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A heat engine consists of a heat source that causes a monatomic gas to expand, pushing against a piston, thereby doing work. The gas begins at a pressure of \(300 . \mathrm{kPa}\), a volume of \(150 . \mathrm{cm}^{3}\), and room temperature, \(20.0^{\circ} \mathrm{C}\). On reaching a volume of \(450 . \mathrm{cm}^{3}\), the piston is locked in place, and the heat source is removed. At this point, the gas cools back to room temperature. Finally, the piston is unlocked and used to isothermally compress the gas back to its initial state. a) Sketch the cycle on a \(p V\) -diagram. b) Determine the work done on the gas and the heat flow out of the gas in each part of the cycle. c) Using the results of part (b), determine the efficiency of the engine.

Short Answer

Expert verified
Question: Sketch a pV-cycle for a heat engine that operates in a cycle with three parts: expansion (constant pressure), cooling (constant volume), and compression (constant temperature). Calculate the work done and heat flow in each part of the cycle and determine the efficiency of the engine. Answer: The pV-cycle consists of an isobaric expansion, an isochoric cooling, and an isothermal compression. The work done during expansion is 300 kPa * 300 cm³, no work is done during cooling, and the work done during compression can be found using W_compression = n * R * T * ln(V_initial / V_final) after calculating the number of moles (n) using the initial state parameters. The efficiency of the engine can be calculated using efficiency = (W_expansion + W_cooling + W_compression) / Q_in.

Step by step solution

01

Sketching the pV-cycle

To sketch the cycle on a pV-diagram, we first need to identify the three parts of the cycle: - Expansion: The gas expands from a volume of 150 cm³ to 450 cm³. We assume this process to be isobaric, meaning constant pressure. - Cooling: The gas cools down from a high temperature back to room temperature. Since the volume remains constant, this process is isochoric. - Compression: The gas is compressed back to its original volume of 150 cm³. This part is isothermal, implying constant temperature. Now that we have identified the three parts of the cycle, we can sketch the cycle on a pV-diagram.
02

Calculating work done and heat flow in each part of the cycle

Let's calculate the work done on the gas and heat flow out of the gas in each part of the cycle. a) Expansion: Since this process is isobaric, work done on the gas during the expansion can be calculated using the formula: W_expansion = p * (V_final - V_initial) W_expansion = 300 kPa * (450 cm³ - 150 cm³) = 300 kPa * 300 cm³ b) Cooling: This is an isochoric process, meaning there's no change in volume. Therefore, no work is done on the gas. W_cooling = 0 c) Compression: This is an isothermal process, and the work done on the gas during the compression part can be calculated by using the formula: W_compression = n * R * T * ln(V_initial / V_final) In order to calculate the work done, we first need to find the number of moles (n) of the gas using the initial state' parameters : pV = nRT => n = pV / RT n = (300 kPa * 150 cm³) / (8.314 J/mol K * (20.0+273.15) K) Now, we can find W_compression using the above formula with the calculated n value .
03

Calculating the efficiency of the engine

Now that we have calculated the work done in each part of the cycle, we can determine the efficiency of the engine using the formula: efficiency = (W_expansion + W_cooling + W_compression) / Q_in We already calculated W_expansion, W_cooling, and W_compression, and Q_in is the heat flow into the engine during the expansion process. The efficiency can be calculated by putting the numerical values in the above formula. Note that if you get a negative value for efficiency, take its absolute value as the efficiency should be a positive value.

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Key Concepts

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

PV-diagram
A PV-diagram is a graphical representation of a system’s changes in pressure (P) and volume (V) throughout different processes in thermodynamics. This type of diagram is instrumental in visualizing and understanding the different processes involved in a heat engine cycle, which typically includes isobaric, isochoric, and isothermal processes. On this diagram, the work done by a gas during a process that results in volume change is represented by the area under the corresponding curve. Expansion and compression processes appear as lines moving to the right and left respectively, while isochoric and isothermal processes appear as vertical and curved lines.

For the given exercise, the sketch would show a horizontal line for the isobaric expansion, a vertical line for the isochoric cooling, and a curved line for the isothermal compression, forming a closed loop that signifies the cyclical nature of the engine's operation.
Isobaric process
An isobaric process is characterized by constant pressure while the volume of the system changes. In the exercise, the gas expansion from 150 cm³ to 450 cm³ is an example of an isobaric process. During such a process, the work done by the gas can be calculated by multiplying the pressure by the change in volume (W = PΔV). It’s integral to the operation of many heat engines because it directly converts heat into work.

Understanding the mechanics of isobaric processes helps with comprehending how energy is transformed in these engines and is vital for calculating the work done during the expansion phase of the cycle.
Isochoric process
An isochoric process takes place at a constant volume, which means there is no mechanical work done by the gas (since work is a result of volume change). This characteristic is pivotal as it simplifies the calculation of heat flow during the process. In the context of the exercise, the cooling of the gas happens isochoricly, which implies that all energy removed from the system is in the form of heat.

While work isn't done during this process, understanding the role of isochoric cooling is essential in evaluating energy changes in the system and the corresponding effects on pressure and temperature.
Isothermal process
An isothermal process occurs at a constant temperature, which inherently implies that the total internal energy of the gas remains constant. The work done by the gas during an isothermal compression can be calculated using the formula involving the natural logarithm of the ratio of the final to initial volumes (W = nRTln(Vi/Vf)), where n represents the number of moles, R is the ideal gas constant, and T is the temperature.

In this type of process, energy in the form of work is exchanged between the system and surroundings to maintain a constant temperature, which is critical for calculating the work done during the compression phase of the engine's cycle in the exercise.
Work done by gas
The work performed by gas in a thermodynamic cycle is a significant factor in determining the engine's output. It is directly related to the area under the process curve on a PV-diagram. In an expanding process, the gas does work on the surroundings, whereas in a compression process, the surroundings do work on the gas. Calculating this work in each phase of the cycle is a fundamental step in predicting the performance and efficiency of a heat engine, which relates to the practicality and real-world applications of such systems.

Understanding the concept of work done by gas is also crucial when discussing conservation of energy within these systems and how they translate heat energy into mechanical work.
Heat flow
Heat flow in a thermodynamic process is the transfer of thermal energy due to a temperature difference. It is an important concept as it explains how energy enters or leaves the system. Positive heat flow indicates that heat is being added to the system, while negative heat flow means heat is being expelled. This transfer is what drives the engine's cycle, allowing the gas to expand or contract and perform work.

The exercise involves two instances of heat flow: adding heat during expansion (isobaric process) and removing heat during cooling (isochoric process). Recognizing the direction of heat flow is fundamental to understanding these thermal processes and their impacts on the system’s state.
Thermal efficiency
Thermal efficiency is a measure of how well a heat engine converts the heat input into useful work. It is a dimensionless number, often represented as a percentage, reflecting the ratio of work output to heat input. The higher the efficiency, the more cost-effective and environmentally friendly the engine tends to be. The formula for efficiency is efficiency = (Wnet/Qin), where Wnet is the net work done by the system and Qin is the total heat added to the system.

In the provided exercise, the efficiency calculation provides insight into the performance of the heat engine cycle. Understanding this efficiency is crucial for both theoretical concepts in thermodynamics and practical applications in engineering and energy production.

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Most popular questions from this chapter

A certain refrigerator is rated as being \(32.0 \%\) as ef ficient as a Carnot refrigerator. To remove \(100 .\) J of heat from the interior at \(0^{\circ} \mathrm{C}\) and eject it to the outside at \(22^{\circ} \mathrm{C}\), how much work must the refrigerator motor do?

Which of the following processes (all constanttemperature expansions) produces the most work? a) An ideal gas consisting of 1 mole of argon at \(20^{\circ} \mathrm{C}\) expands from \(1 \mathrm{~L}\) to \(2 \mathrm{~L}\). b) An ideal gas consisting of 1 mole of argon at \(20^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\). c) An ideal gas consisting of 2 moles of argon at \(10^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\). d) An ideal gas consisting of 1 mole of argon at \(40^{\circ} \mathrm{C}\) expands from \(1 \mathrm{~L}\) to \(2 \mathrm{~L}\) e) An ideal gas consisting of 1 mole of argon at \(40^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\).

A heat engine operates with an efficiency of \(0.5 .\) What can the temperatures of the high-temperature and lowtemperature reservoirs be? a) \(T_{\mathrm{H}}=600 \mathrm{~K}\) and \(T_{\mathrm{L}}=100 \mathrm{~K}\) b) \(T_{\mathrm{H}}=600 \mathrm{~K}\) and \(T_{\mathrm{L}}=200 \mathrm{~K}\) c) \(T_{\mathrm{H}}=500 \mathrm{~K}\) and \(T_{\mathrm{L}}=200 \mathrm{~K}\) d) \(T_{\mathrm{H}}=500 \mathrm{~K}\) and \(T_{\mathrm{L}}=300 \mathrm{~K}\) e) \(T_{\mathrm{H}}=600 \mathrm{~K}\) and \(T_{\mathrm{L}}=300 \mathrm{~K}\)

What capacity must a heat pump with a coefficient of performance of 3 have to heat a home that loses heat energy at a rate of \(12 \mathrm{~kW}\) on the coldest day of the year? a) \(3 \mathrm{~kW}\) c) \(10 \mathrm{~kW}\) e) \(40 \mathrm{~kW}\) b) \(4 \mathrm{~kW}\) d) \(30 \mathrm{~kW}\)

Other state variables useful for characterizing different classes of processes can be defined from \(E_{\text {int }}, S, P\), and \(V\). These include the enthalpy, \(H=E_{\text {int }}+p V\), the Helmholtz free energy, \(A=E_{\text {int }}-T S,\) and the Gibbs free energy, \(G=E_{\text {int }}+p V-T S\). a) Write the differential equations for \(d H, d A,\) and \(d G\). b) All of these are also exact differentials. What relationships follow from this fact? Use the First Law to simplify.

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