Consider one mole of an ideal monatomic gas of constant C_v of 12.47 J mol^-1 K^-1 confined to an adiabatic container sealed by a frictionless weightless piston. This is container A and T_Ai = 1500 K, V_Ai = 123.086 L, and P_Ai = 101.33 kPa.

Container A is connected to container B by a rigid adiabatic wall. Container B is also adiabatic and rigid and initially contains one mole of an ideal monatomic gas. T_Bi = 373 K, P_Bi = 101.33 kPa, and V_Bi = 30.607 L.

Replace the wall between A and B with a massless rigid diathermic partition. Now, reversibly and isothermally compress the gas in A until the temperature in B reaches 1500 K. The pressure in B is now 407.49 kPa. The heat absorbed by gas B (from A) is given by Delta U = nC_v Delta T = 14,054 J. Since all of the walls (except the separating partition) are adiabatic, then the 14,054 J must come from A. A simple calculation shows that V_Af = 39.883 L and P_Af = 312.68 kPa derived from the work done on A to produce 14,054 J.

The entropy change in A is Delta S_A = {14,054/1500 = -9.37 J mol^-1 K^-1. The entropy change in B_i is obtained from Delta S_B = nC_vln(T_Bf /T_Bi) = 17.35 J mol^-1 K^-1. The entropy change of the universe is Delta S_univ = Delta S_A + Delta S_B = +7.98 J mol^-1 K^-1 for the work done on compressing the gas in A plus heating the gas in B.

Please note that the only way the compression in A may occur isothermally is via a reversible process. This is the minimum work of compression. And, entropy changes must be calculated along reversible paths. If you assume the final state for A indicated by the author then 17,289 J are generated. Continuing, if you assume the author's final state for B, then 14,054 J are absorbed. What happens to the excess heat? The author's numbers are incompatible. There are no surprises here and no exceptions to the laws of thermodynamics.

Literature Cited

1. Belandria, J. I. J. Chem. Educ. 1995, 72, 116-118.