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## 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.(e) For the path ibf, Q = 36 cal so Qib = Q – Qbf = 36.6 cal – 19.3 cal = 17.3 cal.