# Jhgkjh

Submitted By LimHyeung
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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 = 101.325 ; ü 1. Reversible, Isothermal Process In an isothermal process for an ideal gas, DU = 0 ; DH = 0 ; thus heat and work are equal and given by: P2 q = w = n R T1 LogA ÅÅÅÅÅÅÅ E J ê. subs P1 -23330.9 J

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ü 2. Reversible Adiabatic Expansion In an adiabatic expansion q = 0; and P V g is a constant. Thus the final state has
1êg g P2 V2 i P1 V1 y Å ; T2 = ÅÅÅÅÅÅÅÅÅÅÅÅÅ ê. g -> 5 ê 3 Å V2 = j ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ z j z n Rla k P2 { P1 V 1 P2 I ÅÅÅÅÅÅÅÅÅÅÅÅÅ M P ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 ÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ Å n Rla
5ê3

3ê5

For an ideal gas cv = 3R/2; thus 3 DU = ÅÅÅÅ n R HT2 - T1 L ê. subs 2 -9147.99 or we can use 3 DU = ÅÅÅÅ HP2 V2 - P1 V1 L laToJ ê. Append@subs, g -> 5 ê 3D 2 -9148.02 For some numeric results, the final temperature and volumes were ad2 = N@8V2 , T2 < ê. Append@subs, g -> 5 ê 3DD 839.8107, 118.636< The work done is dw = -DU 9148.02 For an ideal gas c p = 5R/2; thus the enthalpy change is 5 DH = ÅÅÅÅ HP2 V2 - P1 V1 L laToJ ê. Append@subs, g -> 5 ê 3D 2 -15246.7 or

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5 DH = ÅÅÅÅ n R HT2 - T1 L ê. subs 2 -15246.7 For numerical results in the subsequent...

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