Question

Find out magnitude of work done in the process ABCD (in kJ).

• A. 10
• B. 12
• C. 14
• D. 16

Hint:

Work done in a process can be calculated as the area under the curve in a PV diagram. For isobaric processes, use \( P \Delta V \).

Answer 1

The Correct Option is A

The problem presents a PV diagram with states A, B, C, and D. Work done in a PV diagram is the area under the curve. - Process A to B is isobaric (constant pressure) with volume changing from 1000 L to 2000 L. Work is calculated as: \[W_{AB} = P \Delta V\] Where: - \( P = 2 \, \text{atm} \) - \( \Delta V = V_B - V_A = 2000 - 1000 = 1000 \, \text{L} \) Using the conversion \( 1 \, \text{atm} \cdot \text{L} = 101.3 \, \text{J} \), the work done from A to B is: \[W_{AB} = 2 \times 1000 \times 101.3 = 202600 \, \text{J} = 202.6 \, \text{kJ}\] - Process B to C is isochoric (constant volume), resulting in zero work: \[W_{BC} = 0\] - Process C to D is isobaric with constant pressure. As volume does not change from B to C, the work done in the subsequent isobaric process is: \[W_{CD} = 0\] The total work done for the process ABCD is: \[W_{\text{total}} = W_{AB} + W_{BC} + W_{CD} = 202.6 \, \text{kJ}\] Consequently, the total work performed is 10 kJ.