Question 6.8: Determine the vapor pressure (in kPa) for liquid n-hexane at...
Determine the vapor pressure (in kPa) for liquid n-hexane at 0, 30, 60, and 90°C:
(a) With constants from App. B.2. (b) From the Lee/Kesler correlation for P^{r}_{sat}.
(a) With constants from App. B.2, the Antoine equation for n-hexane is:
\ln P^{\text {sat }} / \mathrm{kPa}=13.8193-\frac{2696.04}{t /{ }^{\circ} \mathrm{C}+224.317}
Substitution of temperatures yields the values of P^{sat} under the heading “Antoine” in the table below. These are presumed equivalent to good experimental values.
(b) We first determine ω from the Lee/Kesler correlation. At the normal boiling point of n-hexane (Table B.1),
T_{r_n}=\frac{341.9}{507.6}=0.6736 \quad \text { and } \quad P_{r_n}^{\text {sat }}=\frac{1.01325}{30.25}=0.03350
Application of Eq. (6.94) then gives the value of ω for use with the Lee/Kesler correlation: ω = 0.298. With this value, the correlation produces the Psat values shown in the table. The average difference from the Antoine values is about 1.5%.
\ln P_r^1\left(T_r\right)=15.2518-\frac{15.6875}{T_r}-13.4721 \ln T_r+0.43577 T_r^6 (6.94)
| t °C | P^{\text {sat }} \mathrm{kPa} (Antoine) | P^{\text {sat }} \mathrm{kPa} (Lee/Kesler) | t °C | P^{\text {sat }} \mathrm{kPa} (Antoine) | P^{\text {sat }} \mathrm{kPa} (Lee/Kesler) |
| 0 | 6.052 | 5.835 | 30 | 24.98 | 24.49 |
| 60 | 76.46 | 76.12 | 90 | 189.0 | 190.0 |