Chapter 3

# Marandsnoteschapter3thesecondlawofthermodynamics 118

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Unformatted text preview:  = ‐Pext ΔV  Therefore:  3R (TB ‐ TA) = ‐Pext (VB ‐ VA)  Since VB = n R TB / PB and PB = Pext we get:  TB = (TA + Pext VA / 3R) / (1 + n / 3) which leads to TB = 192.44 K and   VB = 32 L.  Now that we know the final and the initial states, we can design any  number of reversible processes between A and B (see figure below).  Marand’s Notes: Chapter 3 ‐ The Second Law of Thermodynamics  108  1)  We can envision, an isobaric slow heating from state A to a state C,  characterized by: PC = PA and VC = VB. The temperature at state C is defined  by TC = PC VC / nR .    Therefore, TC = PA VB / nR = 1924.3 K. Then, we carry out an isochoric slow  cooling from state C to state B.  The change in entropy between A and B, ΔSA‐B = ΔSA‐C + ΔSC‐B is given by:  ΔSA‐C  = n CPm ln(TC / TA) = 2 (5R/2) ln (1924.3/300.68) = 77.16 J/K and  ΔSC‐B  = n Cvm ln(TB / TC) = 2 (3R/2) ln (192.44/1924.3) = ‐ 57.43 J/K.  The change in entropy for the A‐B process is therefore equal to 19.73 J/K    2)  We could have also envisioned the path A  D  B, which consists  in a slow isochoric cooling from A to D followed by a slow isobaric heating  from D to B (see figure next page).  State D is characterized by PD = PB and VD = VA. In this case TD = PD VD / n R  which is to say TD = PB VA / n R = 30.07 K  The entropy change between A and B is therefore the sum of the entropy  changes between A and D and between D and B. Therefore:  ΔSA‐D  = n CVm ln(TD / TA) = 2 (3R/2) ln (30.07/300.68) = ...
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