Calculating the specific heat of a simple liquid by the number of elastic oscillators.
Calculate the specific heat of a simple liquid using the number of elastic oscillators
Each liquid molecule has an average of 8 elastic oscillators around it, and the specific heat contributed by the elastic energy is 4R。Therefore, near the three phase points, the specific heat at constant pressure of a single atomic liquid is 5.5R, and the specific heat at constant pressure of a diatomic liquid is 6.5R. Low temperature liquids such as Ar, Kr, Xe, O2, N2, F2, etc. conform to this conclusion.
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The argument is not quite correct. The isochoric heat capacity of liquid argon at its triple point is about 2.8 R. Assuming that the argon atoms are more or less vibrating in cages formed by their neighbours, we have to compare this with the heat capacity of a 3-dimensional harmonic oscillator, which is 3 R.
I do not question the values of the heat capacity (I used the official reference equation of state). I question your argument. Incidentally, the coordination number of liquid argon (close to Ttrp) is about 10.
As I wrote previously, there is no doubt about the NIST data. Cp is the sum of the isochoric heat capacity, CV, and a term containing the compressibility. For liquid argon close to the triple point Cp = 5.36 R, CV = 2.64 R. The latter value is close, but not quite the same, as the value for a system of 3-dimensional harmonic oscillators (3 R). An amount of 2.72 R is due to compression work.
It is a mistake to explain the value of Cp with vibrational movements only.
“… not the compression work, because the pressure at the three-phase point is close to 0.”
No. What matters is the pressure difference upon switching from isochoric to isobaric conditions (per Kelvin), which is not negligible.
What matters is betaV = (dp/d T)V, which is about 2.24 MPa/K for liquid argon at the triple point.
To obtain Cp, you first heat your sample at constant volume by dT, for which you need the energy CV dT. The isochoric heating lets the pressure increase by dp = betaV dT. In the second step you expand the sample to the original pressure. The volume work associated with this step is usually not zero – even if the final pressure is zero.
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