274 4 Thermodynamics and Statistical Physics
4.31 TakingTandPas independent variables
S=f(T,P)dS=(
∂S
∂T
)
PdT+(
∂S
∂P
)
TdPorTdS=T(
∂S
∂T
)
PdT+T(
∂S
∂P
)
TdP=CPdT+T(
∂S
∂P
)
TdPorTdS=CPdT−T(
∂V
∂T
)
PdP=CPdT−TVαdP4.32 TakingPandVas independent variables,
S=f(P,V)dS=(
∂S
∂P
)
VdP+(
∂S
∂V
)
PdVTdS=T(
∂S
∂P
)
VdP+T(
∂S
∂V
)
PdV=T
(
∂S
∂T
)
V(
∂T
∂P
)
VdP+T(
∂S
∂T
)
P(
∂T
∂V
)
PdV=CV
(
∂T
∂P
)
VdP+CP(
∂T
∂V
)
PdV4.33 In the Joule–Thompson effect heat does not enter the expanding gas, that is
ΔQ=0. The net work done by the external forces on a unit mass of the gas
is (P 1 V 1 −P 2 V 2 ), whereP 1 andP 2 refer to higher and lower pressure across
the plug respectively.
ΔW=P 1 V 1 −P 2 V 2
If the internal energy of unit mass isU 1 andU 2 before and after the gas
passes through the plug
ΔU=U 1 −U 2
By the first law of Thermodynamics
ΔQ= 0 =ΔW+ΔU
or U 2 −U 1 =P 1 V 1 −P 2 V 2
or Δ(U+PV)= 0
or ΔH= 0