2.5 Enthalpy
79
where V L1 is the initial volume of the left side of the apparatus and V L2 is the final
volume of the left side of the apparatus. The work done on the gas on the right side is
given by
w R −
V R2
V R1
−P R (V R2 − V R1 )
(2.5-21)
Prior to the transfer, the pressure P 1 of the system was equal to P L , and the initial
volume of the system must have been equal to the magnitude of the change in volume
of the left side:
V 1 V L1 − V L2
(2.5-22)
The final pressure P 2 must be equal to P R , and the final volume V 2 must be equal to
the change in volume of the right side:
V 2 V R2 − V R1
(2.5-23)
From Eqs. (2.5-20) through (2.5-23), the total work done on the system is
w w L + w R P 1 V 1 − P 2 V 2 −∆(PV )
(2.5-24)
Exercise 2.25
a. Show that for any change in state
∆(PV ) P 1 ∆V + V 1 ∆P + (∆P∆V )
(2.5-25)
b. When can ∆(PV ) equal P∆V ? When can it equal V ∆P? When can it equal P∆V + V ∆P?
Because the apparatus is adiabatically insulated from the laboratory, no heat is
transferred to or from the laboratory. Also, no heat is transferred from the system
to the apparatus after the steady state is established, because the chamber on
the right is then at the same temperature as the gas that exits from the plug.
Therefore,
q 0
(2.5-26)
∆U q + w w −∆(PV )
(2.5-27)
∆H ∆U + ∆(PV ) 0
(2.5-28)
The Joule–Thomson process therefore occurs at constant enthalpy, and the
Joule–Thomson coefficient is equal to a partial derivative at constant H and n:
µ JT
∂T
∂P
H,n
(2.5-29)
79
where V L1 is the initial volume of the left side of the apparatus and V L2 is the final
volume of the left side of the apparatus. The work done on the gas on the right side is
given by
w R −
V R2
V R1
−P R (V R2 − V R1 )
(2.5-21)
Prior to the transfer, the pressure P 1 of the system was equal to P L , and the initial
volume of the system must have been equal to the magnitude of the change in volume
of the left side:
V 1 V L1 − V L2
(2.5-22)
The final pressure P 2 must be equal to P R , and the final volume V 2 must be equal to
the change in volume of the right side:
V 2 V R2 − V R1
(2.5-23)
From Eqs. (2.5-20) through (2.5-23), the total work done on the system is
w w L + w R P 1 V 1 − P 2 V 2 −∆(PV )
(2.5-24)
Exercise 2.25
a. Show that for any change in state
∆(PV ) P 1 ∆V + V 1 ∆P + (∆P∆V )
(2.5-25)
b. When can ∆(PV ) equal P∆V ? When can it equal V ∆P? When can it equal P∆V + V ∆P?
Because the apparatus is adiabatically insulated from the laboratory, no heat is
transferred to or from the laboratory. Also, no heat is transferred from the system
to the apparatus after the steady state is established, because the chamber on
the right is then at the same temperature as the gas that exits from the plug.
Therefore,
q 0
(2.5-26)
∆U q + w w −∆(PV )
(2.5-27)
∆H ∆U + ∆(PV ) 0
(2.5-28)
The Joule–Thomson process therefore occurs at constant enthalpy, and the
Joule–Thomson coefficient is equal to a partial derivative at constant H and n:
µ JT
∂T
∂P
H,n
(2.5-29)