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UFID AN 3L3 (33K Key EMA 4314
Fall 2010
Exam #1 (100 points) Gas Equations (30 points)
a. Define equation of state (5 points) than 52.51 Mufﬁn“) :8 (317’ «(it deﬁnes Oi K(m%'l€?ﬂ$le3 arm/n3 games, ire LU we; (m3 ﬁlmy/km (EEK if) S: “A? ( T; V; n) b. State the criteria that must be obeyed for an equation of state to show a phase
transition (5 points) 23?: ”Q 2r?
M ,1 ‘ 37” or it: i c. The Berthelot equation is: RT (1 :Vm—b T143; P Show that the Berthelot equation displays a phase transition and determine the critical volume, temperature and pressure in terms of the constants a, b and R. (20 points) Qt: _ ' “2;: Cl £1 “w gill wt» «91$
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2*“ W «an, 21% “W Cwa ﬁggxﬂ amwj x%_ 2. Gas Equations (20 points)
a. On the axes below, sketch isotherms for the van der Waals gas; include temperatures
above, below and at the critical temperature l3 b. Describe what is meant by van der Waals loops; illustrate these in the figure Wm WW [Wu CW 31(3er (7% m 49?th Cﬁiﬁ‘ggmm OR or mm WEN» LEMR {MK Cm two Wm ‘3; WM lb m 51“” 1% c. What is the Maxwell construction; relate this to van der Waals loops “\5 Wﬁcwdll (Km f’mfi‘ 1M ﬁmw‘w m MW lows L3 Wlﬁﬂﬂr’)
4mm 94th at UN c& f) :: (mi‘m‘f .) gird.“ 4H0 yttrium? “:3 A (Mai 13 Q“ £09,th a ,
gym 3. First Law and related topics (25 points)
a. Define the term state function (5 points)
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SrMeW (tb +M ”firm b. State the first law of thermodynamics in the form (5 points): AU= in“ as; c. A particular gas can be described by Ple—b) = RT. Determine the amount of work done
by this gas as it expands isothermaliy from a molar volume of 3b to a molar volume of 21b (5 points) W 2.1!: wijriv d. The volume of a particular liquid varies with temperature as V=V’(O.75 + 3.9x10"”’T +
1.48x10‘6T2) where V’ is the volume at 300K. Determine the coefficient of thermal
expansion at 320K (5 points) e. The isothermal compressibility of copper at 293K is 7.35 x 10'7 atm"1. Calculate the
pressure that must be applied in order to decrease the density by 0.08 percent. (5
points) is wﬁéﬂf) 1 lMW‘ «,5
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v QR ”' 4. Thermochemistry (25 points)
a. What is Hess’s Law? (5 points) ’ TR $1361.41 dong triﬂﬁtLp? (61 an mkﬂL rmcﬁm it HM .3 Lem
’H‘k ﬁrm tiara 3m, {”haiilow;3 7m mammal; Maelstrom? mwéﬁ Mwh i ﬁaﬁwﬁ mag) that an Viciw Consider the foilowing three chemical equations (1) H2(g)+l2~)2H(g) ArH0=+52.96 kimorl «QM a; {AM/u) Waugh
(2) 2Hz(g)+oz(g)92H20(g) ArHoz—483.64 rumor1 [Si {VLQMQ CAM/1‘
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(3) 4Hl<gi+ozigi9212(5)+2H20(g> WWW WM is c» gasa
K LﬁlU Mei h (WW9? MAM}; .q Given the reactions (1) and (2), determine at 298K (5 points each) W Q‘¢V"@m~3/§>«r 1%ﬁ‘ ’
b. A,H0 for reaction (3): Ag H DC “6“th 3) :. A? (HMC WAJWM :2) W 9 [Ar H§(F‘*5’\Qv€"ﬁm i) 2:: w 482%; 4! w Qumﬂf‘it) Haw“!
1: Mg Skirt 3L kSMQJ I c. ArUO for reaction (3) Arm Ago Mi PM
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Ly U ... $5? SQ, gt} {LT (g) _,, . d. AFHO for Hl(g) AYHGCHii/Qiy): LCEQSTQ) ,9 e. ArHO for HZO(g) Qwiwigi) «2 5(«~4—87,é4») ...
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 Fall '10
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