Problem 47 - University Physics with Modern Physics

Question

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 \(\mathrm{K} ?\)

Step-by-Step Solution

Answer

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 1: Calculating State 1

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

Step 2: 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:\[ V_2 = V_1 \left(\frac{T_2}{T_1}\right) = 4.00\, \text{L} \left(\frac{450\, \text{K}}{300\, \text{K}}\right) = 6.00\, \text{L} \]

Step 3: 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.

Step 4: 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:\[ P_3 = P_2 \left(\frac{T_3}{T_2}\right) = 2.00\, \text{atm} \left(\frac{250\, \text{K}}{450\, \text{K}}\right) \approx 1.11\, \text{atm} \]

Step 5: 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.

Step 6: 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:\[ P_4 = P_3 \left(\frac{V_3}{V_4}\right) = 1.11\, \text{atm} \left(\frac{6.00\, \text{L}}{4.00\, \text{L}}\right) = 1.67\, \text{atm} \]

Step 7: State 4 Description

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

Step 8: 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:\[ P_1 = P_4 \left(\frac{T_1}{T_4}\right) = 1.67\, \text{atm} \left(\frac{300\, \text{K}}{250\, \text{K}}\right) = 2.00\, \text{atm} \]

Step 9: 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).

Step 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.

Step 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.

Step 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 \(T_{hot} = 450\, \text{K}\).

Key Concepts

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.