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62 CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy

Dividing through by dP and making the condition dH = 0 explicit,
TABLE 3.3 Joule-Thomson
Coefficients for Selected C a 0T b + a 0H b
Substances at 273 K and 1 atm P 0P H 0P T = 0
Gas m J-T (K>MPa) 0H
giving a b =-C m (3.53)
P J-T
Ar 3.66 0P T
Equation (3.53) states that (0H>0P) T can be calculated using the measurement of
C 6 H 14 –0.39
material-dependent properties C and m J-T . Because m J-T is not zero for a real gas,
P
CH 4 4.38 the pressure dependence of H for an expansion or compression process for which the
pressure change is large cannot be neglected. Note that (0H>0P) T can be positive or
CO 2 10.9
negative, depending on the value of m J-T at the P and T of interest.
H 2 –0.34 If m J-T is known from experiment, (0U>0V) T can be calculated as shown in Example
Problem 3.10. This has the advantage that a calculation of (0U>0V) based on measure-
He –0.62 T
k
ments of C , m J-T and the isothermal compressibility is much more accurate than a
P
N 2 2.15 measurement based on the Joule experiment. Values of m J-T are shown for selected gases
in Table 3.3. Keep in mind that m is a function of P and ¢P , so the values listed in the
Ne –0.30 J-T
table are only valid for a small pressure decrease originating at 1 atm pressure.
NH 3 28.2
O 2 2.69 EXAMPLE PROBLEM 3.10
Source: Linstrom, P. J., and Mallard, W. G., Using Equation (3.43), (0H>0P) T = [(0U>0V) T + P](0V>0P) T + V , derive an
eds. NIST Chemistry Webbook: NIST
Standard Reference Database Number 69. expression giving (0U>0V) T entirely in terms of measurable quantities for a gas.
Gaithersburg, MD: National Institute of
Standards and Technology. Retrieved from Solution
http://webbook.nist.gov.
a 0H b = B a 0U b + PR a 0V b + V
0P T 0V T 0P T
0H
a b - V
0U 0P T
a b = - P
0V T a 0V b
0P T
C m + V
P J-T
= - P
kV
In this equation, is the isothermal compressibility defined in Equation (3.9).
k

EXAMPLE PROBLEM 3.11
Using Equation (3.43),
0H 0U 0V
a b = ca b + Pda b + V
0P T 0V T 0P T
show that m J-T = 0 for an ideal gas.
Solution
1 0H 1 0U 0V 0V
m J-T =- a b =- ca b a b + Pa b + Vd
C P 0P T C P 0V T 0P T 0P T
1 0V
=- c0 + Pa b + Vd
C P 0P T
1 0[nRT>P] 1 nRT
=- cPa b + Vd =- c - + Vd = 0
C P 0P T C P P
In this calculation, we have used the result that (0U>0V) = 0 for an ideal gas.
T