Special Case of the first law of thermodynamics

[aub]

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 ?


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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.

 

[anathema]

Hi, great job on solving parts (a), (b), and (c)! For part (d), you are correct that the work Wbf is zero, but keep in mind that the change in internal energy ΔEint is not just the difference between Eint,f and Eint,b. It also includes any heat absorbed or work done by the system. So for path bf, we can write:
ΔEint = Qbf – Wbf
Since Wbf = 0, this simplifies to:
ΔEint = Qbf
Therefore, for path bf, Qbf = ΔEint = Eint,f – Eint,b = 41.4 cal – 22.1 cal = 19.3 cal.
For part (e), you are correct that Qib = Q – Qbf, but the value for Qbf should be 19.3 cal, not 19.6 cal. So the correct answer for Qib would be:
Qib = Q – Qbf = 36.6 cal – 19.3 cal = 17.3 cal.
Great job on your attempt! Keep up the good work!