# Special Case of the first law of thermodynamics

## Homework Statement

When a system is taken from state i to state f along path iaf in Fig. below, Q = 50.6 cal and W = 20.3 cal. Along path ibf, Q = 36.6 cal. (a) What is W along path ibf? (b) If W = -13.7 cal for the return path fi, what is Q for this path? (c) If Eint,i = 11.1 cal, what is Eint, f? If Eint,b = 22.1 cal, what is Q for (d) path ib and (e) path bf ?

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c18/pict_18_49.gif

## The Attempt at a Solution

(a), (b), (c) worked. (d) and (e) didnt.
(a) The change in internal energy ΔEint is the same for path iaf and path ibf. According to the first law of thermodynamics, ΔEint = Q – W, where Q is the heat absorbed and W is the work done by the system. Along iaf
ΔEint = Q – W = 50.6 cal – 20.3 cal = 30.3 cal.
Along ibf ,
W = Q – ΔEint = 36.6 cal – 30.3 cal = 6.6 cal.
(b) Since the curved path is traversed from f to i the change in internal energy is –30 cal and Q = ΔEint + W = –30.3 cal – 13.7 cal = – 44 cal.
(c) Let ΔEint = Eint, f – Eint, i. Then, Eint, f = ΔEint + Eint, i = 30.3 cal + 11.1 cal = 41.4 cal.
(d) The work Wbf for the path bf is zero, so Qbf = Eint, f – Eint, b = 41.4 cal – 22.1 cal = 19.3 cal.
(e) For the path ibf, Q = 36 cal so Qib = Q – Qbf = 36.6 cal – 19.3 cal = 17.3 cal.