Partial pressure calculation

Table VII shows a comparison with experimental data by Leyko and Piatkiewicz (1 5) at 80 to 110 °C. At high temperatures partial pressures calculated from the BR- and EMNP-methods deviate by up to 20 per cent from the experimental results, whereas van Krevelen s method - extrapolated to 110 °C - yields partial pressures of hydrogen sulfide which are only about 1/4 to 1/5 of the measured values.

All the partial pressures calculated from the equilibrium constants assume unit activity for the condensed-phase components. This assumption is good when they are solid. Above the melting points of the salts, however, continued decomposition of the salt will result in a solution containing dissolved oxide and the partial pressures will depend on the melt composition, and will therefore change as the decomposition proceeds. Because of the form of Kt, the partial pressure calculation will be worst for small oxide concentrations. An examination of the various tables shows that 02 and NO are the major products of nitrate decomposition, the concentration of N02 being rather minor. This results from the fact that the equilibrium 2 N02 = 2 NO + 02 lies to the right for low pressures. 

Fig. 21.11. Partial pressures calculated from the total pressure curve by Boissonnas method (points) compared with observed partial pressures (full lines) for the system ethanol (1) + chloroform (2) at 45 °C.

Some results are shown in figure 9.17 and 9.18. It is important to note that the bulk resistivity for a given temperature can be adjusted by the SiH2Cl2 or the WF6 flow to an acceptable value (normally close to 800 jtflcm). Also, there are clear differences between this work and that of Price and Wu (for instance discrepancies in the deposition rate dependence on the WF6 and SiH2CI2 flows). It is obvious that all studies were done while there were strong concentration gradients in the reactor. This condition makes partial pressure calculations uncertain and direct comparison between studies difficult. Nevertheless, process optimization is very well possible as shown by the study of Selbrede. 

In an attempt to unravel the kinetics of the SiH2Cl2/WF6 chemistry, Srinivas et al.237, set up their experimental conditions such that the reactant conversion would be less than 10%. In this situation one can assume that the reactor is gradientless and inlet partial pressures (calculated from inlet flow ratios and the pressure) will be close to true wafer surface partial pressures. In calculating the conversion degree, however, one should keep... 

The exhaust to the atmosphere from an incinerator has a SO2 concentration of 0.12 mm Hg partial pressure. Calculate the parts per million of SO2 in the exhaust. 

Two Independent methods have been utilized to examine the nature of the sorption Isotherm. An analysis of the experimental Isotherm compared to an extrapolation of the Infinite dilution behavior allows calculation of an enhancement number for any of the polymers at any given partial pressure. Calculation of a cluster number based on an Independent method shows very close concordance with the enhancement ntimber, providing strong support for the postulate that associated groups of water molecules sorb In the polymer, and account for the anomolous sorption. 

Using these ratios of partial pressures, calculate the values that K would have for each value of x. 

In contrast to the binary Langmuir or SSTM models, the ideal adsorbed solution theory does not lead to a simple explicit relation for the adsorbed-phase composition and loading in terms of the partial pressures. Calculation of the equilibrium for a particular gas-phase composition therefore requires a trial and error procedure. 

In the previous section some criteria were established which determine whether Raoult s law can be apphed for a binaiy mixture over the whole range of concentration. The lesser the extent to which these criteria are true the greater the deviations of the partial pressures from their linear dependency on the mole fractions x,. These deviations can be positive as well as negative. In the first case the partial pressure is higher and in the second case it is lower than the partial pressure calculated with Raoult s law. If the attractive interactions between the two mixture components are much smaller than the ones between identical molecules, positive deviations from Raoult s law result The stronger the interactions between heterogeneous molecules the smaller the partial pressures compared to Raoult s law. 

Product Vapor to be Neglected In Partial-Pressure Calculations... 

The remaining sidestream draw trays are calculated by the same procedure as that outlined in the previous step. Remember that, in making partial pressure calculations, the presence of the next higher product vapor in the total vapor leaving the draw tray must be neglected. This principle is summarized in Table 2.4. 

To calculate partial pressures calculate the total pressure and then use the mole fractions from the previous part to calculate the partial pressures. Calculate the total pressure from the sum of the moles of both components. (Alternatively, you can calculate the partial pressures of the components individually, using the number of moles of each component. Then you can sum them to obtain the total pressure.)... 

Write an equilibrium expression forthis reaction in terms of partial pressures. Calculate a value for/fp forthis reaction, includingthe units. 

Carbon activity calculations and oxygen partial pressure calculations for the gas mixtures used in the experiments are presented in Table 5.1 and plotted in Fig. 5.1. Also shown is the coexistence Cr-Cr203 oxygen pressure. 

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