Each of the four cylinders of a new type of combustion engine has a displacement of \(3.60 \mathrm{~L}\). (The volume of the cylinder expands \(3.60 \mathrm{~L}\) each time the fuel is ignited.) (a) If each piston in the four cylinders is displaced against a pressure of \(1.80 \mathrm{kbar}\) and each cylinder is ignited once per second, how much work can the engine do in \(1.00\) minute? (b) Is the work positive or negative with respect to the engine and its contents?

Short Answer

Expert verified
The engine can do 155520 joules of work in one minute, and the work is positive with respect to the engine.

Step by step solution

01

Convert the pressure to a common unit

First, we need to convert the pressure from kilobars to pascals. Remember that 1 bar is equal to 100000 pascals (Pa). So we have: Pressure in pascals = 1.80 kbar * 100000 Pa/kbar = 180000 Pa.
02

Convert the displacement to cubic meters

Since work is generally computed in joules in the International System of Units (SI), we should convert the volume of displacement from liters to cubic meters: Volume in cubic meters = 3.60 L * (1 m^3 / 1000 L) = 0.0036 m^3.
03

Calculate the work done for one cylinder per ignition

Work is calculated using the formula Work = Pressure * Volume. On one ignition of one cylinder: Work = 180000 Pa * 0.0036 m^3 = 648 J (joules).
04

Calculate the total work for all cylinders over a minute

Each cylinder does 648 J per ignition, and there are four cylinders ignited once per second. Thus, in one second, the engine does 648 J/cylinder * 4 cylinders = 2592 J. In one minute (60 seconds), the total work done is 2592 J * 60 s = 155520 J.
05

Determine the sign of the work

Considering the definition of work from the point of view of the engine, if the system (the engine) does work on the surroundings, then the work is positive. In this case, the engine is doing work by expanding against the external pressure, therefore, the work is positive.

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Key Concepts

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

Pressure Unit Conversion
Understanding pressure unit conversion is vital for solving problems in various scientific disciplines, including physics and engineering. Pressure, which is the force exerted per unit area, can be expressed in multiple units such as pascals (Pa), bars, atmospheres (atm), and pounds per square inch (psi). In the context of combustion engines, pressure often needs to be converted to pascals because the International System of Units (SI) uses pascals as the standard unit for pressure measurements.

For instance, the conversion from kilobars (kbar) to pascals (Pa) is a straightforward process. Since 1 bar is equivalent to 100,000 Pa, to convert kilobars to pascals, you multiply the value in kilobars by 100,000. This step is crucial for calculating the work done by the engine in a combustion process, as seen in our exercise where a pressure of 1.80 kbar converts to 180,000 Pa.
Volume to Cubic Meters Conversion
When working with engine displacement and calculating work, volume often needs to be converted into cubic meters. Volume can be measured in liters, gallons, cubic inches, among others, but for work calculations in the SI system, cubic meters (m³) is the preferred unit of volume.

The conversion from liters to cubic meters is quite simple: since 1 cubic meter is equivalent to 1,000 liters, you divide the volume value in liters by 1,000 to obtain the volume in cubic meters. In our exercise example, the engine has a displacement of 3.60 L which is equivalent to 0.0036 m³. This conversion is vital to ensure that we use consistent SI units to calculate work accurately.
Work-Energy Principle in Physics
The work-energy principle in physics states that the work done on an object is equal to the change in its kinetic energy. Work is defined as the force applied over a distance and, in the case of a constant force, is the product of the force and the distance moved in the direction of the force.

In thermodynamics, work is calculated as the product of pressure and the change in volume of the system. In our exercise, the work done by each cylinder during one ignition is computed using the formula: Work = Pressure × Volume. Thus, when an engine's piston is displaced against a certain pressure over a certain volume, it does work on the gas within the cylinder, which is essential for the engine's operation.
Thermodynamics in Combustion Engines
Thermodynamics plays a fundamental role in understanding combustion engines. These engines convert the chemical energy stored in fuel into mechanical work through a series of thermodynamic processes. The key concept in thermodynamics that applies to engines is the first law, which is a statement of the conservation of energy.

In the context of a combustion engine, when fuel is ignited in the cylinder, it expands against the piston, and as a result, performs work. The first law of thermodynamics, when applied to this scenario, helps us calculate the total work output of the engine. As the engine operates, each cycle within the cylinder involves compressing the air-fuel mixture, igniting it, and allowing it to expand and do work. The exercise example illustrates these principles as it calculates the positive work produced by the engine based on the pressure exerted and the volume expansion in the cylinders.

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Most popular questions from this chapter

A student rides a bicycle to class every day, a 10.-mile round trip that takes 30 . minutes in each direction. The student burns \(1250 \mathrm{~kJ} \cdot \mathrm{h}^{-1}\) cycling. The same round trip in an automobile would require \(0.40\) gallons of gasoline. Assume that the student goes to class 150 days per year and that the enthalpy of combustion of gasoline can be approximated by that of octane, which has a density of \(0.702 \mathrm{~g}^{\cdot \mathrm{cm}^{-3}}(3.785 \mathrm{~L}=1.000 \mathrm{gal})\). What is the yearly energy requirement of this journey by (a) bicycle and (b) automobile?

(a) Describe three ways in which you could increase the internal energy of an open system. (b) Which of these methods could you use to increase the internal energy of a closed system? (c) Which, if any, of these methods could you use to increase the internal energy of an isolated system?

Rank the following compounds in order of increasing enthalpy of vaporization: \(\mathrm{CH}_{4}, \mathrm{H}_{2} \mathrm{O}, \mathrm{N}_{2}, \mathrm{NaCl}, \mathrm{C}_{6} \mathrm{H}_{6}\), and \(\mathrm{H}_{2}\). Do not look up the enthalpy of vaporization. Explain your ordering.

Determine the reaction enthalpy for the hydrogenation of ethyne to ethane, \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\), from the following data: \(\Delta H_{c}{ }^{\circ}\left(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{~g}\right)=-1300 . \mathrm{kJ} \cdot \mathrm{mol}^{-1}, \Delta H_{\mathrm{c}}{ }^{\circ}\left(\mathrm{C}_{2} \mathrm{H}_{6}, \mathrm{~g}\right)\) \(=-1560 . \mathrm{kJ} \cdot \mathrm{mol}^{-1}, \Delta H_{\mathrm{c}}{ }^{\circ}\left(\mathrm{H}_{2}, \mathrm{~g}\right)=-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\).

Barium metal is produced by the reaction of aluminum metal with barium oxide. From the standard reaction enthalpies $$ \begin{array}{ll} 2 \mathrm{Ba}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{BaO}(\mathrm{s}) & \Delta H^{\circ}=-1107 \mathrm{~kJ} \\ 2 \mathrm{Al}(\mathrm{s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s}) & \Delta H^{\circ}=-1676 \mathrm{~kJ} \end{array} $$ calculate the reaction enthalpy for the production of metallic barium in the reaction: $$ 3 \mathrm{BaO}(\mathrm{s})+2 \mathrm{Al}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{Ba}(\mathrm{s}) $$

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