In a certain reversible expansion, a system at 300. K absorbs exactly 6.00 *10² J of heat. In the irreversible recompression to the original state of the system, twice as much work is done on the system as is performed on the surroundings in the expansion. What is the entropy change of the system in the recompression step?

Question

Quizplus · Accepted Answer

Since the expansion is reversible, we know that the entropy change of the system during the expansion is simply given by ΔS = q/T = 6.00 × 1010 J / 300 K = 2.00 × 108 J/K. In the irreversible recompression, we know that the work done on the system is twice as much as the work performed on the surroundings during the expansion. Since the expansion work is given by w = -q = -6.00 × 1010 J, the irreversible recompression does -2w = 1.20 × 1011 J of work on the system. Since the process is irreversible, we cannot use ΔS = q/T to calculate the entropy change directly, but we can use the fact that the total entropy change of the system and surroundings is zero for a cyclic process. Therefore, the entropy change of the surroundings in the recompression is -ΔS = -2.00 × 108 J/K, and the entropy change of the system in the recompression must be the same magnitude but opposite sign: ΔS = 2.00 × 108 J/K. Thus, the correct answer is (D).

In a Certain Reversible Expansion, a System at 300