For ideal gases, the second derivatives in Eqs (6.26) and (6.27) are zero, and the heat capacities are independent of pressure or specific volume.
Example: What change in Cp is caused by increasing the pressure of water at 20oC by 10 MPa? Assuming a to be constant over the pressure range considered, Eq (6.27) simplifies to:
s -a2Tv = -(2 x 10-4)2(293)(1.04 x 10-3) = -1.2 x 10-8 J/mole - K [ dp )T N/m2
For a pressure increase of 10 MPa (10' n/m2), Cp decreases from 75 to 74.9 J/mole-K.
Although the derivatives of CP and CV with respect to pressure can be determined, it is not possible to relate (dCV/dT)V or (5CP/5T)p to the EOS of the substance. These partial derivatives are available only from data such as shown in Figs. 2.4 or 2.6. Example: Application to a nonideal gas
This example demonstrates the method of integrating the differentials represented by Eqs (6.21) - (6.24) between two states of different temperature and pressure. The gas (N2) obeys the Van der Waals equation of state and serves as the vehicle for highlighting the importance of nonideality corrections if the pressures are sufficiently high.
The path of the process is A ^ 2 shown in the figure below. The enthalpy difference h2 - hi is to be calculated from the two-step process: an isobaric step from 1 to A followed by an isothermal step from A to 2.
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