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20.42 A cylinder contains oxygen at a pressure of 2.00 atm. The volume is 4.00 \(\mathrm{L}\) , and the temperature is 300 \(\mathrm{K}\) . Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes: (i) Heated at constant pressure from the initial state (state 1) to state \(2,\) which has \(T=450 \mathrm{K}\) . (ii) Cooled at constant volume to 250 \(\mathrm{K}\) (state 3). (iii) Compressed at constant temperature to a volume of 4.00 \(\mathrm{L}\) (state 4\()\) . (iv) Heated at constant volume to 300 \(\mathrm{K}\) , which takes the system back to state 1. (a) Show these four processes in a \(p V\) -diagram, giving the numerical values of \(p\) and \(V\) in each of the four states. (b) Calculate \(Q\) and \(W\) for each of the four processes. (c) Calculate the net work done by the oxygen. (d) What is the efficiency of this device as a heal engine? How does this compare to the efficiency of a Carnot-cycle engine operating between the same minimum and maximum tem- peratures of 250 \(\mathrm{K}\) and 450 \(\mathrm{K} ?\)

Short Answer

Expert verified
The processes form a closed cycle on the PV diagram; state's pressures and volumes are calculated by the ideal gas law. Efficiency is lower than a Carnot cycle between 250 K and 450 K.

Step by step solution

01

Calculating State 1

In state 1, the oxygen is initially at 2.00 atm pressure, 4.00 L volume, and 300 K temperature.
02

Process 1 (Constant Pressure Heating)

During this process, the gas is heated from 300 K to 450 K at constant pressure. Using the ideal gas law, the final volume at state 2 is calculated as follows:\[ V_2 = V_1 \left(\frac{T_2}{T_1}\right) = 4.00\, \text{L} \left(\frac{450\, \text{K}}{300\, \text{K}}\right) = 6.00\, \text{L} \]
03

State 2 Description

In state 2, the gas is at 450 K, 2.00 atm pressure, and a volume of 6.00 L as calculated from Process 1.
04

Process 2 (Constant Volume Cooling)

The gas is cooled at constant volume from 450 K to 250 K. Using the ideal gas law to find the pressure at state 3:\[ P_3 = P_2 \left(\frac{T_3}{T_2}\right) = 2.00\, \text{atm} \left(\frac{250\, \text{K}}{450\, \text{K}}\right) \approx 1.11\, \text{atm} \]
05

State 3 Description

In state 3, the gas has a volume of 6.00 L, pressure of approximately 1.11 atm, and temperature of 250 K.
06

Process 3 (Constant Temperature Compression)

The gas is compressed from 6.00 L to 4.00 L at constant temperature of 250 K. Using the ideal gas law:\[ P_4 = P_3 \left(\frac{V_3}{V_4}\right) = 1.11\, \text{atm} \left(\frac{6.00\, \text{L}}{4.00\, \text{L}}\right) = 1.67\, \text{atm} \]
07

State 4 Description

In state 4, the gas is at 1.67 atm pressure, 4.00 L volume, and 250 K temperature.
08

Process 4 (Constant Volume Heating)

The gas is heated at constant volume from 250 K back to 300 K. This returns the system to state 1. Using ideal gas law:\[ P_1 = P_4 \left(\frac{T_1}{T_4}\right) = 1.67\, \text{atm} \left(\frac{300\, \text{K}}{250\, \text{K}}\right) = 2.00\, \text{atm} \]
09

PV Diagram

The cycle on a PV diagram consists of the following vertices: (4.00 L, 2.00 atm), (6.00 L, 2.00 atm), (6.00 L, 1.11 atm), (4.00 L, 1.67 atm), back to (4.00 L, 2.00 atm).
10

Calculating Q and W

Each process is analyzed for heat transfer (Q) and work done (W):- Process 1: \( W = P(V_2 - V_1) \), calculate using pressure and volume change.- Process 2: \( Q = nC_v(T_3 - T_2) \), since volume is constant (Cv specific heat at constant volume).- Process 3: \( W = P_3(V_4 - V_3) = 0 \), temperature constant.- Process 4: \( Q = nC_v(T_1 - T_4) \), volume constant.
11

Net Work Done

Net work done by the gas over one cycle:\[ W_{net} = W_{1} + W_{2} + W_{3} + W_{4} \] Consider the sign of work and add contributions from each process.
12

Efficiency Comparison

Efficiency of the heat engine:\[ \eta = \frac{W_{net}}{Q_{in}} \]Compare with Carnot efficiency:\[ \eta_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}} \] where \(T_{cold} = 250\, \text{K}\) and \(T_{hot} = 450\, \text{K}\).

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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 the changes in pressure and volume for a gas during different processes. Understanding this diagram helps visualize what happens during the thermodynamic cycle. In our exercise, we plot the pressure (\( P \)) against the volume (\( V \)) for each state of the oxygen.

The cycle begins at state 1 with 2.00 atm and 4.00 L. During Process 1 (heating at constant pressure), the volume increases from 4.00 L to 6.00 L at the same pressure of 2.00 atm. In Process 2 (cooling at constant volume), the pressure decreases to approximately 1.11 atm while the volume remains at 6.00 L. Process 3 involves compressing the gas at a constant temperature, leading to a decrease in volume back to 4.00 L and an increase in pressure to 1.67 atm. Finally, Process 4 returns the gas back to the initial state (state 1) at constant volume.

Through these plotted points, we form a loop that represents one complete cycle of the gas processes. The vertices of this loop offer valuable information about each state of the gas:
  • State 1: 4.00 L, 2.00 atm
  • State 2: 6.00 L, 2.00 atm
  • State 3: 6.00 L, 1.11 atm
  • State 4: 4.00 L, 1.67 atm
  • Return to State 1: 4.00 L, 2.00 atm
Visualizing these changes in a PV diagram simplifies understanding how the system behaves throughout the cycle.
Heat Transfer
Heat transfer in thermodynamic processes refers to the energy exchange due to temperature differences. It's important to analyze this in each phase of the cycle.

In Process 1, heat is added to the system to increase its temperature from 300 K to 450 K at constant pressure, leading to an increase in volume. This means that heat transfer to the system, denoted by \( Q \), is positive.

Process 2 involves cooling the system from 450 K to 250 K while maintaining a constant volume. During this cooling phase, heat is removed from the system, so \( Q \) is negative.

During Processes 3 and 4, it's also necessary to account for heat transfer. In Process 3, the system is compressed at a constant temperature. For Process 4, heat is again added, raising the temperature back to 300 K while the volume remains fixed. The amount of heat transferred in these processes can be calculated using specific heat capacities and the ideal gas law formulas.

Understanding heat transfer is crucial, as it directly impacts the work done and the efficiency of heat engines like the one in this exercise.
Carnot Efficiency
Carnot Efficiency sets the upper limit for the efficiency of any heat engine operating between two temperatures. It is a theoretical concept based on a reversible engine using two thermal reservoirs.

In our problem, the Carnot cycle would operate between the minimum and maximum temperatures: 250 K (cold reservoir) and 450 K (hot reservoir). The Carnot efficiency \( \eta_{Carnot} \) is calculated by the formula:\[\eta_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}}\]Substituting the given temperatures, we find:\[\eta_{Carnot} = 1 - \frac{250 \text{ K}}{450 \text{ K}} \approx 0.4444\]This means the maximum possible efficiency of a heat engine working between these two temperatures is about 44.44%.

This theoretical efficiency provides a benchmark to evaluate the actual efficiency of our heat engine cycle. Real-world engines will have efficiencies lower than this due to practical limitations and irreversible processes often encountered in real engines.
Work Done in Thermodynamics
Work done in thermodynamics is the energy transferred by the system to its surroundings because of volume change. It is an important concept when analyzing the efficiency and performance of a thermodynamic cycle.

In our exercise, work is done differently depending on the process type. For Process 1, the work done is calculated from the constant pressure heating formula:\[W = P(V_2 - V_1)\]where \( P \) is pressure, and \( V_1, V_2 \) are initial and final volumes respectively.

In Process 3, work is also done although the temperature is constant. Typically, work for processes where volume remains unchanged, like Process 2 and Process 4, isn’t output to the surroundings because the system exerts equal force to the volume.

The net work done over one complete cycle (sum of work in each individual process) is crucial because it indicates the energy output of the cycle. When heat engines are analyzed, the net work done is compared against the total heat input to ascertain the efficiency of the cycle. Understanding how work varies in different types of processes is essential to mastering thermodynamics.

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Most popular questions from this chapter

20.36 . A lonely party balloon with a volume of 2.40 \(\mathrm{L}\) and containing 0.100 \(\mathrm{mol}\) of air is left behind to drift in the temporarily uninhabited and depressurized Intermational Space Station. Sunlight coming through a porthole heats and explodes the balloon, causing the air in it to undergo a free expansion into the empty station, whose total volume is 425 \(\mathrm{m}^{3}\) . Calculate the entropy change of the air during the expansion.

20.22. A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation, the heat rejected by the engine melts 0.0400 \(\mathrm{kg}\) of ice. During this time, how much work \(W\) is performed by the engine?

20.63. An object of mass \(m_{1}\) , specific heat capacity \(c_{1}\) , and temperature \(T_{1}\) is placed in contact with a second object of mass \(m_{2},\) specific heat capacity \(c_{2},\) and temperature \(T_{2}>T_{1} .\) As a result, the temperature of the first object increases to \(T\) and the temperature of the second object decreases to \(T^{\prime} .\) (a) Show that the entropy increase of the system is $$ \Delta S=m_{1} c_{1} \ln \frac{T}{T_{1}}+m_{2} c_{2} \ln \frac{T^{\prime}}{T_{2}} $$ and show that energy conservation requires that $$ m_{1} c_{1}\left(T-T_{1}\right)=m_{2} c_{2}\left(T_{2}-T^{\prime}\right) $$ (b) Show that the entropy change \(\Delta S\) , considered as a function of \(T,\) is a maximum if \(T=T,\) which is just the condition of thermodynamic equilibrium. (c) Discuss the result of part (b) in terms of the idea of entropy as a measure of disorder.

A diesel engine performs 2200 \(\mathrm{J}\) of mechanical work and discards 4300 \(\mathrm{J}\) of heat each cycle. (a) How much heat must be supplied to the engine in each cycle? (b) What is the thermal efficiency of the engine?

20.38. You are designing a Carnot engine that has 2 \(\mathrm{mol}\) of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 \(\mathrm{atm}\) . With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 kg block of ice originally at \(0.0^{\circ} \mathrm{C}\) , using only the beat rejected by the engine?

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