ENGINEERING MATERIAL - EXAMPLE 12.3 : Let a ^ 2 = a x a and a ^ 3 = a x a x a where ^ is power function. Niobium is a metal with a body-centered cubic structure. The length of the unit cell structure is b = 0.3349 nm. (a) Find the volume for a unit cell structure for niobium. (b) There are 2 atoms per unit cell structure of niobium. The metal has a molar mass of 92.9 g / mol. One mole of the metal consists of 6.02 x 10 ^ 23 atoms. Find the mass of niobium per unit cell and the density of niobium.
ENGINEERING MATERIAL - ANSWER 12.3 : (a) V = 0.3349 x 0.3349 x 0.3349 = 0.0376 cubic nanometers. (b) Mass per unit cell = 2 x 92.9 (g / mol) [ 1 mol / (6.02 x 10 ^ 23 atoms) ] = 3.086 x 10 ^ (-22) g = m. V = 0.0376 x [ 10 ^ (-7) ] ^ 3 = 0.0376 x 10 ^ (-21) cubic cm. Then density is m / V = 8.207 g / cubic cm.
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THERMODYNAMIC - EXAMPLE 10.2 : A cylinder with a movable piston contains 0.1 mole of a monoatomic ideal gas. The piston moves through state a, b and c. The heat Q, changes from state c to a is + 685 J. The work W, changes from state c to a is - 120 J. The work, W performed from state a to b then to c is 75 J. By using the first law of thermodynamic, U = Q + W where U is the internal energy : (a) Determine the change in internal energy between states a and c. (b) Is heat added or removed from the gas when the gas is taken along the path abc? (c) Calculate the heat added or removed when the gas is taken along the path abc?
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ENGINEERING MATERIAL - EXAMPLE 12.2 : At 150 degree Celsius, a mixture of 40 wt % Sn and 60 wt % Pb present, forming phases of alpha and beta. Chemical composition of Sn at each phase : CO (overall) : 40 %, CA (alpha) : 11 %, CB (beta) : 99 %. (a) State 2 reasons for the existences of alpha and beta phases for the mixture of Sn - Pb at 150 degree Celsius. (b) By using Lever Rule, calculate the weight fraction of each phase for alpha, WA = Q / (P + Q) and beta, WB = P / (P + Q) where Q = CB - CO and P = CO - CA.
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