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Chemistry 531 Spring 2009 Problem Set 5 Solutions

1. For the reaction

CH4(g) + 2O2(g) −→ CO2(g) + 2H2O(l) use Table 4-2 on page 50 and Table 7-2 on page 171 of your text books to determine at 25◦C

(a) the amount of heat liberated per at constant pressure; Answer:

◦ −1 ∆rHm = [2(−285.83) + (−393.509) − (−74.81)]kJ mol

= −890.359kJ mol−1 (b) the maximum amount of work obtainable per mole for the reaction; Answer:

◦ −1 −1 ∆rGm = [2(−237.129) + (−394.359) − (−50.72)]kJ mol = −817.897kJ mol

A = G − pV ◦ ◦ ∆rAm = ∆rGm − RT ∆ngas = −817897J mol−1−(8.3144J mol−1K−1)(298K)(−2) = −812.941kJ mol−1 = max. work (c) the maximum amount of non-pV work obtainable from the reaction per mole. Answer: ◦ −1 max. non-pV work = ∆rGm = −817.897kJ mol 2. A certain obeys the

pVm = RT + βp

1 where β depends only on the . For this gas show that ∂U ! RT 2 dβ ! = 2 ∂V T (Vm − β) dT

Answer: ∂U ! ∂p ! = T − p ∂V T ∂T V RT p = Vm − β(T ) ∂p ! R RT dβ = + 2 ∂T V Vm − β (Vm − β) dT Then ∂U ! RT RT 2 dβ = + 2 − p ∂V T Vm − β (Vm − β) dT RT RT 2 dβ RT = + 2 − Vm − β (Vm − β) dT Vm − β RT 2 dβ = 2 (Vm − β) dT 3. Show that ! 2 ! ∂Cp ∂ V = −T 2 ∂p T ∂T p Answer: ∂H ! Cp = ∂T p !  !  ∂Cp ∂ ∂H =   ∂p ∂p ∂T T p T ∂ ∂H ! !

∂T ∂p T p But dH = T dS + V dp ∂H ! ∂S ! = T + V ∂p T ∂p T dG = −SdT + V dp ∂S ! ∂V ! = − ∂p T ∂T p

2 Then ∂H ! ∂V ! = V − T ∂p T ∂T p and !   !  ∂Cp ∂ ∂V =  V − T  ∂p ∂T ∂T T p p ∂V ! ∂V ! ∂2V ! = − − T 2 ∂T p ∂T p ∂T p ∂2V ! = −T 2 ∂T p 4. For an ideal gas, show that ∂p !  p  = −γ ∂V S V Answer: Simple Solution: For an ideal gas along an adiabatic reversible path

pV γ = constant

or constant p = (constant S) V γ ∂p ! = −γ(constant)V −γ−1 ∂V S γpV γ = − V γ−1 p = −γ V Long Solution: dH = T dS + V dp ∂H ! ∂p ! = V ∂V S ∂V S For an ideal gas dH = CpdT so that ∂H ! ∂T ! = Cp ∂V S ∂V S

3 or ∂p ! C ∂T ! = p ∂V S V ∂V S But ∂T ! ∂V ! ∂S ! = −1 ∂V S ∂S T ∂T V so that ∂T ! (∂S/∂V ) = − T ∂V S (∂S/∂T )V Now dA = −SdT − pdV ∂S ! ∂p ! = ∂V T ∂T V For an ideal gas C dT dS = V constantV T so that ∂S ! C = V ∂T V T Then ∂T ! (∂p/∂T ) = −T V ∂V S CV and ∂p ! C T ∂p ! = − p ∂V S V CV ∂T V But for an ideal gas ∂p ! nR = ∂T V V and ∂T ! C T nR p = − p = −γ ∂V S CV V V V 5. The work done on a rubber band when it is stretched by an infinitesimal length d` owing to an applied force, F , is given by dw = F d`.

(a) Neglecting the volume change on stretching, show that

∂U ! ∂F ! = F − T ∂` T ∂T `

4 Answer: dU = T dS + F dl ∂U ! ∂S ! = T + F ∂l T ∂l T As usual define A = U − TS. Then

dA = −SdT + F dl

and ∂S ! ∂F ! = − ∂l T ∂T l so that ∂U ! ∂F ! = F − T ∂l T ∂T l (b) Experimentally, the force at constant length is given by

F = γT

where γ is a constant. Show that

∂U ! = 0 ∂` T

Answer: F = γT (constant l) ∂F ! = γ ∂T l ∂U ! = γT − γT = 0 ∂l T (c) Since stretching orders the rubber polymer, it can be shown that

(∂S/∂F )T < 0

i.e. there is an entropy decrease upon stretching. From this sign deduce the sign of (∂`/∂T )F . Use this to predict what will happen to a rubber band with a weight hanging from it, if the temperature of the rubber is raised. Answer: Define G = A − F l

5 dG = dA − F dl − ldF = −SdT + F dl − F dl − ldF = −SdT − ldF ∂S ! ∂l ! = ∂F T ∂T F Because ∂S ! < 0 ∂F T it follows that ∂l ! < 0 ∂T F Then if T increases, l decreases and the rubber band contracts. (Try it - Hang a weight at the end of a rubber band to keep F constant, and use a hair dryer to increase the temperature.)

6. Find an expression for ∆G and ∆A for the isothermal expansion of a van der Waals gas from V1 to V2. Answer: dA = −SdT − pdV = −pdV at constantT

Z V2 ∆A = − pdV V1 " 2 # Z V2 nRT an = − − 2 dV V1 V − nb V V − nb!  1 1  = −nRT ln 2 + n2a − V1 − nb V1 V2 ∆G = ∆A + ∆(pV ) Now for a Van der Waals gas n2a! p + (V − nb) = nRT V 2

n2a n3ab pV + − − nbp = nRT V V 2 or n2a n3ab " nRT n2a# pV = nRT − + + nb − V V 2 V − nb V 2 n2a n2bRT = nRT − + V V − nb

6 Then  1 1   1 1  ∆(pV ) = −n2a − + n2bRT − (constant T ) V2 V1 V2 − nb V1 − nb Then V − nb!  1 1   1 1   1 1  ∆G = −nRT ln 2 +n2a − −n2a − +n2bRT − V1 − nb V1 V2 V2 V1 V2 − nb V1 − nb

V − nb!  1 1   1 1  = −nRT ln 2 + 2n2a − + bn2RT − V1 − nb V1 V2 V2 − nb V1 − nb 7. One mole of is originally contained in a capsule that is enclosed in an evacuated tank of fixed volume that is thermostated to 100. ◦C. The capsule is broken and the liquid water is allowed to evaporate into the previously evacuated tank. After equilibrium is established at 100. ◦C, the pressure of the water in the tank is found to be 0.5 atmospheres. Assuming the water vapor to be an ideal gas and neglecting the volume of the liquid water compared to that of the vapor, calculate ∆U, ∆H, ∆S, ∆A, ∆G, Q and W for the process. The of of water at 100. ◦C and 1. atmospheres pressure is 44. kJ mol−1. Answer: W = 0 ∆H = 44000J ∆U = ∆H − ∆(pV ) ∼ = ∆H − (pvVv − plVl) = ∆H − pvVv = ∆H − nRT = 44000J − (1 mol)(8.3144J mol−1K−1)(373K) = 40899J Q = ∆U = 40899J ∆H p ∆S = + nR ln i T pf 44000J 1 = + (1 mol)(8.3144J mol−1K−1) ln 373K .5 = 123J K−1 ∆G = ∆H − T ∆S = 44000J − (123J K−1)(373K) = −1879J ∆A = ∆U − T ∆S = 40899J − (123J K−1)(373K) = −4980J

7 8. In the container pictured above, the left hand side initially consists of water in equi- librium with its vapor and the right hand side is evacuated. The two sides, each of volume 20. liters, are connected with a closed stop-cock and the entire system is placed in a heat bath thermostated to 25. ◦C. The of water at 25. ◦C is 24. mm and the of water is 44. kJ mol−1. If the stopcock is opened, calculate ∆U, ∆H, ∆S, ∆A, ∆G, Q and W when equilibrium is reached. You may assume some liquid water remains on the left hand side at the end of the process,

8 and you may neglect the volume of the liquid compared to the vapor. You may also assume the vapor obeys the ideal gas law. Answer: pV (0.03158atm)(20 l) n = = = 0.02582mol RT (0.0821 l atm mol−1K−1)(298K) ∆H = (0.02582mol)(44000J mol−1) = 1136J ∆U = ∆H − ∆(pV )  8.3144J  = 1136J − (20 l)(0.03158atm) = 1072J 0.0821 l atm ∆H 1136J ∆S = = = 3.81J K−1 T 298K ∆G = ∆H − T ∆S = 0 ∆A = ∆U − T ∆S = 1072J − (298K)(3.81J K−1) = −63J (∆A < 0 as it should be for a spontaneous process at constantT andV ) W = 0 Q = ∆U = 1072J 9. The vapor pressure of ammonia is given by 3754 ln p = 23.03 − T and that of liquid ammonia is given by 3063 ln p = 19.49 − T where the pressure is expressed in millimeters of . (a) What is the temperature of ammonia at the ? Answer: 3063 3754 19.49 − = 23.03 − T T T = 195.2K (b) What are the latent of sublimation and vaporization of ammonia at the triple point? Answer: ∆H ln p = − + constant RT ∆H sub = 3754K R −1 −1 −1 ∆Hsub = (3754K)(8.3144J mol K ) = 31.14kJ mol −1 −1 −1 ∆Hvap(3063K)(8.3144J mol K ) = 25.47kJ mol

9 (c) What is the of fusion of ammonia at the triple point? Answer: solid → liquid → vapor → solid so that ∆Hfusion = ∆Hsub − ∆Hvap = (31.14 − 25.47)kJ mol−1 = 5.67kJ mol−1

10. The density of liquid is 10. kg liter−1, and the density of solid bismuth is 9.67 kg liter−1. The normal point of bismuth is 270. ◦C and the enthalpy of fusion of bismuth is 52.718 kJ kg−1. Calculate the of bismuth under 100. atmospheres pressure. Answer: ∆Hm T2 p2 − p1 = ln ∆Vm T1

T2 ∆Vm ln = (p2 − p1) T1 ∆Hm (0.1 l kg−1 − 0.1034 l kg−1)(100atm − 1 atm) = 52718J kg−1 l atm  8.3144J  = −6.409 × 10−6 J 0.0821 l atm = −6.49 × 10−4 −4 T2 = 543K exp(−6.49 × 10 ) = 542.6K

11. The enthalpy of vaporization of water is 44. kJ mol−1, and the enthalpy of fusion of is 6. kJ mol−1. If the vapor pressure of ice at 0. ◦C is 6.016 × 10−3 atmospheres, what is the vapor pressure of ice at -20. ◦C? Answer: ∆Hsub = ∆Hvap + ∆Hfusion = (6 + 44)kJ mol−1 = 50kJ mol−1 p 50000J mol−1  1 1  ln = − 6.016 × 10−3atm 8.3144J mol−1K−1 273K 253K p = 1.06 × 10−3atm

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