Internal energy E = i/2 ν R T = i/2 p V, internal energy increment? δE = I/2(？ p2 V2？ = p 1 V 1), so?

δEAB = 5/2( 100 * 3-300 * 1)= 0

Δ ebc = 5/2 (100 *1-100 * 3) =-500 joules.

ECA =？ ? 5/2 ( 300* 1 - 100* 1) = ? 500 joules

Regarding the cycle efficiency, this problem is complicated, and the key lies in the AB process, which is neither purely endothermic nor purely exothermic. There is a turning point of heat absorption and heat release, which is troublesome to calculate and cannot be solved by a sub-problem of 10.

In general college physics textbooks, the computational cycle efficiency is avoided for this problem.