Question

A thermodynamic system is taken from state a to state c in the figure below along either path abc or path adc. Along path abc the work W done by the system is 450 J. Along path adc, W is 120 J. The internal energies of each of the four states shown in the figure are Ua = 150 J, Ub = 240 J, Uc = 685 J, and Ud = 355 J. Calculate the heat flow Q for each of the four processes ab, bc, ad, and dc. In each process, does the system absorb or liberate heat?

Processab Q=J absorbliberate	Processbc Q=J absorbliberate
Processad Q=J absorbliberate	Processdc Q=J absorbliberate

Answer 1

ΔQ = ΔU + ΔW

Process ab

V is constant , ΔW = 0

ΔQ =ΔU = Ub - Ua = 240-150 = 90 J : absorbed

Process bc

ΔQ = (Uc-Ub) +ΔW = (685-240)+450 = 895 J : absorbed

Process ad

ΔQ = (Ud-Ua) +ΔW = (355-150)+120 = 325 J : absorbed

Process dc

V is constant , ΔW = 0

ΔQ = (Uc-Ud) = (685-355) =330 J : absorbed