H as function of P and T. Integration of gives:
(101)
For an ideal gas:
(102)
As shown in (103) the integral in (101) is zero and H for an ideal gas is not a function of P
(103)
Expanding the lower integration boundary we get:
(104)
Combination of the first two terms of the right hand side of (104) yields H for a hypothetical ideal gas. In addition, for an ideal gas mixture there is no excess H
(105)
(106)
Notice
This website processes personal data (e.g. browsing data or IP addresses) and use cookies or other identifiers, which are necessary for its functioning and required to achieve the purposes illustrated in the cookie policy.
You accept the use of cookies or other identifiers by closing or dismissing this notice, by clicking a link or button or by continuing to browse otherwise.
This site is GPDR compliant.
Privacy policy: no personal data and no comments are collected. No data are collected for statistical purposes.