Answer
a. Work done by the system on the surrounding under given conditions is $-3.52 × 10^{3} J.$
b. Volume will decrease as total internal energy of system is decreased and heat is evolved from the system.
Work Step by Step
a)
data:
decrease in internal energy = ΔU = 1.65KJ
energy transferred in the form of heat to the surroundings = q = -1.87KJ
to find:
work done by the system on the surrounding = -W = ?
formula used:
at constant pressure
ΔU= q + W
solution:
according to sign conventions under given conditions
ΔU = -q + (-W)
W = -q - ΔU
by putting the values in formula applied
W = -1.87KJ - 1.65KJ
W = - 3.52 KJ
$as 1 KJ = 10^{3} J $ so
$W = -3.52 × 10^{3} J $
so work done by the system on the surrounding under given conditions is $-3.52 × 10^{3} J.$
b)
volume will decrease as total internal energy of system is decreased and heat is evolved from the system.
