Chemistry and Chemical Reactivity (9th Edition)

$4.55 × 10^2 J$.
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: $4.55 × 10^2 J$.