A closed system receives 200kJ of heat at constant volume. It then rejects 100kJ of heat while it has 50kJ of work done on it at constant pressure. If an adiabatic process can be found which will restore the system to its initial state, the work done by the system during this process is

Options

A. (a) 100kJ
B. (b) 50kJ
C. (c) 150kJ
D. (d) 200kJ

Explanation

To solve the problem step by step, we will analyze the processes involved and apply the first law of thermodynamics.

### Step 1: Identify the Processes
The system undergoes three processes:
1. **Process A to B (Isochoric)**: The system receives 200 kJ of heat.
2. **Process B to C (Isobaric)**: The system rejects 100 kJ of heat and has 50 kJ of work done on it.
3. **Process C to A (Adiabatic)**: We need to find the work done by the system during this process.

### Step 2: Apply the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta U = Q - W \]
Where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat added to the system,
- \(W\) is the work done by the system.

Since the system returns to its initial state (cyclic process), the change in internal energy over the entire cycle is zero:
\[ \Delta U = 0 \]

### Step 3: Calculate Net Heat and Work
For the entire cycle, we can express the net heat and work:
\[ Q_{net} = Q_{AB} + Q_{BC} + Q_{CA} \]
\[ W_{net} = W_{AB} + W_{BC} + W_{CA} \]

From the problem:
- \(Q_{AB} = 200 \text{ kJ}\) (heat added)
- \(Q_{BC} = -100 \text{ kJ}\) (heat rejected)
- \(Q_{CA} = 0 \text{ kJ}\) (adiabatic process)

So,
\[ Q_{net} = 200 - 100 + 0 = 100 \text{ kJ} \]

For work:
- \(W_{AB} = 0 \text{ kJ}\) (no work done in isochoric process)
- \(W_{BC} = -50 \text{ kJ}\) (work done on the system)
- \(W_{CA} = W_{C \to A}\) (work done by the system, which we need to find)

Thus,
\[ W_{net} = 0 - 50 + W_{CA} \]

### Step 4: Set Up the Equation
Since \(\Delta U = 0\), we have:
\[ Q_{net} = W_{net} \]
Substituting the values:
\[ 100 = -50 + W_{CA} \]

### Step 5: Solve for Work Done in the Adiabatic Process
Rearranging the equation gives:
\[ W_{CA} = 100 + 50 = 150 \text{ kJ} \]

### Final Answer
The work done by the system during the adiabatic process (C to A) is **150 kJ**.

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