Thermodynamic Relations
By: S K Mondal
Chapter 11
⎛ ∂V ⎞ TdS = CpdT − T ⎜ ⎟ dp ⎝ ∂T ⎠ p as ∴ ∴
β=
1 ⎛ ∂V ⎞ V ⎜⎝ ∂T ⎟⎠ p
⎛ ∂V ⎞ ⎜ ∂T ⎟ = Vβ ⎝ ⎠p TdS = CpdT − TVβ dp
proved
Let S is a function of p, V ∴ S = S(p, V) ∴
⎛ ∂S ⎞
⎛ ∂S ⎞
dS = ⎜ ⎟ dp + ⎜ ⎟ dV ⎝ ∂V ⎠ p ⎝ ∂p ⎠ V
Multiply both side by T ⎛ ∂S ⎞ ⎛ ∂S ⎞ TdS = T ⎜ ⎟ dp + T ⎜ ⎟ dV ⎝ ∂V ⎠ p ⎝ ∂p ⎠ V or
⎛ ∂S ∂T ⎞ ⎛ ∂S ∂T ⎞ ⋅ TdS = T ⎜ ⎟ dp + T ⎜ ∂T ⋅ ∂V ⎟ dV ∂ ∂ T p ⎝ ⎠p ⎝ ⎠V
or
⎛ ∂S ⎞ ⎛ ∂T ⎞ ⎛ ∂S ⎞ ⎛ ∂T ⎞ TdS = T ⎜ ⎟ ⋅ ⎜ ∂p ⎟ dp + T ⎜ ∂T ⎟ ⋅ ⎜ ∂V ⎟ dV ∂ T ⎝ ⎠V ⎝ ⎝ ⎠p ⎝ ⎠p ⎠V ⎛ ∂S ⎞ Cp = T ⎜ ⎟ ⎝ ∂T ⎠ p
∴
and
⎛ ∂S ⎞ CV = T ⎜ ⎟ ⎝ ∂T ⎠ V
⎛ ∂T ⎞ ⎛ ∂T ⎞ TdS = Cv ⎜ dp + Cp ⎜ ⎟ dV ⎟ ⎝ ∂V ⎠ p ⎝ ∂p ⎠ V β ⎛ ∂p ⎞ = k ⎜⎝ ∂T ⎟⎠ V
From first
or
k ⎛ ∂T ⎞ = β ⎜⎝ ∂p ⎟⎠ V
k ⎛ ∂T ⎞ dp + Cp ⎜ ⎟ dV β ⎝ ∂V ⎠ p
∴
TdS = Cv
∴
β=
∴
1 ⎛ ∂T ⎞ ⎜ ∂V ⎟ = βV ⎝ ⎠p
∴
TdS =
1 ⎛ ∂V ⎞ V ⎜⎝ ∂T ⎟⎠ p
Cv k dp Cp + dV β βV
proved.
(iv) Prove that 2
⎛ ∂V ⎞ ⎛ ∂p ⎞ Cp − Cv = − T ⎜ ⋅⎜ ⎟ ⎟ ⎝ ∂T ⎠ p ⎝ ∂V ⎠T We know that
Page 183 of 265
[IAS-1998]