gamma = \frac{C_p}{C_v}\]
This ratio affects how gases behave when compressed or expanded without heat exchange. For diatomic gases, such as the nitrogen and oxygen primarily found in air, \(\gamma\) is approximately 1.4.

The value of \(\gamma\) is crucial when applying the adiabatic equation, as it directly influences the final temperature and pressure of the compressed gas in the diesel engine. Knowing \(\gamma\) helps predict how efficiently the engine compresses air and, consequently, its ability to reach the temperature required for fuel ignition.
Compression Ratio
The compression ratio is a key concept in the design and function of diesel engines. It's defined as the ratio of the maximum volume in the cylinder when the piston is at the bottom of its stroke (the bottom dead center, BDC) to the minimum volume when the piston is at the top of its stroke (the top dead center, TDC):
\[\text{Compression ratio} = \frac{V_{\text{BDC}}}{V_{\text{TDC}}}\]
In diesel engines, a high compression ratio is used to compress the air so significantly that the temperature rises to a level that can ignite diesel fuel without the need for a spark plug. As demonstrated in the exercise, a compression ratio of 20 to 1 means the volume of air is reduced by a factor of 20 during the compression stroke. A higher compression ratio typically results in higher efficiency and more power output from the engine, but it also requires stronger engine components to withstand the greater pressures produced.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Hot air is less dense than cold air and therefore experiences a net buoyant force and rises. Since hot air rises, the higher the elevation, the warmer the air should be. Therefore, the top of Mount Everest should be very warm. Explain why Mount Everest is colder than Death Valley.

Show that the adiabatic bulk modulus, defined as \(B=-V(d P / d V),\) for an ideal gas is equal to \(\gamma P\).

Consider a box filled with an ideal gas. The box undergoes a sudden free expansion from \(V_{1}\) to \(V_{2}\). Which of the following correctly describes this process? a) Work done by the gas during the expansion is equal to \(n R T \ln \left(V_{2} / V_{1}\right)\) b) Heat is added to the box. c) Final temperature equals initial temperature times \(\left(V_{2} / V_{1}\right)\). d) The internal energy of the gas remains constant.

Air in a diesel engine cylinder is quickly compressed from an initial temperature of \(20.0^{\circ} \mathrm{C}\), an initial pressure of \(1.00 \mathrm{~atm}\), and an initial volume of \(600 . \mathrm{cm}^{3}\) to a final volume of \(45.0 \mathrm{~cm}^{3}\). Assuming the air to be an ideal diatomic gas, find the final temperature and pressure.

6.00 liters of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergo the following processes: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the pressure, volume, and temperature of the gas in states 2 and \(3 .\) How many moles of the gas are there?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free