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c). A substance undergoes the cyclic process shown in Figure 4.2. Work output occurs along path 𝐴𝐵, whereas work input is required along path 𝐵𝐶, and no work is involved in the constant-volume process 𝐶𝐴. Energy transfers by heat occur during each process involved in the cycle.

Calculate the work output during process 𝐴𝐵.

How much work input is required during process 𝐵𝐶?

Determine the net energy input 𝑄 during the cycle.

To solve this problem, we can break it down step by step based on the graph provided:

### 1. **Work Output during Process AB (Isobaric Process)**
Process \( AB \) is an **isobaric expansion** (constant pressure) from volume \( V_A \) to \( V_B \).

- From the graph, the pressure \( P_A = 1 \, \text{Pa} \).
- The volumes \( V_A = 10 \, \text{m}^3 \) and \( V_B = 50 \, \text{m}^3 \).

For an isobaric process, the work \( W \) done by the gas is given by the formula:
\[
W = P \Delta V = P (V_B - V_A)
\]
Substituting the values:
\[
W = 1 \, \text{Pa} \times (50 \, \text{m}^3 - 10 \, \text{m}^3) = 1 \, \text{Pa} \times 40 \, \text{m}^3 = 40 \, \text{J}
\]

So, the **work output during process AB is \( 40 \, \text{J} \)**.

---

### 2. **Work Input during Process BC (Isobaric Process)**
Process \( BC \) is an **isobaric compression** from volume \( V_B \) to \( V_C \).

- The pressure \( P_B = 5 \, \text{Pa} \).
- The volumes \( V_B = 50 \, \text{m}^3 \) and \( V_C = 10 \, \text{m}^3 \).

Again, the work done on the gas in an isobaric process is:
\[
W = P \Delta V = P (V_C - V_B)
\]
Since this is compression, \( V_C - V_B \) will be negative (work is input, not output):
\[
W = 5 \, \text{Pa} \times (10 \, \text{m}^3 - 50 \, \text{m}^3) = 5 \, \text{Pa} \times (-40 \, \text{m}^3) = -200 \, \text{J}
\]

So, the **work input during process BC is \( -200 \, \text{J} \)** (negative, indicating work is done on the system).

---

### 3. **Net Energy Input \( Q \) During the Cycle**
The net energy input \( Q \) is related to the **net work** done during the entire cycle. The first law of thermodynamics states:
\[
\Delta U = Q - W_{\text{net}}
\]
Since this is a **cyclic process**, the change in internal energy \( \Delta U = 0 \). Hence, the net heat \( Q \) is equal to the net work done:
\[
Q = W_{\text{net}} = W_{\text{AB}} + W_{\text{BC}} + W_{\text{CA}}
\]
- Work during AB: \( W_{\text{AB}} = 40 \, \text{J} \)
- Work during BC: \( W_{\text{BC}} = -200 \, \text{J} \)
- Work during CA (constant volume, no work): \( W_{\text{CA}} = 0 \)

Therefore, the net work done is:
\[
W_{\text{net}} = 40 \, \text{J} + (-200 \, \text{J}) + 0 = -160 \, \text{J}
\]
Thus, the **net energy input \( Q = -160 \, \text{J} \)**.

This means that 160 J of energy is lost from the system in the form of heat.

---

Would you like any further details on these steps?

Here are 5 follow-up questions you could explore:
1. What is the significance of a negative net energy input \( Q \) in a cyclic process?
2. How does the first law of thermodynamics apply to cyclic processes in general?
3. Can the work done in an isochoric (constant volume) process ever be non-zero? Why or why not?
4. What real-world applications might involve cyclic processes similar to the one described in the diagram?
5. How would the problem change if the process AB were adiabatic instead of isobaric?

**Tip**: In cyclic thermodynamic processes, the area enclosed by the process on a PV diagram gives the net work done during the cycle.

A substance undergoes the cyclic process shown in Figure 4.2. Work output occurs along path AB, whereas work input is required along path BC, and no work is involved in the constant-volume process CA. Energy transfers by heat occur during each process involved in the cycle. Calculate the work output during process AB. How much work input is required during process BC? Determine the net energy input Q during the cycle.

Learn how to calculate the work output and input during a cyclic process shown in a PV diagram. This problem covers key thermodynamic concepts such as work in isobaric processes and the first law of thermodynamics. Suitable for undergraduate students.

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