CHEM 217 Problems Set 5 Solution, Fall 2011

Chemistry 217 Problem set 5 solution 4.2. step 1: Isothermal Expansion pdV dw = − 2 2 1 1 1 2 2 1 1 1 2 1 1 1 2 2 2 1 1 1 1 1 ln ln V V V V v T dV dV w pdV RT a V b V V b w RT a V b V V E dE dQ dw dQ dE dw C dT dV PdV V RT RT a a dQ dV dV V V b V V b V b Q RT V b = − = − + − ⎛ ⎞ ⎛ ⎞ − ⇒ = − − − ⎜ ⎟ ⎜ ⎟ − ⎝ ⎠ ⎝ ⎠ ⎛ ∂ ⎞ = + ⇒ = − = + + ⎜ ⎟ ⎝ ∂ ⎠ ⎛ ⎞ ⇒ = + − = ⎜ ⎟ − − ⎝ ⎠ ⎛ ⎞ − ⇒ = ⎜ ⎟ − ⎝ ⎠ ∫ ∫ ∫ step 2: Adiabatic Expansion ( ) 2 3 1 2 2 2 2 2 2 2 1 3 2 0 1 1 T V v v v T V T v Q E a a dw dE C dT dV C dT dV w C dT dV V V V w C T T a V V = ⎛ ∂ ⎞ = = + = + ⇒ = + ⎜ ⎟ ⎝ ∂ ⎠ ⎛ ⎞ ⇒ = − − − ⎜ ⎟ ⎝ ⎠ ∫ ∫ step 3: Isothermal Compression ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ − ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − − = b V b V RT Q V a V b V b V RT w 3 4 2 3 3 4 3 4 2 3 ln 1 1 ln step 4: Adiabatic Compression ( ) ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − − = = 4 1 2 1 4 4 1 1 0 V a V T C T w Q v Total work: 4 3 2 1 w w w w w + + + = ( ) ( ) ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − ⎟⎟ − ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = − ⇒ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − − ⎟⎟ + ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ − ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − ⎟⎟ − ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − − ⎟⎟ + ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ − ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − − = b V b V RT b V b V RT w V V a T C T V a V b V b V RT V a V T C T V a V b V b V RT w v v 3 4 2 1 2 1 4 1 2 1 3 4 3 4 2 2 3 1 2 1 2 1 2 1 ln ln 1 1 1 1 ln 1 1 1 1 ln ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = b V b V RT Q 1 2 1 1 ln ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = + − = b V b V b V b V T T Q w y 1 2 3 4 1 2 1 ln ln 1 Find a relation between b V b V − − 3 4 and b V b V − − 1 2 from adiabatic steps: ( ) 2 2 3 2 2 1 2 3 2 1 1 4 1 ln ln From step 2: ln ln ln From step 4: ln ln which means that v v v v v v a RT a dw dE pdV C dT dV dV V V b V C RT dV C dT dV dT R C d T Rd V b V b T V b V b T V b C R R T V b V b T V b C R T V b V ⎛ − ⎞ = ⇒ − = + = + ⎜ ⎟ − ⎝ ⎠ − ⇒ = ⇒ = − ⇒ = − − − − ⎛ ⎞ ⎛ ⎞ − − = − = ⎜ ⎟ ⎜ ⎟ − − ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ − = − ⎜ ⎟ − ⎝ ⎠ 3 2 2 4 3 4 1 4 3 2 2 3 4 2 1 2 2 1 1 1 1 ln ln 1 1 1 ln ln V b b V b V b V b V b V b V b V b V b T T V b V b T T V b V b T T V b V b η − − − − = ⇒ = − − − − ⎛ ⎞ ⎛ ⎞ − − ⎜ ⎟ ⎜ ⎟ − − ⎝ ⎠ ⎝ ⎠ ⇒ = + = − = − ⎛ ⎞ ⎛ ⎞ − − ⎜ ⎟ ⎜ ⎟ − − ⎝ ⎠ ⎝ ⎠ 4.4. a. Chain rule: T S V S T V S V T ⎛ ∂ ⎞ ⎜ ⎟ ∂ ∂ ⎛ ⎞ ⎝ ⎠ = − ⎜ ⎟ ∂ ∂ ⎛ ⎞ ⎝ ⎠ ⎜ ⎟ ⎝ ∂ ⎠ V v V V C S S C T T T T ∂ ∂ ⎛ ⎞ ⎛ ⎞ = ⇒ = ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ dA SdT PdV = − − V A S T ⎛ ∂ ⎞ ⇒ = − ⎜ ⎟ ⎝ ∂ ⎠ and T A P V ⎛ ∂ ⎞ = − ⎜ ⎟ ⎝ ∂ ⎠ V T V T A A V T T V ⎛ ⎞ ⎛ ⎞ ∂ ∂ ∂ ∂ ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ T V S P V T ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⇒ = ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ S V V T T P V C T ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⇒ = − ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ c. T S P S T P S P T ⎛ ∂ ⎞ ⎜ ⎟ ∂ ∂ ⎛ ⎞ ⎝ ⎠ = − ⎜ ⎟ ∂ ∂ ⎛ ⎞ ⎝ ⎠ ⎜ ⎟ ⎝ ∂ ⎠ dG SdT VdP = − − P G S T ⎛ ∂ ⎞ ⇒ = − ⎜ ⎟ ⎝ ∂ ⎠ and T G V P ⎛ ∂ ⎞ = ⎜ ⎟ ⎝ ∂ ⎠ T P P T G G T P P T ⎛ ⎞ ⎛ ⎞ ∂ ∂ ∂ ∂ ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ T P S V P T ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⇒ − = ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ P P P P C S S C T T T T ∂ ∂ ⎛ ⎞ ⎛ ⎞ = ⇒ = ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ So S P P T T V P C T ∂ ∂ ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ S V V P S T P P S V T V C C V V P T T P P C C P V T ⎛ ⎞ ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⎝ ⎠ ⎝ ⎠ ⎜ ⎟ ⇒ = = − = ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ f. ( ) ( )E E T G T H G H ∂ ∂ ∂ ⎟ = ∂ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ E JT p p E JT p p T p C C T H dp C C dT dH ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ⇒ ∂ − = µ µ ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( α) α κ α α α κ α α κ α α α κ α κ α α κ α α µ pV C T p S pV C T T p C pV V C T p SV pV C V T T V p C T p V pV C V S T p V pV C C V T C T p V S T p C G H T p V S T G Vdp SdT dG p p p p p p p p p p E E p E E E JT − − − − − − − − = − − − − − − + − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − − + − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − ⋅ − − − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ + − ⎥⎦ ⎤ ⎢⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ = − + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ⇒ ∂ + − = 1 1 1 1 1 4.5. a. 1 P P P E V P V dE TdS PdV T P T S S T S ⎛ ⎞ ∂ ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = − ⇒ = − = ⎜ − ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ S P T V P S ∂ ∂ ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ (Maxwell’s relation) T S P P P S T T V P S P C T T ⎛ ∂ ⎞ ⎜ ⎟ ∂ ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⎝ ⎠ = − = ⎜ ⎟ ⎜ ⎟ ∂ ∂ ∂ ⎛ ⎞ ⎝ ⎠ ⎝ ⎠ ⎜ ⎟ ⎝ ∂ ⎠ 1 P P P E P V T S C T ⎛ ⎞ ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⇒ = ⎜ − ⎟ ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ b. V V G P dG SdT VdP S V T T ∂ ∂ ⎛ ⎞ ⎛ ⎞ = − − ⇒ = − + ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ P V T V P T V T P ⎛ ∂ ⎞ ⎜ ⎟ ∂ ∂ ⎛ ⎞ ⎝ ⎠ = − ⎜ ⎟ ∂ ∂ ⎛ ⎞ ⎝ ⎠ ⎜ ⎟ ⎝ ∂ ⎠ P V V T α ⎛ ∂ ⎞ = ⎜ ⎟ ⎝ ∂ ⎠ because 1 P V V T α ⎛ ∂ ⎞ = ⎜ ⎟ ⎝ ∂ ⎠ and T V V P κ ⎛ ∂ ⎞ = − ⎜ ⎟ ⎝ ∂ ⎠ because 1 T V V P κ ⎛ ∂ ⎞ = − ⎜ ⎟ ⎝ ∂ ⎠ V V P V G V S T V T α α α κ κ κ ∂ − ∂ ⎛ ⎞ ⎛ ⎞ = = ⇒ = − + ⎜ ⎟ ⎜ ⎟ ∂ − ∂ ⎝ ⎠ ⎝ ⎠ 4.6. dA = -­‐SdT – PdV dG= -­‐SdT + VdP ⇒dA = dG – PdV – VdP where –VdP = 0 (isobaric) dH = TdS + VdP = TdS dE = TdS – PdV = 0 (isothermal) ⇒ TdS = PdV ⇒ dA = dG – dH ⇒ ∆A = ∆G -­‐ ∆H ΔH = ( ) ( ) ( ) 2 2 6 6 3 6 f H O f CO f C H H H H Δ + Δ − Δ = 3(-­‐285.8) + 6(-­‐303.5) – 49.1 = -­‐2727.5 KJ/mol ∆A = -­‐3174.9 + 2727.5 = -­‐447.4 KJ/mol. 4.8. T l E f f T l T ∂ ∂ ⎛ ⎞ ⎛ ⎞ = − ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ dE TdS fdl = + T T T l T l E S T f l l S f Where l T E f f T l T ∂ ∂ ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ∂ ∂ ⎛ ⎞ ⎛ ⎞ − = ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ∂ ∂ ⎛ ⎞ ⎛ ⎞ ⇒ = − ⎜ ⎟ ⎜ ⎟ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⇒ ∫ dSTotal = 0 4.9. ( ) ( ) ( ) 60 60 0 0 60 3 5 2 0 60 3 5 2 3 0 0.23 2.5 10 1.9 10 2.5 10 1.9 10 0.23 2 3 16.9 / . Cp S S S dT T S T T dT S T T T S J mol K − − − − Δ = − = Δ = + × − × ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ × × Δ = + − ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ Δ = ∫ ∫ 4.10. dA SdT PdT = − − where SdT = 0 (isothermal) ( )( )( ) 2 2 1 1 2 1 ln ln 1 8.314 300.15 ln 2 1730 / V V V P A PdV nRT dV nRT nRT V V P A J mol ⇒ Δ = − = − = − = − Δ = − = − ∫ ∫ 4.14. ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 2 3 2 2 3 3 2 2 3 3 1 1 1 1 1 2 2 1 1 1 ln ln 2 3 2 2 P P T P T P dH dQ VdP C dT VdP H C dT VdP RT H a bT cT dT B T B T P dP P B B b c H a T T T T T T RT P RT P B TP B T P TP TP = + = + ⇒ Δ = + ⎛ ⎞ Δ = + + + + + ⎜ ⎟ ⎝ ⎠ Δ = − + − + − + − + − + − ∫ ∫ ∫ ∫ [ ] ( ) ( ) 1 1 2 2 1 1 1 ln ln 2 T T P T T C a c S dT b cT dT a T T b T T T T T T ⎛ ⎞ Δ = = + + = − + − + − ⎜ ⎟ ⎝ ⎠ ∫ ∫ 4.16. a. Prove that dE = 0 ∫ Step 1 (isothermal) ⇒ dE = 0 Step 2 (adiabatic) ⇒ ( ) 2 1 V dE dw nC T T = = − Step 3 (isothermal) 0 ⇒ dE = Step 4 (adiabatic) ⇒ ( ) 1 2 V dE nC T T = − ⇒ ∫ dETotal = 0 b. Prove that dH ∫ = 0 ( ) dH dE d PV = + Step 1 : 2 2 1 1 dH PV PV = − Step 2 : ( ) 3 3 2 2 2 1 V dH PV PV nC T T = − + − Step 3 : 4 4 3 3 dH PV PV = − Step 4 : ( ) 1 1 4 4 1 2 V dH PV PV nC T T = − + − dHTotal ∫ = 0 c. Prove that dA ∫ = 0 ( ) dA dE d TS = − dE TdS SdT = − − Step 1 : 1 1 0 0 dA T dS T dS = − − = − Step 2 : ( ) ( ) 2 1 2 1 V dA nC T T S T T = − − − Step 3 : 2 dA = −T dS Step 4 : ( ) ( ) 1 2 1 2 V dA nC T T S T T = − − − ⇒ ∫ dA = −T1 dS −T2dS but 0 CP dS = T dT = (for isothermal process) ⇒ ∫ dATotal = 0 d. Prove that dG = 0 ∫ ( ) dG dH d TS = − dH TdS SdT = − − Step 1 : 1 1 dG dH T dS = − Step 2 : ( ) 2 2 1 dG dH S T T = − − Step 3 : 3 2 dG dH T dS = − Step 4 : ( ) 4 1 2 dG dH S T T = − − ⇒ ∫ dGTotal = 0 e. Prove that dS ∫ = 0 Step 1 : 0 CP dS = T dT = Step 2 : dS = CP T dT ⇒ ΔS = CP ln T2 T1 Step 3 : dS = CP T dT = 0 Step 4 : dS = CP T dT ⇒ ΔS = CP ln T1 T2 ⇒ ∫ dSTotal = 0