**G) FNT 5:****Groups 1-3: Path A****Groups 3-6: Path B**---**Here are the numerical results for FNT 5:**- **A:** \( W_A = 2000 \, \text{J}, \, Q_A = -2000 \, \text{J} \)- **B:** \( W_B = 1000 \, \text{J}, \, Q_B = -1000 \, \text{J} \)---1) **For your path, briefly show how to find the work and heat between I and III.**2) **Why is the work equal and opposite to heat between I and III?**3) **Explain how the change in internal energy could be the same for the two processes, but the work and heat are different. Put your explanation on the board.** ### Graph: Pressure-Volume Cycle This graph depicts a pressure-volume (P-V) cycle on an idealized thermodynamic system. The axes and labeled points provide important information about the process:#### Axes Details:- **Vertical Axis (Y-axis)**: Represents pressure, \( P \), measured in pascals (\(\text{Pa}\)), specifically shown in multiples of \(10^5\) Pa.- **Horizontal Axis (X-axis)**: Represents volume, \( V \), measured in cubic meters (\(\text{m}^3\)), specifically shown in multiples of \(10^{-3}\) \(\text{m}^3\).#### Key Points and Process Flow:1. **Point I**: Initial state on the lower right corner, at approximately \(10.0 \times 10^{-3} \, \text{m}^3\) for volume and \(2.0 \times 10^5 \, \text{Pa}\) for pressure. 2. **Transition from I to IIa**: - **Process IIa**: The system moves vertically upwards, denoting an increase in pressure at a constant volume. 3. **Point IIa**: Marks the top right corner, with pressure at \(4.0 \times 10^5 \, \text{Pa}\) and volume unchanged. 4. **Transition from IIa to III**: - **Process III**: The system moves horizontally to the left, suggesting a decrease in volume at a constant pressure. 5. **Point III**: Located at the top left corner, with volume reduced to approximately \(5.0 \times 10^{-3} \, \text{m}^3\) and pressure remaining at \(4.0 \times 10^5 \, \text{Pa}\). 6. **Transition from III to IIb**: - **Process IIb**: A vertical movement downward, indicating a decrease in pressure at constant volume. 7. **Point IIb**: Corresponds to the bottom left corner, at approximately \(5.0 \times 10^{-3} \, \text{m}^3\) for volume and \(2.0 \times 10^5 \, \text{Pa}\) for pressure. 8. **Transition from IIb to I**:

G) FNT 5: Groups 1-3: Path A Groups 3-6: Path B Here are the numerical results for FNT 5: A: WA = 2000 J, QA= -2000 J B: WB 1000 J, QB = -1000 J 1) For your path, briefly show how to find the work and heat between I and III. 2) Why is the work equal and opposite to heat between I and III? 3) Explain how the change in internal energy could be the same for the two processes, but the work and heat are different. Put your explanation on the board.

G) FNT 5: Groups 1-3: Path A Groups 3-6: Path B Here are the numerical results for FNT 5: A: WA = 2000 J, QA= -2000 J B: WB 1000 J, QB = -1000 J 1) For your path, briefly show how to find the work and heat between I and III. 2) Why is the work equal and opposite to heat between I and III? 3) Explain how the change in internal energy could be the same for the two processes, but the work and heat are different. Put your explanation on the board.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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