Problem 45 - University Physics with Modern Physics

Question

A cylinder contains oxygen at a pressure of 2.00 atm. The volume is 4.00 \(L\), and the temperature is 300 \(K\). 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 K. (ii) Cooled at constant volume to 250 \(K\) (state 3). (iii) Compressed at constant temperature to a volume of 4.00 \(L\) (state 4). (iv) Heated at constant volume to 300 \(K\), which takes the system back to state 1. (a) Show these four processes in a \(pV\)-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 in the complete cycle. (d) What is the efficiency of this device as a heat engine? How does this compare to the efficiency of a Carnot cycle engine operating between the same minimum and maximum temperatures of 250 \(K\) and 450 \(K\)?

Step-by-Step Solution

Answer

(a) Plot points: 1: (4L, 2atm), 2: (6L, 2atm), 3: (6L, 1.11atm), 4: (4L, 1.67atm). (b) Calculate Q & W using thermodynamics equations for each process. (c) Net Work: Process 1 & 3 work sums. (d) Efficiency compares with Carnot (1-250/450).

Step 1: Identify Initial Conditions and Ideal Gas Law Application

The initial state of the gas is given by a pressure \( P_1 = 2.00 \) atm, a volume \( V_1 = 4.00 \ L \), and a temperature \( T_1 = 300 \ K \). Using the ideal gas law \( PV = nRT \), we can calculate the number of moles \( n \), if needed, though it cancels in the process calculation. The universal gas constant \( R \) can be used as \( 0.0821 \frac{L\cdot atm}{mol\cdot K}\).

Step 2: Process 1 (isobaric heating)

In Process 1, the gas is heated at constant pressure from state 1 to state 2 with \( T_2 = 450 \ K \). Since the process is isobaric, the final volume \( V_2 \) can be calculated using \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \). Solve for \( V_2: V_2 = V_1 \cdot \frac{T_2}{T_1} = 4.00 \cdot \frac{450}{300} = 6.00 \ L \).

Step 3: Process 2 (isochoric cooling)

In Process 2, the gas is cooled at constant volume from state 2 to state 3 with \( T_3 = 250 \ K \). The pressure \( P_3 \) can be determined using \( P_2/T_2 = P_3/T_3 \). Calculate \( P_3: P_3 = P_2 \cdot \frac{T_3}{T_2} = 2.00 \cdot \frac{250}{450} = 1.11 \) atm.

Step 4: Process 3 (isothermal compression)

In Process 3, the gas is compressed isothermally from state 3 to state 4, reaching a final volume \( V_4 = 4.00 \ L \). The pressure \( P_4 \) at state 4 can be calculated using \( P_3 V_3 = P_4 V_4 \). Solve for \( P_4: P_4 = P_3 \cdot \frac{V_3}{V_4} = 1.11 \cdot \frac{6.00}{4.00} = 1.67 \) atm.

Step 5: Process 4 (isochoric heating)

In Process 4, the gas is heated at constant volume from state 4 to state 1 (return to initial conditions) with \( T_4 = 300 \ K \). The pressure can be verified: \( P_1/T_1 = P_4/T_4 \) must hold true. Calculate \( P_1: P_1 = P_4 \cdot \frac{T_1}{T_4} = 1.67 \cdot \frac{300}{300} = 2.00 \) atm. It verifies that we're back at the original state.

Step 6: Plot the pV-Diagram

To plot the processes on a \(pV\)-diagram, plot the points \((4.00, 2.00)\), \((6.00, 2.00)\), \((6.00, 1.11)\), \((4.00, 1.67)\), and back to \((4.00, 2.00)\). Connect them using the correct process paths: horizontal line (isobaric) for constant pressure, vertical line (isochoric) for constant volume, and log curve (isothermal) for constant temperature.

Step 7: Calculate Thermodynamic Quantities for Each Process

For each process, calculate \( Q \) (heat) and \( W \) (work):- **Process 1**: \( Q = nC_p\Delta T \), \( W = P\Delta V = 2.00 \cdot (6.00-4.00) \ = 4.00 \ L\cdot atm \).- **Process 2**: \( Q = nC_v\Delta T \), \( W = 0 \) (since volume is constant).- **Process 3**: Use \( W = nRT \ln(\frac{V_f}{V_i}) \).- **Process 4**: \( Q = nC_v\Delta T \), \( W = 0 \) (since volume is constant).

Step 8: Calculate Net Work Done and Efficiency

Add all work done: \( W_{net} = W_1 + W_2 + W_3 + W_4 \). Since Process 2 and 4 have zero work, net work depends on Process 1 and 3. Efficiency \( \eta = \frac{W_{net}}{Q_{in}} \), where \( Q_{in} \) is the heat absorbed during the cycle (heating processes). Compare with Carnot efficiency \( \eta_c = 1 - \frac{T_{min}}{T_{max}} = 1 - \frac{250}{450} \).

Step 9: Calculate and Compare

Use calculated values to compute numerical answers for work, heat, and efficiency. Compare actual cycle efficiency with Carnot efficiency to evaluate performance.

Key Concepts

Isobaric Process

An isobaric process occurs when the pressure of a gas remains constant as its volume or temperature changes. It's commonly seen in real-world scenarios, like heating air in a balloon. For these processes, the ideal gas law can be a helpful tool:

• The equation is given by \( P \Delta V = n R \Delta T \), where \( P \) is the pressure, \( \Delta V \) is the change in volume, \( n \) is the number of moles, and \( R \) is the universal gas constant.
• The work done \( W \) by the system is calculated as \( W = P \Delta V \). In our exercise, heating from 300 K to 450 K at constant pressure (2 atm) resulted in a volume increase from 4.00 L to 6.00 L.

In an isobaric process, since the pressure remains constant, it is easy to calculate the work done if you know the change in volume. Such processes on a pV-diagram are represented as horizontal lines, reflecting the constant pressure.

Isochoric Process

During an isochoric process, a gas's volume remains constant, which means there is no work done by the gas. This occurs, for example, when gas is heated in a rigid, closed container. Key aspects of isochoric processes include:

• According to the ideal gas law \( \frac{P}{T} = \text{constant} \), since volume doesn't change, any change in temperature results in a proportional change in pressure.
• In the exercise, an isochoric cooling occurred from 450 K to 250 K, where the volume is maintained at 6.00 L. The pressure reduced from the newly calculated value at state 2 to 1.11 atm at state 3.

These processes are depicted as vertical lines on a pV-diagram, showcasing how pressure changes while volume stays the same. Even though no work is done, heat can still be absorbed or released, changing internal energy.

Isothermal Process

In an isothermal process, the temperature of the system remains constant. The energy of the system does not change, making it particularly interesting in thermodynamics. For isothermal processes involving an ideal gas:

• The work done by or on the gas is calculated using the formula \( W = nRT \ln\left(\frac{V_f}{V_i}\right) \), where the temperature and moles of gas \( n \) stay unchanged.
• For the exercise, we see a compression at constant temperature (250 K) resulting in pressure changing from 1.11 atm back to a volume of 4.00 L.

On a pV-diagram, isothermal changes are represented by hyperbolic curves. Since the temperature doesn't change, any work done or heat exchanged exactly offsets the other's effect, maintaining energy balance.

pV-Diagram

A pV-diagram, or pressure-volume diagram, provides a graphical representation of a thermodynamic process, illustrating how pressure and volume change. Essential points on a pV-diagram include:

• Each state corresponds to a unique pressure-volume pair \((P, V)\), plotted as points.
• The paths connecting these points correspond to different thermodynamic processes, where each type has a characteristic shape — horizontal lines for isobaric, vertical lines for isochoric, and curves for isothermal processes.

In the exercise, the trajectory of the cycle comprises:

• An isobaric process from point \((4.00, 2.00)\) to \((6.00, 2.00)\).
• Followed by an isochoric process to \((6.00, 1.11)\).
• Next, an isothermal process moving to \((4.00, 1.67)\).
• Finally, an isochoric process back to the initial state \((4.00, 2.00)\).

Understanding a pV-diagram is crucial for visualizing and calculating the work done in cyclic processes and assessing thermodynamic efficiency.