SolutionsHW2

SolutionsHW2 - HW 2 Problem 4.2 a. To Find: Number of...

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HW 2 Problem 4.2 a. To Find: Number of vacancies per cubic meter at a given temperature. b. Given: T = 850 degrees C = 1123 K Q v = 1.08 eV/atom Density of Fe ( Fe ρ ) = 7.65 g/cc Atomic weight of iron ( Fe A ) = 7.65 g/mol c. Assumptions: (i) The question asks for the equilibrium number of vacancies. (ii) The system is always in equilibrium with its surroundings during the process of attaining the said temperature of 1123 K. (Consider a material ‘quenched’, i.e., rapidly cooled from a ‘high’ temperature of say, 1163K to 1123 K. Since the system has no time to attain thermal equilibrium, it will retain the vacancies that it had at 1163 K at 1123 K!) d. Solution: kT Q v v exp = …(1) Step 1: Calculate N using one of two equivalent methods. Method 1 From density data: 7.65 g ≡ 1cc b 1g ≡ (1/7.65) cc …(2) From atomic weight data: 55.85 g ≡ 1 mol b 55.85 g ≡ 6.023 * 10 23 atoms b 1g ≡ 1/(6.023 * 10 23) atoms …(3)
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Hence, from (2) and (3), 6.023 * 10 23 /55.85 atoms ≡ (1/7.65) cc Or, there are 7.65 * 6.023 * 10 23 /55.85 atoms /cc . Method 2 Alternately, the following formula may be used to directly calculate = Fe Fe A A ρ And, = kT Q A v Fe Fe A v exp Step 2: Substitute values in equation (1). v = ( 6.022 × 10 23 atoms /mol )( 7.65 g /cm 3 ) 55.85 g /mol exp 1.08 eV/atom ( 8.62 × 10 5 eV/atom K ) (850 ° C + 273 K) &v = 1.18 × 10 18 cm -3 OR &v = 1.18 × 10 24 m -3
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Problem 4.9 a. To Find: The weight percentages of the constituents of an alloy, given the weights of these individual constituents. b. Given: Mass of titainium, m Ti = 218 kg Mass of aluminum, m Al = 14.6 kg Mass of vanadium, m V = 9.7 kg c. Assumptions: The alloy contains no other alloying agents – the presence of other elements will modify the values of the weight percentages. d. Solution: For this alloy, the concentration of titanium ( C Ti ) : C Ti = m Ti m Ti + m Al + m V × 100 = 218 kg 218 kg + 14.6 kg + 9.7 kg × 100 = 89.97 wt% Similarly, for the concentration of aluminum ( C Al ) : C Al = 14.6 kg 218 kg + 14.6 kg + 9.7 kg × 100 = 6.03 wt% For the concentration of vanadium ( C V ) : C V = 9.7 kg 218 kg + 14.6 kg + 9.7 kg × 100 = 4.00 wt% C Ti = 89.97 wt% ; C Al = 6.03 wt%; C V = 4 wt%
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Problem 4.11 a. To Find: To find the composition, in atom percent, of an alloy, given the weights of the individual constituents of the alloy. b. Given:
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SolutionsHW2 - HW 2 Problem 4.2 a. To Find: Number of...

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