Infer dry gas from water levels and a wet pressure

An observed volume is not yet an amount of dry gas

A fictional record gives 0.250 L of wet gas at 320 K, external absolute pressure 100 kPa, and an inside water surface 20 cm higher than outside. Predict whether using 100 kPa directly overestimates or underestimates the retained dry amount. Explain the sign before opening the calculation.

The record is synthetic. Its water-vapor correction uses the same NIST saturation correlation as the finite vapor inventory investigation. This is an analysis of stipulated observations, not a practical procedure or a new measurement.

First account for the liquid levels

Define h = inside surface height − outside surface height. At a common depth in connected stationary liquid, pressure is the same from either side. If the inside surface is higher, its liquid column is taller, so the gas above it must exert less pressure:

P_wet = P_external − ρgh.

Here ρ is stipulated liquid density 1000 kg/m³, g is 9.80665 m/s², and h must be in metres. A 20 cm difference is 0.20 m, giving 1961.33 Pa = 1.96133 kPa. Therefore wet pressure is 98.03867 kPa. A surface below outside makes h negative and raises wet pressure above external pressure. Equal levels remove this correction.

The fixed vertical scale shows signed height, not the actual size of an apparatus. Volume remains stipulated at 0.250 L for each selected observation: moving the control compares separate records, not a closed process at conserved amount. Review hydrostatic pressure if the common-depth argument is unfamiliar.

Then separate the water contribution

Assume pure liquid water is present and the vapor has equilibrated at the stated temperature. At 320 K the selected correlation gives approximately 10.493171 kPa of water vapor. Dalton’s relation gives

P_dry = P_wet − P_water = 98.038670 − 10.493171 = 87.545499 kPa.

Both components occupy the full gas space. With R = 8.314462618 L kPa/(mol K), use n_dry = P_dry V/(RT) and multiply by 1000 for mmol. The water amount plus dry amount must equal the amount calculated from wet total pressure. The dry mole fraction is P_dry/P_wet; it is not generally a mass fraction.

Compare the corrected result with P_external V/(RT). At 300 K and a sufficiently negative height, the pressure-level correction and water correction can partly cancel. A small difference from a naive calculation does not show that either neglected mechanism was absent. Save contrasting cases and explain this with two separate ledger entries.

Refuse an inconsistent state

Select 20 kPa external pressure and 340 K. The water saturation pressure alone exceeds the available wet total throughout the permitted height range. Subtraction produces a negative residual. There is no nonnegative dry-gas amount consistent with all the assumed observations and saturated-equilibrium conditions.

Do not call the residual negative molecules or silently replace it with zero. Check pressure, temperature, height and saturation assumptions. For example, a nonequilibrated vapor need not be saturated. The calculation identifies inconsistency but cannot determine which observation or assumption failed. The zero residual boundary, by contrast, can represent saturated water vapor with no dry component.

Transfer and competing explanations

For a new fictional record, external pressure is 90 kPa, signed head −2 kPa and water partial pressure 8 kPa. Predict dry pressure before checking the practice feedback. Then suppose a separate generation estimate exceeds the corrected retained amount. A leak is one explanation; dissolution, residual gas accounting, an incorrect generation estimate or a reacting gas could also matter. Propose an independent check that separates at least two explanations rather than asserting a cause from one number.

The model assumes an insoluble nonreacting dry component, no leak, ideal gases, uniform maintained temperature, pure liquid water, negligible capillary pressure and constant stipulated density. It estimates retained inventory, not generated yield or gas identity. Observational uncertainties and coefficient covariance are not supplied, so the displayed digits are computational precision, not an experimental confidence interval.

Sources and connections

• NIST water phase-change data: Gubkov, Fermor et al. (1964), coefficients calculated by NIST, valid 293–343 K; pressure in bar, temperature in kelvin.
• AP Chemistry CED, 3.4.A.1–3: gas and mixture relations and their representations.
• Existing component-pressure investigation supplies the forward relation; this activity solves an inverse observation problem.