To_Even_Numbered_Problems

To_Even_Numbered_Problems - 0 Solutions to Even Numbered...

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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 7-54 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 K-1)(332.5 ~ 273.15 K) = +2.49 x 103 J iw=AE—q=AE=2.49x103J . 8-14 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 8-16 ’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—1mol-1 8-20 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 mol-1 K-1)(100 — 25 K) +(1.00 mol)(40680 J mol‘1 K‘l) +(1.00 mol)(36.0 J mol-1 K-1)(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 K-1 > 0 8-22 (a) AH = ncPAT = (1.00 mol)(25.1 J mol-1 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 K-1 ...
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This note was uploaded on 12/01/2010 for the course APPLIED SC 131 taught by Professor Professornewstead during the Fall '09 term at Queens University.

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To_Even_Numbered_Problems - 0 Solutions to Even Numbered...

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