Answer
work done on the system during compression is:
$ 1.263 × 10^2 J.$
Work Step by Step
initial volume = V1= 2.5 l
final volume = V2 = 1.25 l
change in volume= ΔV= 1.25l
$pressure= P= 1.01 × 10^5 Pa$
to find:
work done in joules
formula used:
as pressure is kept constant, and work is done on the system, so it would be positive. So:
W = PΔ V
solution:
w=PΔV
$w= (1.01 × 10^5 Pa) × (1.25l)$
as
$1pa = 1 Kg/m.s^2
1l = .001 m^3$
so by converting into required units
$W = (1.01 × 10^5 Kg/m.s^2) × (0.00125m^3)
W = 1.263 × 10^2 Kgm^2/S^2$
as
$1Kgm^2/S^2 = 1J
W= 1.263 × 10^2 J$
so work done on the system during compression is:
$ 1.263 × 10^2 J.$
