HW-03%20Solutions - #1 10.2 (a) This problem first asks...

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10.2 (a) This problem first asks that we rewrite the expression for the total free energy change for nucleation (analogous to Equation 10.1) for the case of a cubic nucleus of edge length a . The volume of such a cubic radius is a 3 , whereas the total surface area is 6 a 2 (since there are six faces each of which has an area of a 2 ). Thus, the expression for G is as follows: G = a 3 G v + 6 a 2 γ Differentiation of this expression with respect to a is as d G da = d ( a 3 G v ) da + d ( 6 a 2 γ ) da = 3 a 2 G v + 12 a γ If we set this expression equal to zero as 3 a 2 G v + 12 a γ= 0 and then solve for a (= a *), we have a *= 4 γ G v Substitution of this expression for a in the above expression for G yields an equation for G * as G *=( a *) 3 G v + 6( a* ) 2 γ =− 4 γ G v 3 G v + 6 γ− 4 γ G v 2 = 32 γ 3 ( G v ) 2
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(b) G v for a cube—i.e., (32) γ 3 ( G v ) 2 —is greater that for a sphere—i.e., 16 π 3 γ 3 ( G v ) 2 = (16.8) γ 3 ( G v ) 2 . The reason for this is that surface-to-volume ratio of a cube is greater than for a sphere.
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10.3 This problem states that ice homogeneously nucleates at –40 ° C, and that we are to calculate the critical radius given the latent heat of fusion (–3.1 x 10 8 J/m 3 ) and the surface free energy (25 x 10 -3 J/m 2 ). Solution to this problem requires the utilization of Equation 10.6 as
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HW-03%20Solutions - #1 10.2 (a) This problem first asks...

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