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Question 9

For the given cyclic process CAB as shown for a gas, the work done is:

Solution

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$$W = \text{Area of }\Delta\text{CAB}$$

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 5 = 10\text{ J}$$

Direction of the cycle is clockwise: $$W = +10\text{ J}$$

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