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Thermodynamics of substances with negative thermal expansion and negative compressibility
Journal of Non-Crystalline Solids 356 (2010) 1168–1172
Contents lists available at ScienceDirect
Journal of Non-Crystalline Solids j o u r n a l h o m e p a g e : w w w. e l s e v i e r. c o m / l o c a t e / j n o n c r y s o l
Thermodynamics of substances with negative thermal expansion and negative compressibility I.A. Stepanov ⁎ Department of Engineering Materials, University of Sheffield, Sir Robert Hadfield Building, Mappin Street, Sheffield, S1 3JD, UK
a r t i c l e
i n f o
Article history: Received 4 July 2009 Received in revised form 27 February 2010 Available online 4 May 2010 Keywords: Entropy Internal energy Negative thermal expansion Heat capacity Mayer's relation
a b s t r a c t It is shown that for substances with positive thermal expansion and positive compressibility, and for substances with negative thermal expansion and negative compressibility, δQ = dU + PdV, but for substances with positive thermal expansion and negative compressibility, and for substances with negative thermal expansion and positive compressibility, δQ = dU − PdV. The result obtained helps to calculate processes which do not obey traditional thermodynamics. © 2010 Elsevier B.V. All rights reserved.
1. Introduction According to thermodynamics [1],
∂S ∂V
= T
α : β
ð1Þ
Here, S is the entropy, V is the volume, α is the thermal expansion coefficient, T is the temperature and β is the isothermal com ∂V pressibility − V1 where P is pressure. One can easily obtain this ∂P T result: 1 ∂V 1 ∂V ∂P ∂S =− =β : V ∂T P V ∂P T ∂T V ∂V T ∂S b0. From Eq. (1), when α b 0, ∂V T On the other hand, α=
δQ ≤ TdS = dU + PdV
∂S ∂V
U
P N 0: = T
for all α ⁎ Tel.: +44 1142225951. E-mail addresses: i.stepanov@sheffield.ac.uk, [email protected]. 0022-3093/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.jnoncrysol.2010.03.013
∂V T
ð2Þ
∂V U
2. Theory According to [1], ð3Þ
and
dU ∂U ∂U ∂U However, = ≠0, and generally ≠0 and + ∂T X ∂X T ∂T X dT dX ∂U ≠0 where X=P, V, S, H, F and G [1], H is the enthalpy, F is the ∂X T dT Helmholtz free energy and G is the Gibbs free energy. Therefore, almost always, when U is constant then T is constant (as will be proven in detail below). When U is constant and T is constant then the right hand sides of Eqs. (1) and (4) must have the same sign. So, there is a contradiction in the thermodynamic equations. One has to pay attention that in a heat exchange, one introduces the quantity of heat δQ into the substance, and U and T vary. For this case, when U changes then T changes, and vice versa, thus when U is a constant then T is a constant, and vice versa. For some ∂U ∂T processes, of course, and are not equal to zero.
ð4Þ
∂U ∂T V
= CV ,
∂U = CP −αPV; ∂T P
ð5Þ
∂U C βP = V ; αT ∂T S
ð6Þ
∂U C −αPV−CV βP ; = P 1−αT ∂T H ∂U βSP−αTS + CV βP ; = βP ∂T F
ð7Þ ð8Þ
I.A. Stepanov / Journal of Non-Crystalline Solids 356 (2010) 1168–1172
∂U = CP −αPV−ðαT−βP ÞS ∂T G
ð9Þ
and
1169
and ∂U C βPV−ðCP −αPV ÞV = V : CP ∂P H
ð20Þ
∂U αT−βP ; = β ∂V T
ð10Þ
∂U = ðβP−αT ÞV; ∂P T
For water, with the highest accuracy, dU = 0, then Therefore, if dT
ð11Þ
dP 1 3 Pa⋅kg = = 10 : dH V m3
∂U αT−βP ; = α ∂S T
ð12Þ
∂U ∂H
αT−βP ; αT−1
ð13Þ
∂U βP−αT ; = βP ∂F T
ð14Þ
= T
∂U = βP−αT: ∂G T So, in general
dU dT
ð15Þ
=
∂U ∂T X
+
∂U dX ∂X T dT
prove that for almost all processes,
is not equal to 0. One can
dU ≠0, dT
thus when U is constant
then T is constant and the contradiction does exist. Let us find out what happens when U is constant but T is not constant [1]: ∂U ∂U αT−βP dV: dT + dV = CV dT + dU = β ∂T V ∂V T From this, if
ð16Þ
dU = 0, then dT
1 dV βCV =− : ðβP−αT ÞV V dT
ð17Þ
ð18Þ
dP CP = ∼ −108 Pa/K. For dT αTV a constant pressure, dP = 0 hence Eq. (18) is wrong. For all processes dP with ≠−108 Pa=K, when U is constant then T is constant. dT Another interesting example: recall that [1] For water, with the highest accuracy,
∂U ∂H
= P
CP −αPV : CP
∂U ∂H P
= 1 and
∂U ∂P H
= −V.
ð21Þ
For heat exchange at constant pressure, dH N 0 but dP = 0 thus dP ≠ 103 Pa⋅kg/m3, Eq. (21) is not true. In general, for all processes with dH when U is constant then T is constant. So, it is proven that for condensed matter in almost all cases, when U is a constant then T is a constant, and vice versa. This means, that for these cases
∂S ∂V
α = β
= T
∂S ∂V
:
ð22Þ
U
(There is a general relation
∂S ∂V U
=
∂S ∂U − T1 ∂V T ∂V T
[1]). Using
Eqs. (3), (4) and (22) one must come to the conclusion that for cases with α b 0, δQ ≤ TdS = dU– PdV
ð23Þ
and α = β
∂S ∂V
U
P =− : T
ð24Þ
The following consideration supports the conclusion made in this paper. There is the generalized Mayer's relation [3]:
dU ≠0 and when U For every process which does not satisfy Eq. (17), dT is constant then T is also constant. For water at 273 K and atmospheric pressure, α=−68.05·10− 6 K− 1, CP =4217.6 J/(kgK), CP −CV ≈2.5 J/(kgK) and β=5.09·10− 10 Pa− 1 [2]. 1 dV βCV = ∼−0.1 K− 1 Substituting these values into Eq. (17), one gets V dT αTV 1 dV = with the highest accuracy. At a constant pressure, V dT −5 −1 α ≈ − 7 · 10 K , thus Eq. (17) is not true for water at a constant pressure. And, in general, for all processes with constant U 1 dV ≠−0:1K1 , T is also constant. and V dT dU = 0, then From Eqs. (5) and (11) one can find that, when dT dP CP −αPV = : dT ðαT−βP ÞV
ð19Þ
CP − CV =
Tα2 V : β
ð25Þ
It is derived from the first and second laws of thermodynamics without simplification. Its derivation is given in the Appendix. If one derives it assuming that TdS = dU − PdV for α b 0, then it looks like [2]: CP −CV = −
Tα2 V b 0: β
ð26Þ
Let us consider the thermal expansion of a solid. In the first approximation, V is constant. Then, in the first approximation [1], CP =
∂H αV : = CV + β ∂T V
ð27Þ
From Eq. (27), for α b 0, CP b CV and Eq. (26) is true, not Eq. (25). In the second approximation, P = const and V ≠ const. Let us suppose that in that case CP N CV for α b 0. There is Rolle's lemma: If f(x) is continuous in the interval a ≤ x ≤ b, and if f(a) N 0 and f(b) b 0 then there exists at least one value of x (say x = c) such that f(c) = 0, where a b c b b. As heat capacity is a continuous function, from Rolle's lemma it follows that there exists one value of V [say Vc = V0 + ΔV, V0 is the constant V in Eq. (27)] such that CP(Vc) − CV = 0. This is a contradiction because from Eqs. (25) and (26), at α b 0, CP(V) − CV ≠ 0. Else, when P = const, for V0 − ΔV b V b V0 + ΔV and α b 0, due to the continuity of heat capacity, CP – CV b 0. This is another contradiction to
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I.A. Stepanov / Journal of Non-Crystalline Solids 356 (2010) 1168–1172
Eq. (25). So, it is proven that for negative thermal expansion, Eq. (26) is true. This means that for all processes ∂S ∂S 1 ∂U sign = sign − : ∂V U ∂V T T ∂V T
ð28Þ
Suppose that the quantity of heat δQ = 4217.6 J is introduced at constant atmospheric pressure to a kilogram of water at 273 K. Then if δQ = dU + PdV is true, 4217.6 J = 4220.1 J − 2.5 J where dU = 4220.1 J and PdV = − 2.5 J. This means that CP b CV. However, according to Mayer's relation CP − CV N 0. This is a contradiction. 3. Results One can draw a conclusion: in this paper it is proven that for ∂S is always negative and equals −P/T. This result
negative α,
∂V U
has been obtained previously [2,4–10]. In [2], there is a misprint in k kT
Eq. (11). The correct equation is: for α b 0, CV = CP S . Recently, substances with negative isothermal compressibility β were discovered [11–17]. For them, from Eqs. (1) and (4) it follows (when α N 0 and applying the same arguments as for the derivation of Eq. (24)) that
∂S ∂V
T
α = b0 β
ð29Þ
δQ = dU–PdV:
ð30Þ
Also, one can use the following argument: For adiabatic expansion, ∂U . For a substance with negative from Eq. (3) it follows that P = − ∂V S
β, this formula predicts the wrong sign of pressure and Eq. (3) must be replaced by Eq. (30). When α b 0 and β b 0, ∂S ∂V
ð31Þ
δQ = dU + PdV:
ð32Þ
T
and
So, when α N 0 and β N 0 or α b 0 and β b 0 then δQ = dU + PdV;
ð33Þ
and when α N 0 and β b 0 or α b 0 and β N 0 then δQ = dU–PdV:
ð34Þ
In [18], a result is obtained that fully supports the conclusion of this paper. There, a stretched solid of unit volume is considered. For this sample: dS =
∂S ∂σij
ð37Þ
which agrees strongly with Eqs. (33) and (34). In [9], it is shown that many phase transitions in ice disobey the Clausius–Clapeyron equation very much despite it being obtained from the first and second laws of thermodynamics without simplifications. If one derives the Clausius–Clapeyron equation taking into account that for α b 0, TdS = dU – PdV then it reads: dT T ðsignα2 V2 −signα1 V1 Þ = : dP ΔH
ð38Þ
The ordinary Clausius–Clapeyron equation reads dT T ðV2 −V1 Þ : = ΔH dP
ð39Þ
In [9], Eq. (38) was applied to these phase transitions and good agreement was observed, Table 1.
The reason why it has been supposed that the first law of thermodynamics (3) is tenable is that it agrees with experiments on substances with positive thermal expansion, and with the Clausius– Clapeyron equation which describes the melting of ice with high accuracy. Water has negative expansion at P = 14 MPa [19]. At this pressure ice melts at T ≈ 272 K. In [20] there is an experimental point in the interval 272–273 K: 4.9 MPa and T = 272.8 K. This obeys Eq. (39): Eq. (38)
α N0 β
=
TΔS = ΔU + 3Tασ = ΔU + signðαÞPΔV
4. Discussion
and
where the coefficients are constants. Assuming that the material is isotropic, α ij = αδ i j and σij = σδi j, this equation can be written in the form
!
dσij +
T
∂S ∂T
σ
dT = αij dσij +
Cσ dT: T
ð35Þ
where σij is the stress tensor, αij is the thermal expansion tensor and Cσ is the heat capacity per unit volume at constant stress. Eq. (35) is the exact equation derived without simplifications. For small stresses and temperature changes the properties will always be linear and it turns into: σ
TΔS = Tα ij σij + C ΔT
ð36Þ
dT = − 0.074 K/MPa. The modified Clausius–Clapeyron dP dT = −1.71 K/MPa. However, now that substances gives dP
with negative compressibility have been discovered, one can suggest that water has negative isothermal compressibility in the temperature range where it has negative expansion. In [15] it was shown that negative thermal expansion and negative compressibility can coexist. So, the first law of thermodynamics does not seem to be that unshakable as it was thought earlier. The author also thinks that a very strong proof is provided by Eqs. (35)–(37). An important result which confirms the conclusion of this paper is obtained in [8]. There it is found that the difference between the heats of chemical reactions measured by the Van't–Hoff equation, ΔH0, and obtained by calorimetry, ΔQ0, is PΔV 0. The Van't–Hoff equation must give the same results as calorimetry because it is derived from the first and the second laws of thermodynamics
Table 1 Phase transitions in ice measured experimentally and described by the Clausius– Clapeyron Eq. (39) and by the modified Clausius–Clapeyron Eq. (38) [9]. Pressure, Temperature, Transition K between ice MPa phases
dT/dP, Clausius– dT/dP, Eq. dT/dP, (38), K/MPa experiment, Clapeyron equation, K/MPa K/Mpa
LDA → Ic 0.1 HDA → LDA 0.1 LDA → HDA 600 III → V 344 V → VI 600 VI → VII 2000 VI → VIII 1900
− 0.09 0.18 − 0.26 3.6 − 8.7 3.7 0.21
∼ 140 ∼ 120 ∼77 ∼ 250 ∼ 260 ∼ 320 ∼ 240
0.08 − 0.43 0.28 − 0.26 − 0.26 − 0.30 − 0.23
− 0.08 0.43 − 0.28 7.1 − 8.1 3.5 0.23
I.A. Stepanov / Journal of Non-Crystalline Solids 356 (2010) 1168–1172
without simplifications. However, the difference between the two methods is far beyond the error limits. Nobody has succeeded in explaining this phenomenon. The author in [8] supposed that, for endothermic chemical reactions with increasing volume and for exothermic chemical reactions with decreasing volume, the law of conservation of energy is: ΔU = ΔQ + PΔV + ∑ μ i ΔNi ; (notice the change of sign in front of P and that
∂S ∂V U;Ni
Derivation of the generalized Mayer's relation In general dS =
∂S ∂P
ð41Þ
Here, K is the equilibrium constant of the chemical reaction. The traditional Van't–Hoff equation is: d ΔH0 ΔU 0 + PΔV 0 ln K = = : 2 dT RT RT 2
dT +
P
∂S ∂P
dP;
ðA1Þ
T
ð42Þ
∂V =− ; ∂T P
ðA2Þ
CP dT−αVdP: T
ðA3Þ
T
dS =
0
d ΔQ + PΔV ln K = : dT RT 2
∂S ∂T
and from the definition of CP and the Maxwell relation P T
= − b 0),
here μ i are chemical potentials and ΔNi are changes in the number of moles of the substance i. If one derives the Van't–Hoff equation using Eq. (40) then it is: 0
Appendix
ð40Þ
i
1171
Dividing Eq. (A3) by dT and choosing the direction of V = constant one obtains
∂S ∂T
= V
CV C ∂P = P −αV : T T ∂T V
ðA4Þ
From The Van't–Hoff equation has often been used for the determination of the heat of chemical reactions. It gives correct results for reactions with ΔV → 0. Therefore, it has always been assumed that, for chemical reactions, ΔQ = ΔH = ΔU + PΔV:
ð43Þ
However, from the traditional first law of thermodynamics it follows that ΔQ = ΔU + PΔV− ∑ μ i ΔNi :
ð44Þ
i
From Eqs. (43) and (44), ΔQ = ΔQ− ∑ μ i ΔNi . One may not neglect i
the term ∑ μ i ΔNi . Thus, the Van't–Hoff equation cannot give correct i
results either for ΔV → 0 or for ΔV ≠ 0. For reactions with ΔV ≠ 0 this equation gives the wrong results.
dV = αVdT–βVdP it follows that
∂P ∂T
The result obtained in this paper is of fundamental importance if true. It will lead to a major revision of thermodynamics. Any limitations of this theory have to be properly identified by the scientific community at large. Many equations of thermodynamics will change and some of them will change drastically, for example the Clausius– Clapeyron equation. This equation is used for the calculation of phase diagrams but often gives enormously wrong results [4]: a paradox not explained before. However, its modified form, Eq. (38), produces very reasonable results [4]. The result of this paper is very important in high pressure physics and it has to be checked by high pressure experiments. It already produces good results in the calculation of phase transitions and heats of chemical reactions which do not obey traditional thermodynamics, as shown above. A direct measurement of the compressibility of water is necessary in the interval 273–277 K where it has negative expansion. If it is proven that water has negative compressibility in this interval, then this would prove the theory of this article once and for all. Other high pressure experiments, for example, phase transition studies are also of great importance.
= V
α : β
ðA6Þ
Introducing Eq. (A6) into Eq. (A4) results in the generalized Mayer's relation: CV C α2 V : = P− β T T
ðA7Þ
If one supposes that for α b 0, TdS = dU – PdV, then the Maxwell relation (A2) will look like
5. Conclusion
ðA5Þ
∂S ∂P
= T
∂V ∂T P
ðA8Þ
and the generalized Mayer's relation will be CV C α2 V : = P + β T T
ðA9Þ
References [1] M.Kh. Karapetyants, Chemical Thermodynamics, Vysshaja Shkola, Moscow, 1974, pp. 108–110, (English translation: M. Kh. Karapetyants, Chemical Thermodynamics, (Translated by G. Leib), Mir, Moscow, 1978). [2] I.A. Stepanov, Comput. Modell. New Technol. 4 (2) (2000) 72 (Available at www. tsi.lv). [3] J. Guemez, C. Fiolhais, M. Fiolhais, Am. J. Phys. 67 (1999) 1100. [4] I.A. Stepanov, Physica B 349 (2004) 251. [5] I.A. Stepanov, in: A.N. Linke (Ed.), Progress in Chemical Physics Research, Nova Science Publishers, Inc, Hauppauge, NY, 2005, pp. 107–114. [6] I.A. Stepanov, in: A.N. Linke (Ed.), Progress in Chemical Physics Research, Nova Science, Nova Science Publishers, Inc, Hauppauge, NY, 2005, pp. 115–120. [7] I.A. Stepanov, in: A.N. Linke (Ed.), Chemical Physics: New Research, Nova Science Publishers, Inc, Hauppauge, NY, 2006, pp. 83–87. [8] I.A. Stepanov, in: A.N. Linke (Ed.), Trends in Chemical Physics Research, Nova Science Publishers, Inc, Hauppauge, NY, 2006, pp. 117–127.
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[9] I.A. Stepanov, in: S.V. Arnold (Ed.), Chemical Physics Research Trends. Horizons of World Physics, vol. 252, Nova Science Publishers, Inc, Hauppauge, NY, 2007, pp. 335–341. [10] I.A. Stepanov, Chem. Phys. Res. J. 2 (3/4) (2008) 277. [11] R. Gatt, J.N. Grima, Phys, Stat. Sol. (RRL) 2 (2008) 236. [12] R. Lakes, K.W. Wojciechowski, Phys. Stat. Sol. B 245 (2008) 545. [13] R.H. Baughman, S. Stafström, Ch. Cui, S.O. Dantas, Science 279 (1998) 1522. [14] B. Moore, T. Jaglinski, D.S. Stone, R.S. Lakes, Cell. Polym. 26 (2007) 1.
[15] A.L. Goodwin, D.A. Keen, M.G. Tucker, PNAS 105 (2008) 18708. [16] J.N. Grima, D. Attard, R. Gatt, Phys. Stat. Sol. B 245 (2008) 2405. [17] W. Ren, J.T. Ye, W. Shi, Z.K. Tang, C.T. Chan, P. Sheng, New J. Phys. 11 (2009) 103014. [18] J.F. Nye, Physical Properties of Crystals, Clarendon Press, Oxford, 1985, p. 176. [19] NIST Webbook. Available at http://webbook.nist.gov. [20] S.J. Henderson, R.J. Speedy, J. Phys. Chem. 91 (1987) 3069.
Contents lists available at ScienceDirect
Journal of Non-Crystalline Solids j o u r n a l h o m e p a g e : w w w. e l s e v i e r. c o m / l o c a t e / j n o n c r y s o l
Thermodynamics of substances with negative thermal expansion and negative compressibility I.A. Stepanov ⁎ Department of Engineering Materials, University of Sheffield, Sir Robert Hadfield Building, Mappin Street, Sheffield, S1 3JD, UK
a r t i c l e
i n f o
Article history: Received 4 July 2009 Received in revised form 27 February 2010 Available online 4 May 2010 Keywords: Entropy Internal energy Negative thermal expansion Heat capacity Mayer's relation
a b s t r a c t It is shown that for substances with positive thermal expansion and positive compressibility, and for substances with negative thermal expansion and negative compressibility, δQ = dU + PdV, but for substances with positive thermal expansion and negative compressibility, and for substances with negative thermal expansion and positive compressibility, δQ = dU − PdV. The result obtained helps to calculate processes which do not obey traditional thermodynamics. © 2010 Elsevier B.V. All rights reserved.
1. Introduction According to thermodynamics [1],
∂S ∂V
= T
α : β
ð1Þ
Here, S is the entropy, V is the volume, α is the thermal expansion coefficient, T is the temperature and β is the isothermal com ∂V pressibility − V1 where P is pressure. One can easily obtain this ∂P T result: 1 ∂V 1 ∂V ∂P ∂S =− =β : V ∂T P V ∂P T ∂T V ∂V T ∂S b0. From Eq. (1), when α b 0, ∂V T On the other hand, α=
δQ ≤ TdS = dU + PdV
∂S ∂V
U
P N 0: = T
for all α ⁎ Tel.: +44 1142225951. E-mail addresses: i.stepanov@sheffield.ac.uk, [email protected]. 0022-3093/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.jnoncrysol.2010.03.013
∂V T
ð2Þ
∂V U
2. Theory According to [1], ð3Þ
and
dU ∂U ∂U ∂U However, = ≠0, and generally ≠0 and + ∂T X ∂X T ∂T X dT dX ∂U ≠0 where X=P, V, S, H, F and G [1], H is the enthalpy, F is the ∂X T dT Helmholtz free energy and G is the Gibbs free energy. Therefore, almost always, when U is constant then T is constant (as will be proven in detail below). When U is constant and T is constant then the right hand sides of Eqs. (1) and (4) must have the same sign. So, there is a contradiction in the thermodynamic equations. One has to pay attention that in a heat exchange, one introduces the quantity of heat δQ into the substance, and U and T vary. For this case, when U changes then T changes, and vice versa, thus when U is a constant then T is a constant, and vice versa. For some ∂U ∂T processes, of course, and are not equal to zero.
ð4Þ
∂U ∂T V
= CV ,
∂U = CP −αPV; ∂T P
ð5Þ
∂U C βP = V ; αT ∂T S
ð6Þ
∂U C −αPV−CV βP ; = P 1−αT ∂T H ∂U βSP−αTS + CV βP ; = βP ∂T F
ð7Þ ð8Þ
I.A. Stepanov / Journal of Non-Crystalline Solids 356 (2010) 1168–1172
∂U = CP −αPV−ðαT−βP ÞS ∂T G
ð9Þ
and
1169
and ∂U C βPV−ðCP −αPV ÞV = V : CP ∂P H
ð20Þ
∂U αT−βP ; = β ∂V T
ð10Þ
∂U = ðβP−αT ÞV; ∂P T
For water, with the highest accuracy, dU = 0, then Therefore, if dT
ð11Þ
dP 1 3 Pa⋅kg = = 10 : dH V m3
∂U αT−βP ; = α ∂S T
ð12Þ
∂U ∂H
αT−βP ; αT−1
ð13Þ
∂U βP−αT ; = βP ∂F T
ð14Þ
= T
∂U = βP−αT: ∂G T So, in general
dU dT
ð15Þ
=
∂U ∂T X
+
∂U dX ∂X T dT
prove that for almost all processes,
is not equal to 0. One can
dU ≠0, dT
thus when U is constant
then T is constant and the contradiction does exist. Let us find out what happens when U is constant but T is not constant [1]: ∂U ∂U αT−βP dV: dT + dV = CV dT + dU = β ∂T V ∂V T From this, if
ð16Þ
dU = 0, then dT
1 dV βCV =− : ðβP−αT ÞV V dT
ð17Þ
ð18Þ
dP CP = ∼ −108 Pa/K. For dT αTV a constant pressure, dP = 0 hence Eq. (18) is wrong. For all processes dP with ≠−108 Pa=K, when U is constant then T is constant. dT Another interesting example: recall that [1] For water, with the highest accuracy,
∂U ∂H
= P
CP −αPV : CP
∂U ∂H P
= 1 and
∂U ∂P H
= −V.
ð21Þ
For heat exchange at constant pressure, dH N 0 but dP = 0 thus dP ≠ 103 Pa⋅kg/m3, Eq. (21) is not true. In general, for all processes with dH when U is constant then T is constant. So, it is proven that for condensed matter in almost all cases, when U is a constant then T is a constant, and vice versa. This means, that for these cases
∂S ∂V
α = β
= T
∂S ∂V
:
ð22Þ
U
(There is a general relation
∂S ∂V U
=
∂S ∂U − T1 ∂V T ∂V T
[1]). Using
Eqs. (3), (4) and (22) one must come to the conclusion that for cases with α b 0, δQ ≤ TdS = dU– PdV
ð23Þ
and α = β
∂S ∂V
U
P =− : T
ð24Þ
The following consideration supports the conclusion made in this paper. There is the generalized Mayer's relation [3]:
dU ≠0 and when U For every process which does not satisfy Eq. (17), dT is constant then T is also constant. For water at 273 K and atmospheric pressure, α=−68.05·10− 6 K− 1, CP =4217.6 J/(kgK), CP −CV ≈2.5 J/(kgK) and β=5.09·10− 10 Pa− 1 [2]. 1 dV βCV = ∼−0.1 K− 1 Substituting these values into Eq. (17), one gets V dT αTV 1 dV = with the highest accuracy. At a constant pressure, V dT −5 −1 α ≈ − 7 · 10 K , thus Eq. (17) is not true for water at a constant pressure. And, in general, for all processes with constant U 1 dV ≠−0:1K1 , T is also constant. and V dT dU = 0, then From Eqs. (5) and (11) one can find that, when dT dP CP −αPV = : dT ðαT−βP ÞV
ð19Þ
CP − CV =
Tα2 V : β
ð25Þ
It is derived from the first and second laws of thermodynamics without simplification. Its derivation is given in the Appendix. If one derives it assuming that TdS = dU − PdV for α b 0, then it looks like [2]: CP −CV = −
Tα2 V b 0: β
ð26Þ
Let us consider the thermal expansion of a solid. In the first approximation, V is constant. Then, in the first approximation [1], CP =
∂H αV : = CV + β ∂T V
ð27Þ
From Eq. (27), for α b 0, CP b CV and Eq. (26) is true, not Eq. (25). In the second approximation, P = const and V ≠ const. Let us suppose that in that case CP N CV for α b 0. There is Rolle's lemma: If f(x) is continuous in the interval a ≤ x ≤ b, and if f(a) N 0 and f(b) b 0 then there exists at least one value of x (say x = c) such that f(c) = 0, where a b c b b. As heat capacity is a continuous function, from Rolle's lemma it follows that there exists one value of V [say Vc = V0 + ΔV, V0 is the constant V in Eq. (27)] such that CP(Vc) − CV = 0. This is a contradiction because from Eqs. (25) and (26), at α b 0, CP(V) − CV ≠ 0. Else, when P = const, for V0 − ΔV b V b V0 + ΔV and α b 0, due to the continuity of heat capacity, CP – CV b 0. This is another contradiction to
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Eq. (25). So, it is proven that for negative thermal expansion, Eq. (26) is true. This means that for all processes ∂S ∂S 1 ∂U sign = sign − : ∂V U ∂V T T ∂V T
ð28Þ
Suppose that the quantity of heat δQ = 4217.6 J is introduced at constant atmospheric pressure to a kilogram of water at 273 K. Then if δQ = dU + PdV is true, 4217.6 J = 4220.1 J − 2.5 J where dU = 4220.1 J and PdV = − 2.5 J. This means that CP b CV. However, according to Mayer's relation CP − CV N 0. This is a contradiction. 3. Results One can draw a conclusion: in this paper it is proven that for ∂S is always negative and equals −P/T. This result
negative α,
∂V U
has been obtained previously [2,4–10]. In [2], there is a misprint in k kT
Eq. (11). The correct equation is: for α b 0, CV = CP S . Recently, substances with negative isothermal compressibility β were discovered [11–17]. For them, from Eqs. (1) and (4) it follows (when α N 0 and applying the same arguments as for the derivation of Eq. (24)) that
∂S ∂V
T
α = b0 β
ð29Þ
δQ = dU–PdV:
ð30Þ
Also, one can use the following argument: For adiabatic expansion, ∂U . For a substance with negative from Eq. (3) it follows that P = − ∂V S
β, this formula predicts the wrong sign of pressure and Eq. (3) must be replaced by Eq. (30). When α b 0 and β b 0, ∂S ∂V
ð31Þ
δQ = dU + PdV:
ð32Þ
T
and
So, when α N 0 and β N 0 or α b 0 and β b 0 then δQ = dU + PdV;
ð33Þ
and when α N 0 and β b 0 or α b 0 and β N 0 then δQ = dU–PdV:
ð34Þ
In [18], a result is obtained that fully supports the conclusion of this paper. There, a stretched solid of unit volume is considered. For this sample: dS =
∂S ∂σij
ð37Þ
which agrees strongly with Eqs. (33) and (34). In [9], it is shown that many phase transitions in ice disobey the Clausius–Clapeyron equation very much despite it being obtained from the first and second laws of thermodynamics without simplifications. If one derives the Clausius–Clapeyron equation taking into account that for α b 0, TdS = dU – PdV then it reads: dT T ðsignα2 V2 −signα1 V1 Þ = : dP ΔH
ð38Þ
The ordinary Clausius–Clapeyron equation reads dT T ðV2 −V1 Þ : = ΔH dP
ð39Þ
In [9], Eq. (38) was applied to these phase transitions and good agreement was observed, Table 1.
The reason why it has been supposed that the first law of thermodynamics (3) is tenable is that it agrees with experiments on substances with positive thermal expansion, and with the Clausius– Clapeyron equation which describes the melting of ice with high accuracy. Water has negative expansion at P = 14 MPa [19]. At this pressure ice melts at T ≈ 272 K. In [20] there is an experimental point in the interval 272–273 K: 4.9 MPa and T = 272.8 K. This obeys Eq. (39): Eq. (38)
α N0 β
=
TΔS = ΔU + 3Tασ = ΔU + signðαÞPΔV
4. Discussion
and
where the coefficients are constants. Assuming that the material is isotropic, α ij = αδ i j and σij = σδi j, this equation can be written in the form
!
dσij +
T
∂S ∂T
σ
dT = αij dσij +
Cσ dT: T
ð35Þ
where σij is the stress tensor, αij is the thermal expansion tensor and Cσ is the heat capacity per unit volume at constant stress. Eq. (35) is the exact equation derived without simplifications. For small stresses and temperature changes the properties will always be linear and it turns into: σ
TΔS = Tα ij σij + C ΔT
ð36Þ
dT = − 0.074 K/MPa. The modified Clausius–Clapeyron dP dT = −1.71 K/MPa. However, now that substances gives dP
with negative compressibility have been discovered, one can suggest that water has negative isothermal compressibility in the temperature range where it has negative expansion. In [15] it was shown that negative thermal expansion and negative compressibility can coexist. So, the first law of thermodynamics does not seem to be that unshakable as it was thought earlier. The author also thinks that a very strong proof is provided by Eqs. (35)–(37). An important result which confirms the conclusion of this paper is obtained in [8]. There it is found that the difference between the heats of chemical reactions measured by the Van't–Hoff equation, ΔH0, and obtained by calorimetry, ΔQ0, is PΔV 0. The Van't–Hoff equation must give the same results as calorimetry because it is derived from the first and the second laws of thermodynamics
Table 1 Phase transitions in ice measured experimentally and described by the Clausius– Clapeyron Eq. (39) and by the modified Clausius–Clapeyron Eq. (38) [9]. Pressure, Temperature, Transition K between ice MPa phases
dT/dP, Clausius– dT/dP, Eq. dT/dP, (38), K/MPa experiment, Clapeyron equation, K/MPa K/Mpa
LDA → Ic 0.1 HDA → LDA 0.1 LDA → HDA 600 III → V 344 V → VI 600 VI → VII 2000 VI → VIII 1900
− 0.09 0.18 − 0.26 3.6 − 8.7 3.7 0.21
∼ 140 ∼ 120 ∼77 ∼ 250 ∼ 260 ∼ 320 ∼ 240
0.08 − 0.43 0.28 − 0.26 − 0.26 − 0.30 − 0.23
− 0.08 0.43 − 0.28 7.1 − 8.1 3.5 0.23
I.A. Stepanov / Journal of Non-Crystalline Solids 356 (2010) 1168–1172
without simplifications. However, the difference between the two methods is far beyond the error limits. Nobody has succeeded in explaining this phenomenon. The author in [8] supposed that, for endothermic chemical reactions with increasing volume and for exothermic chemical reactions with decreasing volume, the law of conservation of energy is: ΔU = ΔQ + PΔV + ∑ μ i ΔNi ; (notice the change of sign in front of P and that
∂S ∂V U;Ni
Derivation of the generalized Mayer's relation In general dS =
∂S ∂P
ð41Þ
Here, K is the equilibrium constant of the chemical reaction. The traditional Van't–Hoff equation is: d ΔH0 ΔU 0 + PΔV 0 ln K = = : 2 dT RT RT 2
dT +
P
∂S ∂P
dP;
ðA1Þ
T
ð42Þ
∂V =− ; ∂T P
ðA2Þ
CP dT−αVdP: T
ðA3Þ
T
dS =
0
d ΔQ + PΔV ln K = : dT RT 2
∂S ∂T
and from the definition of CP and the Maxwell relation P T
= − b 0),
here μ i are chemical potentials and ΔNi are changes in the number of moles of the substance i. If one derives the Van't–Hoff equation using Eq. (40) then it is: 0
Appendix
ð40Þ
i
1171
Dividing Eq. (A3) by dT and choosing the direction of V = constant one obtains
∂S ∂T
= V
CV C ∂P = P −αV : T T ∂T V
ðA4Þ
From The Van't–Hoff equation has often been used for the determination of the heat of chemical reactions. It gives correct results for reactions with ΔV → 0. Therefore, it has always been assumed that, for chemical reactions, ΔQ = ΔH = ΔU + PΔV:
ð43Þ
However, from the traditional first law of thermodynamics it follows that ΔQ = ΔU + PΔV− ∑ μ i ΔNi :
ð44Þ
i
From Eqs. (43) and (44), ΔQ = ΔQ− ∑ μ i ΔNi . One may not neglect i
the term ∑ μ i ΔNi . Thus, the Van't–Hoff equation cannot give correct i
results either for ΔV → 0 or for ΔV ≠ 0. For reactions with ΔV ≠ 0 this equation gives the wrong results.
dV = αVdT–βVdP it follows that
∂P ∂T
The result obtained in this paper is of fundamental importance if true. It will lead to a major revision of thermodynamics. Any limitations of this theory have to be properly identified by the scientific community at large. Many equations of thermodynamics will change and some of them will change drastically, for example the Clausius– Clapeyron equation. This equation is used for the calculation of phase diagrams but often gives enormously wrong results [4]: a paradox not explained before. However, its modified form, Eq. (38), produces very reasonable results [4]. The result of this paper is very important in high pressure physics and it has to be checked by high pressure experiments. It already produces good results in the calculation of phase transitions and heats of chemical reactions which do not obey traditional thermodynamics, as shown above. A direct measurement of the compressibility of water is necessary in the interval 273–277 K where it has negative expansion. If it is proven that water has negative compressibility in this interval, then this would prove the theory of this article once and for all. Other high pressure experiments, for example, phase transition studies are also of great importance.
= V
α : β
ðA6Þ
Introducing Eq. (A6) into Eq. (A4) results in the generalized Mayer's relation: CV C α2 V : = P− β T T
ðA7Þ
If one supposes that for α b 0, TdS = dU – PdV, then the Maxwell relation (A2) will look like
5. Conclusion
ðA5Þ
∂S ∂P
= T
∂V ∂T P
ðA8Þ
and the generalized Mayer's relation will be CV C α2 V : = P + β T T
ðA9Þ
References [1] M.Kh. Karapetyants, Chemical Thermodynamics, Vysshaja Shkola, Moscow, 1974, pp. 108–110, (English translation: M. Kh. Karapetyants, Chemical Thermodynamics, (Translated by G. Leib), Mir, Moscow, 1978). [2] I.A. Stepanov, Comput. Modell. New Technol. 4 (2) (2000) 72 (Available at www. tsi.lv). [3] J. Guemez, C. Fiolhais, M. Fiolhais, Am. J. Phys. 67 (1999) 1100. [4] I.A. Stepanov, Physica B 349 (2004) 251. [5] I.A. Stepanov, in: A.N. Linke (Ed.), Progress in Chemical Physics Research, Nova Science Publishers, Inc, Hauppauge, NY, 2005, pp. 107–114. [6] I.A. Stepanov, in: A.N. Linke (Ed.), Progress in Chemical Physics Research, Nova Science, Nova Science Publishers, Inc, Hauppauge, NY, 2005, pp. 115–120. [7] I.A. Stepanov, in: A.N. Linke (Ed.), Chemical Physics: New Research, Nova Science Publishers, Inc, Hauppauge, NY, 2006, pp. 83–87. [8] I.A. Stepanov, in: A.N. Linke (Ed.), Trends in Chemical Physics Research, Nova Science Publishers, Inc, Hauppauge, NY, 2006, pp. 117–127.
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