6. A pure perfect crystal is one in which every molecule is identical, and the molecular alignment is perfectly even throughout the substance. This means that in a perfect crystal, at 0 Kelvin, nearly all molecular motion should cease in order to achieve ΔS=0. The crystal must be perfect, or else there will be some inherent disorder. It also must be at 0 K; otherwise there will be thermal motion within the crystal, which leads to disorder. http://www.allaboutscience.org/third-law-of-thermodynamics-faq.htm
7. For non-pure crystals, or those with less-than perfect alignment, there will be some energy associated with the imperfections, so the entropy cannot become zero. http://www.allaboutscience.org/third-law-of-thermodynamics-faq.htm
8. At a temperature of absolute zero there is no thermal energy or heat. At a temperature of zero Kelvin the atoms in a pure crystalline substance are aligned perfectly and do not move. There is no entropy of mixing since the substance is pure.
9. At absolute zero, the entropy of a pure crystal is also zero. S = 0 at T = 0 K
10. At temperatures near 0 K, nearly all molecular motion ceases and, when entropy = S, ΔS = 0 for any adiabatic process. Pure substances can (ideally) form perfect crystals as T -> 0. Max Planck's strong form of the third law of thermodynamics states the entropy of a perfect crystal vanishes at absolute zero. The original Nernst heat theorem makes the weaker and less controversial claim that the entropy change for any isothermal process approaches zero as T -> 0:The implication is that the entropy of a perfect crystal simply approaches a constant value.http://en.wikipedia.org/wiki/Absolute_zero
11. Perfect crystals never occur in practice; imperfections, and even entire amorphous materials, simply get "frozen in" at low temperatures, so transitions to more stable states do not occur. http://en.wikipedia.org/wiki/Absolute_zero
12. Based on the third law we can set an absolute scale for entropy based on the “perfect crystal”. S = klnW voh.chem.ucla.edu/vohtar/summer04/classes/14B/.../notes81304.pdf
13. Where: K = Boltzman’s constant = 1.381*10-23 J/K ln = natural log = 2.3*log lnX = 2.3logX W = degree’s of molecular orientation raised to the power of the number of molecules. voh.chem.ucla.edu/vohtar/summer04/classes/14B/.../notes81304.pdf
15. Four process cycle: An Ideal Diesel cycle with air as the working fluid has a compression ratio of and a cutoff ratio of 2. At the beginning of the compression process, the working fluid is at 14.7 psia, 80 °F, and 117 in³. Utilizing the cold-air standard assumptions, determine: A. temperature and pressure of the air at the end of each process B. the network output and the thermal efficiency, and C. the mean effective pressure THERMODYNAMICS AN ENGINEERING APPROACH 4th edition YUNUS A. CENGEL and MICHAEL A. BOLES, 465
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