Question

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.

Answer 1

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