Question

would someone please help with these, the answer isnt 2.121 or 2.128kJ.

How much work, w, must be done on a system to decrease its volume from 15.0 L to 6.0 L by exerting a constant pressure of 6.0 atm? 2.121 kJ

Answer 1

1. - work done = – POV 2 w - - P(V2 - V) w – – 6.0 (6.0-15.0) L.atm. w - 54.0 l.atm. As we knowl, l-atm = 101.32500 joules So, we 548101.325005 – 5471.55 I or 5.471 KI

2. Wo – PAV Firstly, convert units of the given data then apply formula 137.15 = 1.353. L-atm. 783 torr = 1.030 ath 64.4 m2 = 0.0644L Also, work done by system is negative we- P (V₂ - V.) (+ 1.353) L-atm = +1.030 atm (v.2 - 0.0644) 1.353 L-atm = V2 - 0.0644 L 1.030 atm 1.313 2 - 0.0644L - 12 V2 = 1.2486 L 1248.6 m2