Let us now look at two entirely different situations, both involving non- PdV work. Many of the examples of thermodynamical calculations have hitherto involved PdV work in a system in which the working substance has been an ideal gas. This slight puzzle will remain with us until Chapter 16, when we meet Nernst’s Heat Theorem and the Third Law of Thermodynamics. The entropy appearing in equations 12.9.9 and 12.9.11 is surely the absolute entropy, and we cannot calculate this unless we know the entropy at T = 0 K. After all, all we have done in this example is to calculate the increase in entropy as we took the sample up to 25 oC and 1 atmosphere – we haven’t really calculated the absolute entropy. But this leaves us in a rather uncomfortable position. Now that we have calculated the absolute entropy at a given temperature and pressure, we can calculate the increase in the Helmholtz and Gibbs functions from equations 12.9.9 and 12.9.11. Hence, taking the entropy to be zero at 0 K, the required entropy is 124000 J K −1 kmole −1. See equation 12.9.4, from which we see that there is a decrease of entropy equal to R ln( P 2/ P 1) = 8314ln(1.103 × 10 5 / 7173) = 22000 J K −1 kmole −1. Increase pressure to 1 atmos = 1.013 × 105 Pa at constant temperature. Assuming that we know C P as a function of temperature in this range, that comes to 70000 J K −1 kmole −1.ĥ. The increase in entropy is ∫ C P d(ln T). Increase temperature to 298.15 K at constant pressure. The molar latent heat of vaporization is 911000 kmole −1. Vaporize it at the same temperature and pressure. The molar latent heat of fusion is 117000 J kmole −1.

Liquefy it at the same temperature and pressure. Entropy is thus the only thermodynamic potential for which we can calculate an absolute value. Assuming that we know C P as a function of temperature in this range, that comes to 2080 J K −1 kmole −1.Ģ. (That’s the triple point.) The increase in entropy is ∫ C P d(ln T). Heat the solid hydrogen from 0 K to 13.95 K at a pressure of 7173 Pa. You will find it helpful to sketch these stages on a drawing similar to figure VI.5.ġ. We can do this in five stages, as follows.

By way of example, assuming that the molar entropy of hydrogen at 0 K is zero, calculate the absolute entropy of a kmole of H2 gas at a temperature of 25 oC (298.15 K) and a pressure of one atmosphere. A simple statement of the law is that heat always flows spontaneously from hotter to colder regions of matter (or 'downhill' in terms of the temperature gradient). We can, of course, calculate the molar entropy of a substance at some temperature provided that we define the entropy at a temperature of absolute zero to be zero. The second law of thermodynamics is a physical law based on universal empirical observation concerning heat and energy interconversions.

Absolute entropy from cp