Some Physical Properties of Compressed Gases. IV. The Entropies of Nitrogen, Carbon Monoxide and Hydrogen

Abstract

Previous work on these gases has yielded values for the expansion coefficients $\left(\frac{T}{v}\right){\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p$, $-\left(\frac{p}{v}\right){\left(\frac{\mathrm{dv}}{\mathrm{dp}}\right)}_T$, and for $-T{\left(\frac{d^{2}v}{d{T}^2}\right)}_p$ at temperatures between -70° and 400° or 500°C and at pressures between 25 and 1200 atmospheres. The evaluation of such derivatives renders possible the calculation of many physical properties of the substance. To those properties that have been reported previously for nitrogen, carbon monoxide and hydrogen, the authors now add the change in entropy $\Delta S=-\int 1^p{\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p\mathrm{dp}$ along isotherms. The integral is evaluated by graphical quadrature from the previously determined expansion coefficient $\left(\frac{T}{v}\right){\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p$. The absolute entropy at various pressures along an isotherm is then obtained by adding $\Delta S$ to the entropy at one atmosphere. The calculated values of $\Delta S$ are shown in a table, and the absolute entropies are shown by isotherms at various temperatures between -75° and 600°C to 1200 atmospheres. With an ideal gas, the integral $\Delta S=-\int 1^p{\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p\mathrm{dp}$ would be simply $-R\ln p$ along all isotherms. The authors' calculations show that $\Delta S$ is always greater in absolute value than $R\ln p$, the difference being more pronounced at low temperatures and high pressures as would be expected. At 1200 atmospheres and at ordinary temperatures, all three gases have lost roughly 15 units of entropy. When an actual gas is compressed along an isotherm, ${\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p$ has at first the value that it would have if the gas were ideal, namely, $\frac{R}{p}$, but it gradually becomes greater than $\frac{R}{p}$. This disparity continues until with sufficient pressure a point is reached where ${\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p$ again becomes equal to, and finally less than, $\frac{R}{p}$. Eventually ${\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p$ must become so small that $\left|p^c{\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p\right|$ with $c>\mathrm{}1\mathrm{}$ remains bounded, in order that $\Delta S$ may be finite when $p$ is infinite. The authors find that ${\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p$ for these gases becomes equal to $\frac{R}{p}$ at about 1200 atmospheres. If ${\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p\leqq \frac{R}{p}$ and $p>\mathrm{}1000\mathrm{}$, the pressure must be raised to 2×${10}^6$ atmospheres or higher to extract an additional 15 units of entropy. Hence it would seem impossible ever to reduce one of these gases above the critical temperature to a state of near zero entropy by the application of pressure alone, unless the graph of ${\left(\frac{\mathrm{dv}}{\mathrm{dT}}\right)}_p$ against $p$ has humps or infinite discontinuities (giving convergent integrals) beyond 1000 atmospheres.