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Unformatted text preview: 0 Solutions to Even Numbered Thermodynamics Problems
0 o w ‘
. 7_52 Ass1gned to Date m 132.  54.0 g
= E = ,
n 39.948 g mol—1 1 352 m0] AB = O (isothermal, reversible) __ V2_ PI
20— nRTln—ﬁ— _nRTln1—__,2 1.50 atm
= — 1.352 1 8.315 J 1'1 K‘1 4 =
( mo mo )( 00 K) In 400 atm +4410 J g = —'w = —4410 J
AH. = 0 because AT = 0 754 For an adiabatic process, q = 0, and E . HW=eW Here we have P1 = 1.00 atm, P2 = 2.00 atm, and ‘ —1 —1
3 V1 = nRT1 = (2.00 mol)(0.08206 L atm mol K )(273.15 K) = 44.83 L
. : P1 1.00 atm
CP CP 29.3 7 = E = op — R = 29.3 .— 8.315 = 1’396_ V2 7 P1
(—) _ E _ 0.500 V2 = 0.6087(44.83 L) = 27.29 L ' _ P2112 _ (2.00 atm)(27.29 L) _
T2 — nR _ (2.00 mol)(0.08206 L atm mol‘lK‘l) — 332.5 K AE = neg/AT = (2.00 mol)(29.3  8.315 J marl K1)(332.5 ~ 273.15 K)
= +2.49 x 103 J
iw=AE—q=AE=2.49x103J . 814 The melting point of tetraphenylgermane is 505.65 K. The molar enthalpy of ‘ fusion is
C 381.03 g mol—1 x 106.7 J g—1 = 4.066 x 104 J mol—1 We combine these numbers to get the molar entropy of fusion as follows: AHfus 4.066 x 104 J. mol‘1
= m AS” = Tm. 505.65 K = +8040 J K'lmol‘l 816 ’I‘routon’s rule states that the molar entropy of vaporization of most substances
is near 88 J K‘1m01”1. Gas and liquid forms of a substance are in equilibrium at the normal boiling point, so we can write: 3 —1
: AHvap = 16.15x 10 J mol = 184 K Tb new, .88JK—1mol1 820 Consider ﬁrst the reversible heating of 1.00 mol water from 25°C to 150°C. This process consists of three steps: heating water to 100°C, evaporating it,
and heating steam to 150°C. q '= (1.00 mol)(75.4 J mol1 K1)(100 — 25 K)
+(1.00 mol)(40680 J mol‘1 K‘l) +(1.00 mol)(36.0 J mol1 K1)(150 — 100 K)
= 43,135 J AS = ncP(liq) 1n + 71%;” + nep(gas)1n 177::
= (1.00 mol)(75.4 J mol"1 K‘l) >< ln(373.15/298.15) mo —1 +(1.00 mol)———4°6:§3’315 Kl +(1.00 mol)(36.0 J mol‘1 K‘l) >< ln(423.15/373.15) = 130.5 J K—1 Now consider the irreversible process in the flash evaporation. Because 5' is a
function of state, A8 for the water is unchanged. ASmte, = 130.5 J K—1 From the point of view of the iron, however, the heat q = 48,135 J is removed
at a constant temperature of 150°C = 423.15 K. Thus the ﬁnal state of the
iron is the same as would be achieved by removing 48,135 J reversibly at ﬁxed temperature. 7
q 48,135 J _1
ASim =—=——_—_~— 1.
n T 423.15K 1 38”:
Astotal = 130.5 — 113.3 = 15.7 J K1 > 0
822 (a) AH = ncPAT = (1.00 mol)(25.1 J mol1 K‘1)(273.15'— 373.15 K) = —2510 J _ Tf _ _1 _1 273.15
AS _ nepln Ti _ (1.00 mol)(25.1 J mol K )In 373.15 (b) Ashen = —7 .83 J K‘1 because S is a function of state and the initial and
ﬁnal states of the iron are the same as in part (a). For the water, heat is being added at a constant temperature of 273.15 K. Its ﬁnal state is the same as would
be found if q1, = 2510 J were added at T = 273.15 K. 2510 J wa er 2 _— =  K—l
AS t 27315 K +9 19 J new = 1.36 J K—1' = —7.83 J K1 ...
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 Fall '09
 ProfessorNewstead
 Thermodynamics, Entropy, mol, atm mol

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