TITLE.PM5

366 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th7-2.pm5

∴ FH∂∂TvIK

s

= −cKT

v

β. (Ans.)

+Example 7.12. Derive the third Tds equation

Tds = cv (^) HF∂∂TpIK
v
dp + cp ∂
∂
F
HG
I
KJ
T
v p
dv
and also show that this may be written as :
Tds =
cv
β
Kdp +
c
v
p
β
dv.
Solution. Let s = f(p, v)
Then ds = HF∂∂psIK
v
dp + FH∂∂svIK
p
dv
or Tds = T FH∂∂psIK
v
dp + T
∂
∂
F
HG
I
KJ
s
v p
dv
= T FH∂∂TsKI FH∂∂TpKI
vv

• T
  ∂
  ∂
  F
  H
  I
  K
  ∂
  ∂
  F
  H
  I
  K
  s
  T
  T
  p v p^ dv
  But
  ∂
  ∂
  F
  H
  I
  K
  s
  T v =
  c
  T
  v and ∂
  ∂
  F
  H
  I
  K
  s
  T p =
  c
  T
  p
  Hence Tds = c Tp
  v
  vpdp c Tv dv
  p
  ∂
  ∂
  F
  H
  I
  K +
  ∂
  ∂
  F
  H
  I
  K ...Proved.
  Also
  ∂
  ∂
  F
  H
  I
  K
  T
  p v =
  −
  ∂
  ∂
  F
  H
  I
  K
  ∂
  ∂
  F
  H
  I
  K
  1
  p
  v
  v
  T T p
  = –
  ∂
  ∂
  F
  H
  I
  K
  T
  v p^
  ∂
  ∂
  F
  H
  I
  K
  v
  pT =
  K
  β
  and
  ∂
  ∂
  F
  H
  I
  K
  T
  v p =
  1
  βv
  Substituting these values in the above Tds equation, we get
  Tds = cKdp
  c
  v
  v pdv
  ββ


• ...Proved.
  Example 7.13. Using Maxwell relation derive the following Tds equation
  Tds = cp dT – T
  ∂
  ∂
  F
  H
  I
  K
  v
  T p^ dp. (U.P.S.C. 1988)
  Solution. s = f (T, p)
  Tds = T (^) HF∂∂TsIK
  p
  dT + T (^) HF∂∂psIK
  T
  dp ...(i)
  where cp = T
  ∂
  ∂
  F
  HG
  I
  KJ
  s
  T p
  Also, FH∂∂spIK
  T
  = – FH∂∂TvIK
  p
  ......Maxwell relation
  Substituting these in eqn. (i), we get
  Tds = cp dT – T HF∂∂TvIK
  p
  dp. (Ans.)