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Unformatted text preview: 1 bar Chemistry 440 Hour exam ...... ......... ..
Department of Chemistry, Oregon State University 21 October 2009 2 8 J/(K  mol) 2 0.08 L ~ atm/(K  mol)
105 Pa 2 1 atm I2 (be + we n = qby — / Peacth U+PV (2% (3—91)]: (gist l
i Q
V 3T
132%,); 1+5v+ pRT __3_V
vaPT )P CPI‘71 —1 = TZf/gy—l
$2 + $3 + . . . 1+sz+133p2+~~ for an adiabatic process; 7 = p: 2
V 0p
Cu 1. Provide the equation(s) and deﬁne terms. (a) (4 pts) State the First Law of Thermodynamics as it applies to the universe and
to a system (twoequations). AUUNV 1‘ AUW: +AUSvrr‘ : 0 WW enmm O’FH‘VL
vmmse ts (ionseNtol A074; :: 3.5g 4—Wm TKLC/‘angl In inferML emy '
Of +M ws—km as Mam of~
M Walesde 4— wovka MW ms. (b) (2 pts) Deﬁne an adiabatic process in terms of the thermodynamic variables of
the system (one equation). 9&4; = O (c) (4 pts) Deﬁne the conditions for the critical point of a single component ﬂuid (two
equations). (garzggrzo N @T;<%9T:O 2. (10 pts) The internal energy per mole U of a model ﬂuid is given by
WAT) = CRT  cm (1) Derive (5—)U if 0., and a are constants. dU:®O\T—O\O\P=O CST) : 0L
Wu 5"; 3. (10 pts) The van der Waals equation, without allowance for attractive forces, is given
by ‘
nRT . V — nb
For C52, by: 0.040 L / mole, and its molecular weight is roughly 80 g/ mole. P: (2) a) If the mass density of liquid carbon disulﬁde is 1.2 g/mL, what is its molar density
in moles / L? mg “000111: x mow: I200 =\5‘mdu/L
m. L 80g 759’ b) What is the lowest molar density (moles / L) at which Eq(2) fails? Vnnbzo, L: I :1S‘nmUL
b 034on2: .—
.— (L
v 4. (10 pts) A Joule—Thomson coefﬁcient has a value, an 2 —1.25 mK/bar. Does the
temperature increase or decrease and by how much when the pressure drop is 1 kbar? Mn :— 61?)” =(~l.2§&§)€103bax) bar ’—
.— AT‘ +1.26 K T In creases 5. Consider a. ﬂuid whose equation of state obeys
P ='pRT(1+ bp)  £192; p = n/V
where a and b are constant. Derive: (a) (12 pts) BAT), Bg(T), the second and third virial coefﬁcients, respectively;
n a.
2'; P/PRT: 1+b9—0LP/RTr—1432P'r83? 1‘" 8320 (b) (6 pts) and the dependence of dU(T, V) on V, note av: _ 8P _
(Wt—487%” (3—1,? = PRch?) so
V‘  @973 PRT(1+bP)"‘{PRTU+Lp) 4192}
r;qu 9. ,_ 2. '
om: OLP dv —or(g_)ow 6. (12 pts) One mole of argon (assumed to be an ideal gas) is compressed reversibly from 6 L to 1 L at 300 K. Calculate AU,won and qby. First, consider an ISOTHERMAL
process and second, an ADIABATIC process. — (m)... (a) isothermal O ’2... H' . 9. .4
CV AT CV AT . 1
(L) dU'r—O nag—{Pow '=— at“ — RT dV/v
.3 e ~Ww= RTlh(VZ/v.)—_ RTflh"'ihe] '= "KT
Wan :1 RT: 8L3’OOKza'4hy (b) adiabatic $4
E. _ yr D’ékCJL—l': 5J2;—
T» ’ V2 CV C"
2’3 943
T2 __ e __
T‘ " 'r) TZ’IR C 7. (10 pts) Prove that 8P
CUP: KT V
0( :— l V — 8V
cychc
W12. 8. (10 pts) Determine if dqby is an exact (differential for an ideal gas which undergoes a
reversible process. _dU=CV fiT =4er ~P0W
' iw: CvdT +Pdv I “W gs“):— @a=%(%1>= %
0 ,~ draw is m+ exact 9. (10 pts) dU(S, V) is an exact differential, where S is the entropy and T is the absolute
temperature. If ' dU(S, V) = TdS — PdV (6) then by the condition of exactness . . .
0V 5 ‘65 V ...
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