• Entropy is that property of a substance which remains constant if no heat enters or leaves the substance, while it does work or alters its volume, but which increases or diminishes should a small amount of heat enter or leave.

• The Third Law of Thermodynamics means that as the temperature of a system approaches absolute zero, its entropy approaches a constant (for pure perfect crystals, this constant is zero).

“The entropy of a substance varies

with the temperature of the substance.

The lower the temperature,the lower the entropy. “

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

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

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.

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

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

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

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

SAMPLE PROBLEM

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

SOLUTION

GIVEN:

R= 0.3704 psia. Ft³/ lbm.R Cp= 0.240 Btu/ lbm . R Cv= 0.171 Btu/ lbm . R k= 1.4

1-2 s=c

2-3 p=c

3-4 s=c

4=1 v=c

a. V2= V1 = 117 in³

r 18

V2= 6.5 in ³

V3= rc V2 = (2)(6.5 in³)

V3= 13 in³

V4= V1 = 117 in³

Process 1-2 (isentropic compression of an ideal gas, constant specific heat)

T2= T1 V1 ^ k-1 = (540 R)(18) ^ 1.4-1 = 1716 °RV2

P2 = P1 V1 ^ k = (14.7 psia)(18)^ 1.4 = 841 psia V2

Process 2-3 (constant- pressure heat addition to an ideal gas)

P3= P2= 841 psiaP2 V2 = P3V3

T2 T3

T3= T2 V3 = (1716 R)(2) = 3432 °R

V2

Process 3-4 ( isentropic expansion of an ideal gas, constant specific heats)

T4 = T3 V3 ^ k-1 = (3432 R) 13 in³ ^ 1.4 – 1 = 1425 °R V2 117in³

P4= P3 V3 ^ k = (841 psia) 13 in³ ^ 1.4 = 38.8 psia

V4 117 in³

b. m = P1V1 = (14.7 psia) ( 117 in³) 1 ft³

RT1 (0.307 psia. ft³/ lbm . R)(540 R) 1728 in³

= 0.00498 lbm

Qin= mCp(T3-T2)

= ( 0.00498 lbm)(0.240 Btu/ lbm . R) [(3432-1716 R)]

Qin= 2.051 Btu

Qout= mCv(T4-T1)

= (0.00498 lbm)(0.171 Btu/ lbm. R)[(1425- 540) R]

Qout= 0.754 Btu

Wnet = Qin – Qout = 2.051 – 0.754 = 1.297 Btu

Thermal efficiency= Wnet x 100% = 1.293 Btu x 100 % Qin 2.051 Btu

Thermal efficiency = 63.2 %

C. Pm = Wnet

Vmax- Vmin

= Wnet

V1- V2

= 1.297 Btu 778.17 lbf. ft 12 in

(117 – 6.5) in³ 1 Btu 1 ft

Pm= 109.6 psia

S 1-3 = 0

S 2-3 = m Cp ln T3

T2

= ( 0.00498 lbm) (0.240 Btu/ lbm . R)ln (3432 °R) – (0.00498 lbm) (0.240 Btu/ lbm . R)ln (1716 R)

S 2-3 = 0.00828

S 3-4 = 0

S 4-1= m Cv ln T4

T1

= ( 0.00498 lbm) (0.171 Btu/ lbm . R) ln (1425) -

(0.00498 lbm) (0.171 Btu/ lbm . R) ln (540 R)

S 4-1 =0.00826

P-v DIAGRAM

P, psia

Qin v=c 841 2 60 3

s=c

s=c

38.8 4 p=c

Qout 14.7 30 1

1 60

6.5 13 117 v(in³)

T-S DIAGRAM

T (°R)

3432 Qin 3 T3

1716 2 30 T2

1425 4 T4

Qout

540 1 60 T1

s

REFERENCES: www.allaboutscience.org/third-law-of-thermodynamics-faq.htm http://en.wikipedia.org/wiki/Absolute_zero voh.chem.ucla.edu/vohtar/summer04/classes/14B/.../

notes81304.pdf www2.ucdsb.on.ca/tiss/stretton/CHEM2/entropy4.htm THERMODYNAMICS AN ENGINEERING APPROACH 4th edition

YUNUS A. CENGEL and MICHAEL A. BOLES, 465 Thermodynamics 1, Hipolita B. Sta. Maria, 46