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Unformatted text preview: (/31) di : + L$Z+xﬂws a» X33 [new aha:Aﬂg d? {243+ am *1 ax; +[3LL + mm 13 AV 0‘96 {3% + $005 $ + m QM
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=5 +ULdinJl
: MSWM \/ wom’r Jm SMow 5 :3 aggputjﬂtsm —.. ﬁx’kguﬂmsﬂj 023L449”: 2$+w33 \/ P27.15) A pellet of Zn of mass 10.0 g is dropped into a ﬂask containing dilute H2804 at a
pressure of P = 1.00 bar and temperature T = 298 K. What is the reaction that occurs?
Calculate w for the process. Zn(s) + HZSO4(aq) —» Zn2+(aq) + 8042‘ (aq) +H2(g) The volume of H2 produced is given by 10.0 lmolH 8.3l4J I‘lK'1 29 K
=_——g——:1—>< 2 x__llio_5__L8__= 3_79X103m3
65.39 g(mol Zn) lmolZn 1X10 Pa
w=—P AV atemal AV z volume of H2 produced.
w: —1X105Pa><3.79><10'3m3 = —379 J P2.17) 1 mol ofan ideal gas, for which Cm: 3/2R, initially at 200°C and 1.00 x 106 Pa
undergoes a twostage transformation. For each of the stages described in the following
list, calculate the ﬁnal pressure, as well as q, w, AU, and AH. Also calculate q, w, AU,
and AH for the complete process. ~ a) The gas is expanded isothermally and reversibly until the volume doubles. b) Beginning at the end of the ﬁrst stage, the temperature is raised to 800°C at constant volume.
a) P2 =5£=5=0500x10615a
V2 2
w=—nRT lnI—fz—=—8.314Jmol"K"x1n2=—1.69><103J 1 AU=O andAH=0 because AT =0 q =—w=1.69x103J EL P = TZP1 = 353 K >< 0.500 X106Pa
P2’ 2 1; I 293 K
AU = nnymAT=1.5x8.3l4Jmol'lK‘1X(353 K—293 K): 748 J
w=0 because AV=0 ' q = AU = 748 J = 6.02 x lOsPa T
b—1=
)1? AH = nprmAT = n(C,,,m + R) AT = %x 8.314 J min1Kl ><(353 K—293 K) =l.25x103J" . 7 For the overall process, q =1.69><103J + 748 J = 2.44x103J w=—1.69><103J+0=——1.69><103J
AU=O+7481=748 J AH =0+1.25X103J=1.25X103J P2.22) A cylindrical vessel with rigid adiabatic walls is separated into two parts by a
frictionless adiabatic piston. Each part contains 50.0 L of an ideal monatomic gas with
C14," = 3/2R. Initially, T , = 298 K and P, = 1.00 bar in each part. Heat is slowly
introduced into the left part using an electrical heater until the piston has moved
sufﬁciently to the right to result in a ﬁnal pressure Pf: 7 .50 bar in the right part.
Consider the compression of the gas in the right part to be a reversible process. a) Calculate the work done on the right part in this process and the ﬁnal temperature in ‘ the right part. b) Calculate the ﬁnal temperature in the left part and the amount of heat that ﬂowed into
this part. The number of moles of gas in each part is given by
PV. 1.00 barX50.0 L
— ' ' — = 2.02 mol ' 1'27. — 8.3145><10'2Lbarrnol‘1K’1 x 298 K n a) We ﬁrst calculate the ﬁnal temperature in the right side. T. V. T. P. T. P, T. P, Tf = 2.24X298 K = 667 K 3 X 8.314 I mol—1K"1
——————X( AU = w = nCVIMAT = 2.02 mol>< 2 667 K— 298 K) = 9.30X1031 b) We ﬁrst calculate the ﬁnal volume of the right part.
RTJ _ 2.02 molx8.3 l4><10'2Lbarmol“K‘1 X 667 K = 14.9 L ’f P1 7.50 bar
Therefore, mf= 100.0 L — 14.9 L = 85.1 L. P V 7.50 hm T = If If =———————= 3 a? s‘w‘uv
’f nR 2.02 molx8.314X10‘2Lbarmol‘1K“ 3'80X10 K Vill Pll ‘
J axos’ram \5 a5? AU=nCMAT=202 mol><8.314Jmol“K"x(3.80x103K—298 K)=58.8x103 . .b .
From Part (a), w=9.30 X103J m “Um
q=AU—w=58.3 x 103J+9.30 x103J=67.3 x 1031 P2.26) An ideal gas undergoes a singlestage expansion against a constant external
pressure Paternal at constant temperature from T, Pi, Vi, to T, Pf, Vf. a) What is the largest mass m that can be lifted through the height h in this expansion?
b) The system is restored to its initial state in a singlestate compression. What is the
smallest mass m’that must fall through the height h to restore the system to its initial
state? c) Ifh = 10.0 cm, P,=1.00X106 Pa, Pf: 0.500x106 Pa, T= 300 K, and n = 1.00 mol
calculate the values of the masses in parts (a) and (b). 3 Consider the expansion —P V —V
m z .4 f .>
gh
for the ﬁnal volume to be Vf, the external pressure can be no bigger than Pf
—P V — V.
mm = ,< f .>
gh b) Consider the compression gh
for the ﬁnal volume to be K, the pressure can be no smaller than P,
—P. V—V
m; : .< , .)
n c) r
1 —1
V = nRT =1.00 mol>< 8.314 Jmaol K X 300 K = 249X104m3
‘ B 1.00X10 Pa
Pfo=P,. i
6 3 3
Vf :_IiL= 1.00 X 10 Pa >< 2.469 X 10 m =4.98x10_3m3
f 0.500 X 10 Pa
— . 6 4. 8 10'3 3—2.49x10'3 3
m = 0500x10 Pax( 9 x m m)=1.27X103kg
m 9.81m S"2 XO.100 m
—1.00><106P X 2.49x10‘3 3—4.98><103m3
m’. = ————a(—2—m——————) =2.54><103kg
m’" 9.81ms‘ XO.100m ' P230) One mole of N2 in a state deﬁned by T, = 300 K and V, = 2.50 L undergoes an
isothermal reversible expansion until Vf= 23.0 L. Calculate w assuming a) that the gas is
described by the ideal gas law and b) that the gas is described by the van der Waals
equation of state. What is the percent error in using the ideal gas law instead of the van
der Waals equation? The van der Waals parameters for N2 are tabulated in Table 7.4. a) for the ideal gas
V
wmmiue = —nRT1n—i = —1mol><8.314 Jmorln‘1 x300 K xln 23'0 L = _5,54x1031
Vi 2.50 L
b) for the van der Waals gas
V’ V’ RT a
w = — .lPwemaldV =_J{Vm _b _;3‘ V; ,
V! I’/
=  j M V+ j 1’; dV
V. V»: _b Vi Vm The ﬁrst integral is solved by making the substitution y = V,,, — b. y/ 1 RT )dV=— {(Ejdy =—RT[ln(Vf +b)1n(Vi +b)] Vmb y! y V: Therefore, the work is given by V—b) 1 1
w=—nRT1n((V:_b) +a[E—7f] = —1 moix8.314 Jmor‘Kl x300 K x In23.0 L—0.0380 L
2.50 L—0.0380 L
105PaX10'6m6[  1 _ 1 ]
bar L2 2.50><10—3m3 23.0><10‘3m3
w=—5.56x10"J+48.7 J=—5.52><103J
—5.52x1031+5.54x103J = —————————— = ——0.4‘V
Percent error 100x _5.52X1031 o +1.366 L2 bar>< 3P 3:7 3T— 313
relationship, show that the isothermal compressibility and isobaric expansion coefﬁcient are related by = .
T P am) = 1(1)
8P 3T P T 3T 8P T P 1 8V 1 8V
Because ,3 — 7[3fjp and K — —7[¥l (8311 “(333541. °r [i—ﬁl “(SEl A P3.7) Because Vis astate function, {ivy/j ] =[ 3 [8V] ] ~ USng this
P T T P P3.9) The molar heat capacity Cg," of 802(g) is described by the following equation
over the range 300 K < 1700 K. CP,m _3 T _7 T2 .9 T3 . . .
=3.093+6.967X10 —45.81><10 —3+l.035X10 7. lnthls equatlon, Tls
R  K K K _ T" n the absolute temperature in Kelvin. The ratios ensure that Cg," has the correct dimension. Assuming ideal gas behavior, calculate q, w, AU, and AH if 1 mol of SOz(g)
is heated from 75° to 1350°C at a constant pressure of 1 bar. Explain the sign of w. 2
1m 3.093+6.967x10‘35—4ss1x10‘71 T! 2
AH = q = 7‘ ICp'de = 8.314 Jmol"K'1 x I T3 K K 61%
T’ 348” +1.035x10‘9ﬁ?
1 T T2 .6 T3 10 T4 1523.15
:1 mol><8.3l4Jmol x 3.093—+0.0034835—7—1527x10 7+2.5875x10 _4
K K K K 343.15 _ 1 molx3.279x1041m01" +7.279X10“Jmol’15.375X10“Jmol‘1
+1.490><104Jmo1"‘K‘1 = 6.67x104J AU=AH—A(PV)=AH—nRAT
= 6.673x104J—1mo1x8.3141m01"K1x(1623.15 K—348.15 K) =5.61><104J
w = AU—q = 5.613x104J—6.673x10“J= —1.06x1o‘J Because the gas is heated, it expands. Therefore AV> 0 and w < O. a a1» 3 ap . .
8—11 y T v and y El, T, determine 1fdPisanexact differential. m (VMR— Therefore, dP is an exact differential. P3.16) A mass of35.0 g ofH20(s) at 273 K is dropped into 180.0 g ofH20(l) at 325 K
in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equilibrium has been reached. Assume that Cpl," for H20 is constant at its values for
298 K throughout the temperature range of interest. i H 0 ice ' H 0 H20 _
niceAHﬁcl:ion + niceCPjn (Tf )+ nHEOCP; )— 0 , Au Au H10 H20_ ice
_ "AuCPmIi +nH10CP,m 71 AH T _ ice ﬁm‘an =
f nAu CRT" + "mac:an
&x75.291 JK‘Imol‘1x273 19%an JK‘lmol‘1x325 K
18.02 g1cemol 18.02 gHZO mol
———35'°. gm lx6008 Jmor1
18.02 glce mol"
35.0 gice 100.0 gHZO ——_——_lx75.291 JK‘lmol‘1+—————_l—x75.291 JK'Imol'l
18.02 glcemol 18.02 gH20m01 rf =303K ...
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 Fall '06
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