4.3 Additional Useful Thermodynamic Identities 169
(
∂U
∂V
)
T,nT
α
κT−P (4.3-5)
whereαis the coefficient of thermal expansion andκTis the isothermal compressibility.EXAMPLE 4.9
Evaluate the internal pressure of liquid benzene at 298.15 K and 1.000 atm.
Solution(
∂U
∂V)T,n(298.15 K)(
1. 237 × 10 −^3 K−^1)9. 67 × 10 −^10 Pa−^1−101325 Pa 3. 81 × 108 PaThis internal pressure is equal to 3760 atm.Exercise 4.6
a.Evaluate the internal pressure of liquid water at 25◦C and 1.000 atm.
b.Calculate the gravitational force per unit area on a column of water 100 m in height. Explain
how the internal pressure relates to the fact that in a giant sequoia tree, sap can be brought
to a height of nearly 100 m in the tree whereas barometric pressure can raise it only to about
10 m against a vacuum. What can you say about the attractive forces between the sap and the
walls of the vessel containing it? If a gas bubble appeared in the sap, what would happen?An equation for (∂H/∂P)T,nthat is analogous to Eq. (4.3-2) can be derived in a
similar way. We convert Eq. (4.2-11) to a derivative equation:
(
∂H
∂P)
T,n−T
(
∂S
∂P
)
T,n+V
(
∂P
∂P
)
T,nT
(
∂S
∂P
)
T,n+V
Using the Maxwell relation of Eq. (4.2-22), we obtain
(
∂H
∂P)
T,n−T
(
∂V
∂T
)
P,n+V (4.3-6)
EXAMPLE4.10
Show that for an ideal gas (∂H/∂P)T,n0, using only the equation of state,PVnRT,
and Eq. (4.3-6).
Solution
(
∂H
∂P)T,n−T(
∂V
∂T)V,nV−T(
∂
∂T(
nRT
P))P,n+V−nRT
P
+V 0