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Unformatted text preview: CHEM 131
MIDTERM EXAM K. Lindenberg Fall 2006 NAME W STUDENTID
1. /25
2. /20
3. /20
4. /25
5. /10
6. /5
TOTAL MIDTERM /105 TOTAL NUMBER OF PAGES INCLUDING THIS COVER PAGE: 14. NOTE: Your answers must be Clear to me. Make sure to offset your answer clearly — if it is
lost amidst a lot of other stuff I may not ﬁnd it. If you have more than one answer I won’t
know which one you want me to use. NOTE: As announced, this exam is based entirely on your homeworks. After
the exam I will point you to the homework problems that appear on the exam.
The numbers may have been modiﬁed from those on the homeworks in some
of the problems. ANOTHER NOTE: Unless told otherwise, you can assume, without further justiﬁcation, that all gases are ideal. CHEM 131 1. TRUE AND FALSE  25 points max, 10 points min
All of these T/F questions were on your homeworks. I am NOT trying to trip you
up or fool you, so I have NOT chosen the really subtle ones. These are ones that you
should have clear and ﬁrm answers to. Also7 they have been listed in the order in which
they appeared on your homeworks. YOU NEED ONLY ANSWER T OR F. Correct
choice = 0.5 points, no choice = 0 points, incorrect choice 2 —0.2 points. 2.4a (1) The P — V work in a mechanically reversible process in a closed
system always equals —PAV. (2) The inﬁnitesimal P — V work in a mechanically reversible process
in a closed system always equals —PdV. (3) For every process, AEsyst = —AES,W. (4) For every cyclic process, the ﬁnal state of the surroundings is
the same as the initial state of the surroundings. (5) AH is deﬁned only for a constant—pressure process.
(6) For a constant—volume process in a closed system, AH 2 AU. (7) A thermodynamic process is deﬁned by the ﬁnal state and the
initial state. (8) AT 2 0 for every adiabatic process in a closed system. (9) If neither heat nor matter can enter or leave a system, that
system must be isolated. (10) Since a Carnot cycle is a cyclic process, the work done in a
Carnot cycle is zero. (11) A change of state from state 1 to state 2 produces a greater increase
in entropy when carried out irreversibly than when done reversibly. (12) The heat q for an irreversible change of state from state 1 to
state 2 might differ from the heat for the same change of state carried out reversibly. (13) The system’s entropy change for an adiabatic process in a closed
system must be zero. (14) For a reversible process in a closed system, dq = TdS. (15) For a closed system, AS can never be negative. lﬂ 11" in lab we HM In W1 l‘“ H 3.114
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3.4 la. (16) For an adiabatic process in a closed system7 AS cannot be negative.
(17) For a process in an isolated system, AS cannot be negative.
(18) For an adiabatic process in a closed system, AS must be zero. (19) For any closed system, equilibrium corresponds to the position
of maximum entropy of the system. (20) The entropy of an isolated system can never decrease.
(21) For every process in an isolated system, AT : 0. (22) If a closed system undergoes a reversible process for which AV = 0,
then the P — V work done on the system in this process must be zero. (23) AS when 1 mol of N2(g) goes irreversibly from 25°C
and 10 L to 25°C and 20 L must be the same as AS when
1 mol of N2(g) goes reversibly from 25°C and 10 L to 25°C and 20 L. (24) For a closed system with P — V work only, the cyclic integral f PdV
is equal to zero. (25) For a closed system with P — V work only, the cyclic integral f dqm, / T
is equal to zero. (26) For a closed system with P — V work only, the cyclic integral f dqm,
is equal to zero. (27) The relation AG = AH GITAS is valid for all processes.
(28) Always, G = A + PV. (29) The chemical potential ,u, is a state function. (30) it, is an intensive property. (31) p, in a phase must remain constant if T, P, and at, remain
constant in a phase. (32) The chemical potential of benzene in a solution of benzene and
toluene must equal the chemical potential of toluene in that solution. (33) The chemical potential of sucrose in a saturated solution of
sucrose in water at 300 K and 1 bar must be equal to the molar Gibbs
energy of solid sucrose at 300 K and 1 bar. 4 in l” H 7+1 ya H H In in WWW“ 1*“ la lii H l—l,§QC (34) AG is always zero for a reversible process in a closed 145%. 4.0m
43% Mt Lark system capable of P — V work only. (35) The work done by a closed system can exceed the decrease in the
system’s internal energy. 36 Gs st + Gsurr is constant for any process.
( y
(37) A353,“ + A33”, is positive for every irreversible process. e term 3 an ar state imp 1es t at t e tem erature lS .
(38)Th t d d ' 1‘ h h p ' 0°C
(39) The term standard state implies that the temperature is 25°C.
(40) Doubling the coefﬁcients of a reaction doubles its AH". (41) The reaction N2 + 3H2 —> 2NH3 has 2, V, = —2. (42) The reaction N2(g) + 3H2(g) —> 2NH3(g) has AH; < AU;. (43) When an exothermic reaction in a closed system with P — V work only is run under isobaric and adiabatic conditions, AH = 0. (44) When a substance is in its thermodynamic standard state, the
substance must be at 25°C. (45) G of an element in its stable form and in its standard state at
25°C is taken to be zero. (46) The chemical potential of an ideal gas 1' in an ideal gas mixture
at temperature T and partial pressure B equals the chemical potential
of pure gas 1' at temperature T and pressure 13,. (47) For an ideal gas reaction, K 2, is always dimensionless. (48) For an ideal gas reaction, K109 for the reverse reaction is
the reciprocal of K; for the forward reaction. (49) For a particular ideal gas reaction, K2, is a function of
temperature but is independent of pressure and of the initial composition of the reaction mixture. (50) For an ideal gas reaction, AH" must be independent of T. f” b1 H ltl‘lldl‘hbblnld It H H l“1 lnH CHEM 131 Moi WM” 2. 20 points total, 4 points each part
Consider a perfect gas contained in a cylinder and separated by a frictionless adiabatic
piston into two sections, A and B. Section B is in contact with a water bath that
maintains it at constant temperature. Initially TA = T3 = 320K, VA = VB = 1.50 L,
and n A 2 113 = 1.75 mol. Heat is supplied to Section A and the piston moves to the
right reversibly until the ﬁnal volume of Section B is 1.00 L. Calculate the following.
Assume CV,m = 20.0 J K‘1 mol‘l. Also7 R = 8.314 J K‘1 mol‘l. (a) The work done by the gas in Section A. ‘ ’ Vg),_ “£1 __
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Data for this problem: CW. = 75.291 J /mol K for water. AH," for melting of ice is
6020 J ﬁnal. The molecular weight of water is 18 g/mol. (8.) Calculate the change in entropy when 95 g of water at. 0°C is added to 195 g
of water at 95°C in an insulated container. mic «n “T: :_(n1C "MOE
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/’M 4. 25 points total 5 point each part For each of the following processes, deduce Whether each of the quantities q, w, AU,
AH, AA, and AG is positive zero, or negative. a)Reversible melting of solid benzene at 1 atm and the normal melting point Heat QM KW MW [E] AHzgh 50
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WM'W’ “Mr 5° (b) Reversible melting of ice at 1 atm and 0°C. 10 CHEM 131 (c) Reversible isothermal expansion of a perfect gas. ﬁ»cm§l‘v So lﬂl—lzol
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ACZrT—Ag $0 91420 (A940! (d) Adiabatic expansion of a perfect gas into a vacuum (Joule experiment). lW’Ol ﬂdzﬁww solAULOl Au: AU +AUOV) =AU+mR£W
Bull Ali—.0 W ﬂf;0 .30 lbl—l70i AS >0 (inmggw What‘s, WM) ﬂairTAS so GEE
DC: ”TBS So 11 CHEM 131 )Joule— Thomson adiabatic throttling of a perfect gas L.“ :51 1*“ <3 \ T— m5‘1’m‘)’ AJLaLmLt'C 9" 12 CHEM 131 5. 10 points
The reaction N2(g) <—+ 2N(g) has K; ; 3 x 10‘6 at 4000K. A certain gas mixture at 4000 K has partial pressures PM = 650 torr, PN = 3.00 torr, and PHe = 300 torr.
Is the mixture in reaction equilibrium? If not, will the amount of N(g) increase or
decrease as the system proceeds to equilibrium at 4000 K in a ﬁxed volume? 1 bar = 750 torr, «Ml“KB Moi. W [email protected] ~13 {um/1A MW ,, 13 W CHEM 131 6. 5 points
The Gibbs free energy of a certain gas is given by 1 1
Gm = RTlnp+A+Bp+ §Cp2+§Dp3 Where p is the pressure and A, B, C, and D are constants (independent of T and p).
Obtain the equation of state of the gas. ’30 , 9F _
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9P. 1' a 9 > 14 ...
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 Fall '08
 Lindenberg

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