# Jhgkjh Essay

8376 Words 34 Pages
Introduction to the Thermodynamics of Materials
Preliminaries
‡ Settings
Off@General::spellD

‡ Physical Constants Needed for Problems ü Heat Capacities The generic heat capcity c 105 bT Å Cp = a + ÅÅÅÅÅÅÅÅÅÅ + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ; T2 103 The heat capacities of various elements and compounds are CpAgs = Cp ê. 8a Ø 21.30, b Ø 8.54, c Ø 1.51 8.3144 , Rla -> 0.082057 < ; The number of moles can be calculated from the starting state: P 1 V1 nmols = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. nums ; Å Rla T1 subs = Append@nums, n -> nmolsD 8V1 Ø 10, T1 Ø 298, P1 Ø 10, P2 Ø 1, R Ø 8.3144, Rla Ø 0.082057, n Ø 4.08948< Finally, this constant will convert liter-atm energy units to Joule energy units. All results are given in Joules: laToJ
Reversible isothermal expansion to 10 atm pressure

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Final volume is P1 V1 P2 = 10 ; V2 = NA ÅÅÅÅÅÅÅÅÅÅÅÅ ê. initE Å P2 22.5 For isothermal process, DU=0 and q=w. They are given by (using PV = nRT): V2 q = w = 101.325 P1 V1 LogA ÅÅÅÅÅÅÅ E ê. init V1 9243.84 For an ideal gas, DU = 0 for an isothermal process (U only a function of T). Finally DH=0 because DU=0 and PV = constant. b. Reversible adiabatic expansion to P=10 atm. The final volume is
1êg g i P1 V1 z j ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ y Å ê. Append@init, g -> 5 ê 3DE V2 = NAj z k P2 {

19.1314 The final temperature is T1 P2 V2 T2 = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. init Å P 1 V1 255.085 The number of moles is P1 V1 n = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. init Å T1 Rla 9.13999 Thus the total change in internal energy is
T2

dU = ‡

T1

n 1.5 R „ T ê. init

-5119.88 The heat work done for his adiabatic process is

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