## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions - Study Questions - Page 217a: 25

#### 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: $1.263 × 10^2 J.$

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