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# 503hw6 - Solution to Homework#6 1 At equilibrium...

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Solution to Homework #6 1 . At equilibrium temperature, T e , D D D G H T S e = - =0, Therefore, D D S H T e = / = -10 10 Jm -3 /1000 K= -10 7 Jm -3 K -1 The free energy change of transformation per unit volume is then D D D G H T S v = - = -10 10 Jm -3 — 700 x (-10 7 ) Jm -3 K -1 = -3x10 9 Jm -3 Assuming the cubic precipitate has an edge of a, the total free energy change for coherent transformation is: ( ) D D G G E a a coh v coh coh = + + 3 2 6 g Let ¶D G a coh = 0 , then the critical nucleus size is: a G E coh coh v coh * = - + 4 g D = -(4 x 0.01 Jm -2 )/(-3 x 10 9 Jm -3 + 10 9 Jm -3 ) = 2 x 10 -11 m The free energy change for forming a critical nucleus is: ( )( ) ( ) D D G G E a a coh v coh coh coh coh * * * = + + 3 2 6 g = (-3 x 10 9 Jm -3 + 10 9 Jm -3 ) x (2 x 10 -11 m) 3 + 6 x (2 x 10 -11 m) 2 x 0.01 Jm -2 = 8 x 10 -24 J The coherent nucleation rate is: I coh = I 0 exp(- D G * coh /kT) = 10 33 exp[-8x10 -24 /(1.381x10 -23 x700 )] = 9.99x10 -32 ( m -3 s -1 ) Similarly, the total free energy change for incoherent transformation is: ( ) D D G G E a a incoh v incoh incoh = + + 3 2 6 g = D G a a v incoh 3 2 6 + g The critical nucleus size is: a G incoh incoh v * = - 4 g D = -(4 x 0.5 Jm -2 )/(-3 x 10 9 Jm -3 ) = 2/3 x 10 -9

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