A gas expands linearly from 4.42 bar to 3.1 bar behind a piston from a volume 0.12 m3 to 0.3 m3. What is the work done during the process? Give your answer in kJ with 3 significant digits.
The work done in the process is the area limited by the process equation (linear function) from V1 to V2. As it follows from the picture, this area is a trapezoid P1-V1-V2-P2. Using the geometric formula, we can find this area as
"\\displaystyle W = \\frac{P_1+P_2}{2} (V_2-V_1) = \\frac{(3.1+4.42)\\cdot 10^5}{2}(0.3-0.12) = 67\\, 680=67.680 \\; kJ"
Answer: 67.68 kJ
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