The air was under a pressure of 10 ^ 5 Pa and occupied a volume of 0.6 m3.

The air was under a pressure of 10 ^ 5 Pa and occupied a volume of 0.6 m3. What kind of robot will be performed when its volume is reduced to 0.2 m3.

P = 10 ^ 5 Pa.
V1 = 0.6 m ^ 3.
V2 = 0.2 m ^ 3.
А – ?
The work of the gas during compression or expansion is determined by the formula: A = F * Δl, where F is the force under which compression or expansion occurs, Δl is the distance the piston has moved.
Pressure force F = P * S, where P is the gas pressure, S is the area of the piston under which compression occurs.
A = P * S * Δl.
The product S * Δl = ΔV by how much the gas volume has changed.
A = P * ΔV.
ΔV = V1 – V2.
A = P * (V1 – V2).
A = 10 ^ 5 Pa * (0.6 m ^ 3 – 0.2 m ^ 3) = 40,000 J.
Answer: to compress the gas, we performed work A = 40,000 J.