Derivation of The Third Tds Equation in Thermodynamics: August 2018
Derivation of The Third Tds Equation in Thermodynamics: August 2018
Derivation of The Third Tds Equation in Thermodynamics: August 2018
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Kebbi State University of Science and Technology
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Abstract
P&V Independent involves the application of T&V independent together with the Application of second law of thermodynamics.
The third TdS equation together with the first and second TdS equations has been known to many as the “tedious equations” due
to a lot of derivations with resemblances involved. The TdS equations enables us to calculate the change of entropy during
various reversible processes in terms of either dV and dT, or dP and dT, or dV and dP, and even in terms of directly measurable
quantities such as the coefficient of expansion and the bulk modulus. This Manuscript involves another way of deriving the
Thirds TdS equation applying the second law of thermodynamics together with equations already derived and introduced from
the derivations of T&V which is also an application of the second law of thermodynamics.
Keywords
Putting, Substituting, Eqn, Constant
Received: May 6, 2018 / Accepted: June 19, 2018 / Published online: August 6, 2018
@ 2018 The Authors. Published by American Institute of Science. This Open Access article is under the CC BY license.
http://creativecommons.org/licenses/by/4.0/
* Corresponding author
E-mail address:
24 Uchenna Okwudili Anekwe et al.: Derivation of the Third TdS Equation in Thermodynamics
enigmatic and provocative fundamental statement of properties other than those introduced in T & V independent
relevance to engineering, physics, chemistry and biology. It is i.e.
instructive to recall that the discovery of this profound
principle of nature originally emerged from the practical ∂u ∂u
du = dT + dv (2)
concerns of nineteenth century industrialists who desired to ∂T v ∂v T
increase the efficiency of steam engines. Although the second
Then since
law of thermodynamics may be stated in a myriad of ways.
The energy of the universe is constant. The entropy of the
∂T ∂T
universe seeks a maximum. Thus, the second law may be dT = dp + dv (3)
∂p v ∂v p
summarized by the deceptively simple inequality, ∆ ≥ 0 S Univ,
pertaining to the increase in the entropy of the universe (or of
Putting eqn. (3) into eqn. (2) we’ve
any isolated entity) produced as a result of spontaneous
(irreversible) processes. Although there is no question
∂u ∂T ∂T ∂u
regarding the wide ranging implications of this and other du = dp + dv + dv (4)
∂T v ∂p v ∂v p ∂v T
statements of the second law, it is important to note that none
of these statements can in themselves be used to quantify the
∂u ∂T ∂u ∂T ∂u
excess entropy produced as the result of a given irreversible du = ∂p dp + ∂T ∂v dv + ∂v dv (5)
process. ∂T v v v p T
2. Methodology ∂u
C v= (9)
∂T
This section involves derivation of p and V independent which
is an application of the second law of thermodynamics. putting eqn (9) into eqn (7) we've
The P & V Independent show that
∂u ∂T
= Cv (10)
Tds = C
∂T
dv + C
∂T ∂p v ∂p v
dp
p ∂V p v ∂P v
recall that
Solution
h = u + pv (11)
The energy difference between two neighboring equilibrium
states in which the pressure and volume differs by dp and dh = du + pdv + vdp (12)
dv is given as at constant volume and constant pressure
∂u ∂u dv = 0 & dp = 0 ∴ eqn(12) becomes
du = dp + dv (1)
∂p v ∂v p
dh = du (13)
In equation (1) the partial derivatives do not involve any
Physics Journal Vol. 4, No. 2, 2018, pp. 23-28 25
putting eqn(13) into eqn (8) we've But we can also write
∂h ∂h ∂T ∂h ∂s ∂s
ds = dp + dv
∂v = ∂T ∂v + ∂v (14)
∂p v ∂v p
(24)
p v p T
∂h ∂h ∂T ∂s C p ∂T
∂v = ∂T ∂v (15) ∂v = T ∂v (26)
p v p p p
∂h ∂2 s C ∂ ∂T Cv ∂ 2T
C p= (16) = v = (27)
∂T v ∂v∂p T ∂v ∂p v T ∂v∂p
Putting eqn (16 ) into eqn (15) we've differentiating eqn (26) partially w.r.t p we've
∂h ∂T ∂ ∂s ∂2s ∂ C p ∂T C p ∂ 2T
∂v = C p ∂v (17) = = =
∂p ∂v p ∂p∂v ∂p T ∂v p T ∂p∂v
(28)
p p
From eqn (13) substituting dh for du ineqn (17 ) we've equating the R.H.S of eqn (27) &(28) we've
∂u ∂T Cv ∂ 2 T C p ∂ 2T
∂v = C p ∂v (18) = (29)
p p T ∂v ∂p T ∂p∂v
Hence putting eqn (10 ) & (18) into eqn (1) we've From eqn ( 29 )
∂T ∂T Cv C p
du = Cv dp + C p dv (19) ⇒ =
∂p v ∂v p T T
∴ Cv = C p (30)
From the combined first and second law of thermodynamics
we’ve
From eqn ( 23) we've
1
ds = ( du + pdv ) (20)
Cv ∂T Cp ∂T
T
ds = dp + ∂v dv (31)
T ∂p v T p
At constant volume process dv = 0 hence eqn ( 20 ) becomes
Multiplying through eqn ( 31) by T we've
1
ds = du (21)
T
∂T ∂T
Tds = Cv dp + C p dv (32)
Putting eqn (19 ) into eqn ( 21) we've ∂p v ∂v p
∂T ∂T ∂T ∂T
ds =
1 Tds = C p dv + Cv dp
Cv dp + C p dv (22) ∂v p ∂p v
(33)
T ∂p v ∂v p
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