This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: exists. At the crack tip R 1 = , R 2 has small negative value. Consequently, ln(S 1 /S ) would have a large negative value. S 1 /S &lt;&lt;1. Solubility at the crack tip is small. 3.Calculate the solubility change S 1 /S for silica spheres with radius 1000A and 100A at room temperature in water. Assume that the surface energy is 0.5J/m 2 , density is 2.3g/cm 3 . The molecular weight of SiO 2 is 60 g/mol. ln(S 1 /S ) = (1/R 1 +1/R 2 )(M/RT ) = (2 /r)(M/RT ) =(1/r)(0.060/8.314x 300x2.3x10 3 ) =1.05x10-8 /r For r=1000A = 10-7 m, ln (S 1 /S ) = 0.105 S 1 /S 0 = 1.11 r=100A = 10-8 m , ln (S 1 /S ) = 1.05 S 1 /S 0 = 2.86 4. Calculate the excess pressure a sphere of SiO 2 with radius 100 A is subjected to. P = 2 /r P = 2x 1 (J/m 2 )/ 10-8 m = 2x10 8 (J/m 3 ) = 2x10 8 (N/m 2 ) = 200MPa....
View Full Document
This note was uploaded on 04/29/2008 for the course MTLE 2100 taught by Professor Tomozawa during the Spring '08 term at Rensselaer Polytechnic Institute.
- Spring '08