INSTRUCTOR’S MANUAL
86
Comment. As shown by Problem 5.1, increasing the temperature does not necessarily increase the − − and r H − is the determining maximum non-expansion work. The relative magnitude of r G− factor. P5.31
The Gibbs–Helmholtz equation is ∂ G H =− 2 ∂T T T so for a small temperature change − − − − − r G2− r G1− r G− r H − r H − T and = − = T T2 T1 T 2 T T2 − − − − − − dT r G190 r G220 r G− 1 1 r H − − − so d and = + r H − =− T T190 T220 T190 T220 T2 T190 − − − − T190 − r G190 1− = r G220 + r H − T220 T220 For the monohydrate − − r G190
= (46.2 kJ mol
−1
)×
190 K 220 K
+ (127 kJ mol
−1
+ (188 kJ mol
−1
190 K , )× 1− 220 K
− − r G190 = 57.2 kJ mol−1
For the dihydrate − − r G190
= (69.4 kJ mol
−1
)×
190 K 220 K
190 K )× 1− , 220 K
− − r G190 = 85.6 kJ mol−1
For the monohydrate − − r G190 = (93.2 kJ mol−1 ) ×
190 K 220 K
190 K + (237 kJ mol−1 ) × 1 − , 220 K
− − r G190 = 112.8 kJ mol−1
P5.32
The change in the Helmholtz energy equals the maximum work associated with stretching the polymer. Then dwmax = dA = −f dl For stretching at constant T ∂A ∂U ∂S f =− =− +T ∂l T ∂l T ∂l T assuming that (∂U/∂l)T = 0 (valid for rubbers) ∂S 3kB l 2 ∂ f =T − =T +C ∂l T ∂l T 2N a 2 3kB l 3kB T =T − =− l N a2 N a2 This tensile force has the Hooke’s law form f = −kH l with kH = 3kB T /N a 2 .
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