## Heat Capacities

Subtracting Eq (6.23b) from Eq (6.24b) gives:

Dividing by dv and holding p constant gives:

Inverting the partial derivative on the left hand side yields:

For an ideal gas, the product of T and the two partial derivatives is equal to the gas constant. For nonideal gases, on the other hand, the two heat capacities can differ significantly from R (see problem 6.8).

For condensed phases, the first partial derivative in Eq (6.25) is replaced by av and the second by Eq (6.8), yielding:

For solids or liquids with a small coefficient of thermal expansion, the heat capacities are approximately equal, as shown by the following example.

Example: CP - CV for liquid water at 20oC and 10 MPa

The pertinent properties are: a = 2.0x10-4 K-1; p = 4.4x10-4 MPa-1; v = 1.04x10-3 m3/kg With these values, Eq (6.25b) gives:

Cp - Cv = 0.5 J/mole-K = 0.06R By way of comparison, CP of water is 75 J/mole-K or 9.0R. The difference in the heat capacities of this substance is clearly negligible.

Equation (6.23b) provides the starting points for calculating the effect of specific volume on CV. As a thermodynamic property, ds must be an exact differential, so Eq (6.4) applies to the coefficients of dT and dv in Eq (6.23b). The terms in Eq (6.1) can be paired with the corresponding terms in Eq (6.23b) by: x = T, y = p, M = CV/T, N = (5p/5T)v, Using these identifications in Eq (6.4) yields: 