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.