An ideal gas undergoes the cyclic process shown in the original P-V graph. The cycle is a circle spanning P = 300 to 500 kPa and V = 150 to 350 cm³. The work done by the gas in one cycle is N×10⁻¹ J. Take π = 3.14. Find N.

Question

G1 Physics · Accepted Answer

Net work done in a cyclic P-V process equals the signed area enclosed by the loop. The graph is an ellipse/circle in scaled coordinates with pressure diameter ΔP = 500-300 = 200 kPa = 200×10³ Pa, and volume diameter ΔV = 350-150 = 200 cm³ = 200×10⁻⁶ m³. Area enclosed is W = (π/4)(ΔP)(ΔV). Thus, W = (3.14/4)(200×10³)(200×10⁻⁶) = 31.4 J. Write 31.4 J = 314×10⁻¹ J. Hence N = 314.