4.3 Additional Useful Thermodynamic Identities 171
EXAMPLE4.11
CalculateCV, m for liquid water at 0.00◦C and 1.000 atm. The constant-pressure heat
capacity is equal to 75.983 J K−^1 mol−^1. The coefficient of thermal expansion is equal to
− 68. 14 × 10 −^6 K−^1 at this temperature (this is one of the few cases in which this quantity is
negative). The compressibility is equal to 5. 098 × 10 −^10 Pa−^1. The molar volume is equal
to 18.012 cm^3 mol−^1.
SolutionCP, m−CV, m( 273 .15 K)(
18. 012 × 10 −^6 m^3)(
− 68. 14 × 10 −^6 K−^1) 2(
5. 098 × 10 −^10 Pa−^1) 0 .04481 J K−^1 mol−^1CV, m 75 .983JK−^1 mol−^1 − 0 .04481 J K−^1 mol−^1 75 .938JK−^1 mol−^1Exercise 4.8
a.Find the value ofCV, mfor liquid water at 25.00◦C and 1.000 atm. The coefficient of thermal
expansion is equal to 2. 07 × 10 −^4 K−^1 , the molar volume is equal to 18.0687 cm^3 mol−^1 ,
and the compressibility is equal to 45. 24 × 10 −^6 bar−^1 .CP, mis equal to 75.351 J K−^1 mol−^1.
b.At 3.98◦C liquid water has a maximum density and the coefficient of thermal expansion
vanishes. What is the value ofCP, m−CV, mat this temperature?
c.Show that Eq. (4.3-11) leads to the expression for an ideal gas obtained in Chapter 2:CP, m−CV, mR (ideal gas)d.Calculate the value of the ratioγCP, m/CV, mfor liquid water at 25.00◦C and 1.000 atm
and compare it to the value of the same ratio for argon gas at the same temperature and
pressure.There are a number of organic liquids that have fairly large coefficients of thermal
expansion. In these cases the difference betweenCV, mandCP, mis somewhat larger than
is the case with water in the preceding example. For almost all solids the difference
betweenCP, mandCV, mis quite small.EXAMPLE4.12
Calculate the value ofCV, mfor liquid benzene at 20◦C and 1.00 atm. The density of benzene
at 20◦C is 0.8765 g cm−^3. The value of the isothermal compressibility in the appendix applies
at 25◦C. Assume that you can use this value at 20◦C.
Solution
From Eq. (4.3-11)CPCV+TV α^2
κT