materials 1m03 – assignment 2 - summer 2013 -...
TRANSCRIPT
Due Date: July 10th 2013 Name: ____________________ Student #: _________________
Materials 1M03 – Assignment 2 – Answer Sheet
1. (3 marks) Calculate the energy for vacancy formation (in eV/atom) in Nickel given that the equilibrium number of vacancies at 500°C is 3x10-4 per m3. The density for this metal is 8.90 g/cm3.
Use 3 significant figures in the answer.
Answer:
!! = ! !(!!!!" )
−kT ln!!! = !!
Solve for N, look up AW of Ni: 58.69 g/mol
! =!!!!"
! =6.022×10!" !"#$%!"# × 8.90 ×10
! !!!
58.69 !/!"#
! = 9.13×10!" !"#$%!!
Plug back into first equation and solve for Qv
−(8.62×10−5!"
!"#$ ∙ !)(500 + 273K) ln(
3×10!! !!!
9.13×10!" !!!) = !!
!! = 4.98 !"/!"#$
2. (1 mark) For a Pb-Sn alloy that consists of 81 wt% Pb, what is the composition of Sn (in at%)? Use 4 significant figures in the answer
Answer: !!! = !!!!
!!!!!!!!!
!!! = (100− 81)!!
(100− 81)!! + 81!! ×100%
!!! = (19 × 207.2 !/!"#)
(19 × 207.2 !/!"#)+ (81 × 118.71 !/!"#) ×100%
Due Date: July 10th 2013 Name: ____________________ Student #: _________________
!!! = 29.05%
3. (1 mark) Which of these compounds would most likely result in an interstitial
solid solution?
A) Copper and Nickel B) Carbon and Iron C) Titanium and Chromium D) Sodium and Chlorine E) Tin and Silver F) B, D, E G) All of the above H) None of the above Answer: B
4. (3 marks) The activation energy for the diffusion of atomic species A in metal B is 111 kJ/mol. It is also known that the value of D at 1127 °C is 7x10-11 m2/s. Calculate the diffusion coefficient (in m^2/s) at 827 °C.
Answer: ! = !!exp (!!!!")
7x10!!!!! ! = !!exp (−111000 !
(8.314 !!"# ∙ !)(1400 !)
)
!! = 9.70×10!!!! !
! = 9.70×10!!!! ! exp (−111000 !
(8.314 !!"# ∙ !)(1100 !)
)
! = 5.20×10!!"!!/!
5. (2 mark) Define Steady-State Diffusion
Answer: Where there is no net accumulation or depletion of diffusing series. The diffusion flux is independent of time.
6. (3 marks) It's possible to purify hydrogen gas by diffusing it through a thin sheet of palladium.
Due Date: July 10th 2013 Name: ____________________ Student #: _________________
Calculate the number of kilograms of hydrogen that pass through a 1.5 mm thick sheet of palladium per hour under the following conditions: The palladium sheet has an area of 0.8 m2. The temperature during the diffusion is 300 °C. The diffusion coefficient is 1.33 m2/s. The concentrations on the high and low-pressure sides of the sheet are 4 and 0.6 kg of hydrogen per cubic meter of palladium. Steady-state conditions have been attained.
(Give your answer in kg/hr)
Use scientific notation, significant figures in answer: 3
Answer: ! = −! !"!"
! = −(1.33!!
! )(0.6− 4) !"!!
(1.5×10!! !)
! = 3,014.67!"!! ∙ !
#kg/hr = J*t*area
#kghr = 3,014.67
!"!! ∙ !× 3600
!ℎ! × 0.8!
!
#kghr = 8.68 ×10! !"/ℎ!
7. (1 mark) The ____________________ of a diffusing species is the amount of
energy required to cause 1 mole of atoms to diffuse within a host metal.
Answer: Activation Energy
8. (1 mark) True or False
The diffusion flux is the driving force for diffusion (the force that compels a reaction to occur).
Answer: False, the concentration gradient is the driving force
9. (3 marks) If an evil doctor came up with an evil plan to booby-trap the classroom with an arsenal of 2L diet cola containers and a spherical mint candy to drop in
Due Date: July 10th 2013 Name: ____________________ Student #: _________________
each. How small in diameter (in cm) would each spherical mint candy need to be for the reactions not to occur thus saving the class!
Assume the surface energy of the mint candy is 1,560 J/m2 and the volume free energy of solution is -300,000 J/m3.
Answer:
∆!!"! = 43!!
!∆!!"# + 4!!!!!
!∆!!"!!" = 0 = 4!!!∆!!"# + 8!"!!
r∗ = −2!!∆!!"#
r∗ = −2(1560)−300,000
r∗ = 1.04 cm
d < 2.08 cm
10. (2 marks) What are the two conditional terms that must be met for an atom to diffuse from one lattice site to another?
Answer:
a. There must be an empty adjacent site b. The atom must have sufficient energy
Mark: / 20