This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 12.5 kJ of heat flows into the system and 4.1 kJ of work is done by the system
on the surroundings.
∆E = q + w = (+12.5 kJ) + (-4.1 kJ) = +8.4 kJ 6-1 Chapter 6 Thermochemistry PV Work
work = force x distance w = F∆h P = F/A so F = PA so P = F/A P = F/A W = P x A x ∆h
but ∆V = Vfinal – Vinitial = A x ∆h, therefore
∆h w = P x A x ∆h = P∆V
since ∆V is positive if the system does work on the surroundings, we must
change the sign so w is positive when done on the system
w = -P∆V
Calculate w for a gas compressed from 5.0 L to 1.0 L at a constant external pressure of 5.0 atm.
w = -P∆V = -(5.0 atm)(1.0 L – 5.0 L) = +20 L atm
to convert L atm to J, remember that 1 atm = 101,325 pascals 2 1 Pa = 1 N/m 1 J = 1 Nm 1 L atm = (1 dm3) x (101,325 N/m2) = (1 x 10-3 m3) x (101,325 N/m2) = 101.325 Nm = 101.3 J
A balloon absorbs 2.0 kJ of heat and expands from 2.0 L to 3.2 L against a constant pressure of 0.950 atm.
w = -P∆V = -(0.950 atm) x (3.2 L – 2.0 L) x 10...
View Full Document