#### Answer

$4.55 × 10^2 J$.

#### Work Step by Step

data:
initial volume = V1= 0.75 l
final volume = V2 = 1.20 l
change in volume= ΔV= 0.45l
$pressure= P= 1.01 × 10^5 Pa$
to find:
work done in joules
formula used:
As pressure is kept constant, and work is done by the system, it would be negative.
W = -PΔ V
solution:
w=PΔV
$w= -(1.01 × 10^5 Pa) × (0.45)$
as
$1pa = 1 Kg/m.s^2
1l = .001 m^3$
so by converting into required units
$W =- (1.01 × 10^5 Kg/m.s^2) × (0.0045m^3)
W = -4.4545 × 10^2 Kgm^2/S^2$
as
$1Kgm^2/S^2 = 1J$ and limiting it to two significant figures
$W= -4.55 × 10^2 J$
so work done on the system during compression is:
$4.55 × 10^2 J$.