J.R. S. answered 11/15/23
Ph.D. University Professor with 10+ years Tutoring Experience
At constant pressure, work (w) is equal to the pressure times the change in volume, and if the change in volume is positive, w will be negative, and that means work was done BY the system ON the surroundings.
w = -P∆V
So, for this problem we want to see which reactions have a POSITIVE change in volume:'
a). 2A(g) + 3B(g) ==> 4C(g) has 5 mols on left and 4 mols on right = reduction in volume = no
b). A(s) + B(g) ==> 2C(g) has 1 mol gas on left and 2 on right = increase in volume = yes
c). 2C(g) + B(g) ==> 3C(g) has 3 mol gas on left and right = no change in volume = no
d). A(s) + 2B(g) ==> C(g) has 2 mol gas on left and 1 on right = reduction in volume = no
So, the only system that does work on the surroundings would be system (b).