Explanation Given: Aluminum silicate has three different crystalline state − kyanite, andalusite and sillimanite. The molar volume of kyanite is 44.09 × 10 − 6 m 3 , that of andalusite is 51.53 × 10 − 6 m 3 and that of sillimanite is 49.90 × 10 − 6 m 3 . Formula used: Write the expression for Gibbs free energy. G = − T S + P V + μ N …… (1) Here, G is Gibbs free energy, T is the absolute temperature, S is the entropy, P is the pressure, V is the volume, μ is the chemical potential and N is the number of molecules. Write the expression for the final temperature of the material. T 2 = T 1 + Δ G Δ S …… (2) Here, T 2 is the final temperature, T 1 is the initial temperature, Δ G is the change in Gibbs energy and Δ S is the change in entropy. Write the expression for the slope of phase boundary. d P d T = S 1 − S 2 V 1 − V 2 …… (3) Calculation: Write the expression for the infinitesimal change in G . d G = − S d T + V d P + μ d N Rearrange the above expression at constant number of molecules N and at constant temperature T , for which d N = d T = 0 . ( ∂ G ∂ P ) T , N = V Rearrange the above expression. d G = V d P …… (4) Integrate both sides of expression (4) for kyanite. G k = G k 0 + V k P …… (5) Integrate both sides of expression (4) for andalusite. G a = G a 0 + V a P …… (6) Integrate both sides of expression (4) for sillimanite. G s = G s 0 + V s P …… (7) Refer to the text at the back of the book. Substitute − 2443.88 × 10 3 J for G k 0 , 10 5 Pa and 44.09 × 10 − 6 m 3 for V k in expression (5). G k = − 2443875.591 J Substitute − 2442.66 × 10 3 J for G k 0 , 10 5 Pa and 51.53 × 10 − 6 m 3 for V a in expression (6). G a = − 2442654.847 J Substitute − 2440.99 × 10 3 J for G k 0 , 10 5 Pa and 49.90 × 10 − 6 m 3 for V s in expression (7). G s = − 2440985.01 J Refer to the back of the book. Substitute 298 K for T 1 , − 2442654.847 J for G a , − 2443875.591 J for G k , 93.22 J ⋅ K − 1 for S a and 83.81 J ⋅ K − 1 for S k in expression (2). T k a = 298 K + − 2442654.847 J + − 2443875.591 J 93.22 J ⋅ K − 1 − 83.81 J ⋅ K − 1 = 428 K Substitute 298 K for T 1 , − 2440985.01 J for G a , − 2443875.591 J for G k , 96.11 J ⋅ K − 1 for S s and 83.81 J ⋅ K − 1 for S k in expression (2). T k s = 298 K + − 2440985.01 J + − 2443875.591 J 96.11 J ⋅ K − 1 − 83.81 J ⋅ K − 1 = 533 K Substitute 298 K for T 1 , − 2440985.01 J for G a , − 2442654.847 J for G a , 96.11 J ⋅ K − 1 for S s and 93.22 J ⋅ K − 1 for S a in expression (2). T a s = 298 K + − 2440985.01 J + − 2442654.847 J 96.11 J ⋅ K − 1 − 93.22 J ⋅ K − 1 = 876 K Refer to the back of the book

1st Edition

ISBN: 9780201380279

Author: Daniel V. Schroeder

Publisher: Addison Wesley

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Question

Chapter 5.3, Problem 39P

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The slope of all three phase boundaries and the variation of the phase diagram.

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Chapter 5.3, Problem 38P

Chapter 5.3, Problem 40P

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