THIRD LAW
H. THIRD LAW
The Third law states that a crystalline solid substance at an absolute temperature of zero (i.e., 0 K) possesses zero entropy. This implies that the substance exists in a state of per- fect order at that temperature in the absence of energy, a condition that is not particularly use- ful, since it is no longer possible to extract work from it. In other words, the entropy of any crystalline matter tends to zero as ∂U/∂S → 0. We will see in Chapter 7 that s(0 K) is inde-
pendent of pressure, i.e., s(0 K, P =1 bar) = s(0 K, P). Entropy values are tabulated for most substances using the datum s = 0 at 0 K.
In general, substances at low temperatures exist in the condensed state so that for an incompressible substance
ds = c s dT/T. At very low temperatures the specific heat–temperature relation for a solid, c m
s = αT can be applied, so that
(s – s m ref (0)) = αT /m, m ≠0 (61a) For Debye solids m = 3, and α = (1944/θ 3
(61b) where θ D is a constant dependent upon the solid.
D ) kJ kmole –1 K –4 , at T < 15 K,
In summary, according to the Third law s = s ref (0) = 0 at an absolute temperature of zero.
v. Example 22 The specific heat of a Debye solid (for temperatures less than 15 K) is represented by the relation c = (1944 T 3 / 3 ) kJ kmole s –1 θ K –1 . Obtain a relation for the entropy with respect to temperature for cyclopropane C 3 H 6 for which θ = 130. What are the values
of the entropy and internal energy at 15 K. (cf. also Figure 33 ). Solution The molar specific entropy (0 K, 1 bar) = 0 kJ kmole –1 K –1 (i.e., point A in Figure 33a ). Since s =∫ s c dT T /,
s = (1944 T 3 / 3 θ )/3 kJ kmole –1 K –4 .
At 15 K, s (15 K, 1 bar) = (1944 × 15 3 /130 3 )/ 3 = 0.995 kJ kmole –1 K –1 . The internal energy
u = ∫c s dT so that at
15 K, u 4 = (1944 × 15
11.2 kJ kmole –1 . w. Example 23
At a pressure of 1 bar, evaluate the entropy and internal energy of
cyclopropane C 3 H 6
when it exists as (a) saturated solid; (b)
saturated liquid; and (c) saturated vapor
given that the specific heat c s follows the Debye equation with
θ D = 130 K and m = 3 when T < 15 K, c s =
28.97 kJ kmole –1 K –1 Figure 33: Illustration of a: a: P–T diagram; b. s–T diagram. for 15 K < T < T MP , where the melting point T
MP = 145.5 K at P = 1 bar, h sf = 5442 kJ k mole , and the normal boiling point
T –1
= 240.3 K; and for the liquid c l = 76.5 kJ kmole K , and h fg = 20,058 kJ kmole –1 .
From Example 22, –1 s (s, 15 K) = 0.99 kJ kmole K (at point B in Figure 33 ). Hence, s (s, 145.5) – s (s,15) = 28.97 ln (145 ÷15) (i.e., point C, saturated solid in Figure 33a
and b which represents a saturated solid that is ready to melt). Therefore, s (s, 145.5) = 65.72 + 0.99 = 66.71 kJ kmole –1 K –1 , and u (s, 145.5) = 28.97
× (145 – 15) + 11.2 = 377.7 kJ kmole –1 . The liquid entropy may be evaluated as follows:
s (l, 145.5) – s(s,145.5) = 5442 ÷145.5 = 37.40 kJ kmole –1 K –1 . Therefore,
s (l, 145.5) = 37.4 + 66.71 = 104.11 kJ kmole –1 K –1 (point D in Figure 33 ) so that s (l, 145.5) = 37.4 + 66.71 = 104.11 kJ kmole –1 K –1 (point D in Figure 33 ) so that
Since h = u + Pv, and for solids and liquids Pv « u, it follows that for these substances
h ≈ u or h sf ≈u sf and h fg ≈u fg . Hence,
u(l,145.5) = 377.7 + 5442 =5819.9 kJ kmole –1 , and u(l,240.3) = 5819.9 + 76.5
× (240.3 – 145.5) = 13132 kJ kmole –1 . In the gaseous state
s (g, 240.3) – s (l,240.3) = 20,058 ÷240.3 = 83.5 kJ kmole –1 K –1 , i.e., s (g,240.3) = 83.5 + 142.51 = 226.01 kJ kmole –1 K –1 (point G in Figure 33 ).
Therefore, s (g,240.3) = 226 kJ kmole –1 K –1 , and h(g, 240.3) = 13132 + 20.058 = 33,190 kJ kmole –1 . u(g, 240.3) = 33,190 - 8.314 –1 × 240.3 = 31,192 kJ kmole .
Remarks If the reference condition for the entropy is selected at the saturated liquid state (i.e., at point D), we can arbitrarily set s = 0 there. Therefore, at point C, saturated solid, s C
= –37.40 kJ kmole –1 K –1 , and at point B, s B = –37.40 – 65.72 = –103.12 kJ kmole –1 K –1 . Such a procedure is generally used for water, since the reference condition with respect to its entropy is based on the saturated liquid state at its triple point tempera- ture of 0.01ºC. At this state it is usual to set s = h ≈ u = 0. Methods for evaluating at any pressure and temperature will be discussed in Chapter 7. Recall from the First law that δQ - δW =dU. Therefore, if a process involves irre-
versibility, then dS = δQ/T b + δσ so that dS = dU/T b + δW/ T b + δσ.
In case the process is mechanically reversible, then the entropy balance equation for a closed system can also be written as
dS = dU/ T b + P dV/ T b + δσ.
where δσ >0 for irreversible processes and equals zero for reversible processes.