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. 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. 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.
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