Water-vapour concentration above an open tank: 23-Chem-A3 Heat and Mass Transfer, December 2014, Q1

Question 1: Water-vapour concentration above an open tank (Part A — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Humid air sits above a water surface; only the two boundary vapour concentrations are asked — the driving conditions are the air temperature, the total pressure, the relative humidity at the open top and saturation at the wet interface.

Quantity	Value
Air temperature $T$	310 K
Total pressure $P$	1 bar $=1.0\times10^5$ Pa
Relative humidity at top	$\phi = 40\%$
Water vapour pressure $p^\text{sat}$	0.06221 bar $=6221$ Pa

Find. The molar concentration of water vapour $c_A$ at the open top ($z=0$) and at the air–water interface ($z=L$), in mol/m³.

Figure 1 — Water evaporates upward through the 3-ft stagnant air column. The interface is saturated (100% RH at the liquid temperature); the open top is fixed by the ambient 40% relative humidity. Only these two end concentrations are asked, so the column geometry (the sketch’s taper) is immaterial.

Approach. Convert each boundary partial pressure to a molar concentration with the ideal-gas law $c_A = p_A/RT$; the interface partial pressure is the saturation value, the top is that saturation value scaled by the relative humidity.

Partial pressure at the interface ($z=L$). The gas in contact with liquid water is saturated, so $p_{A,L}=p^\text{sat}=6221$ Pa.
Partial pressure at the open top ($z=0$). Relative humidity scales the saturation pressure: $$p_{A,0}=\phi\,p^\text{sat}=0.40\times6221 = 2488\ \text{Pa}.$$
Molar concentration at the top. Apply the ideal-gas law with $R=8.314$ J/mol·K and $T=310$ K: $$c_{A,0}=\frac{p_{A,0}}{RT}=\frac{2488}{(8.314)(310)}=\boxed{0.966\ \text{mol/m}^3}.$$
Molar concentration at the interface. Same law at the saturated face: $$c_{A,L}=\frac{p_{A,L}}{RT}=\frac{6221}{(8.314)(310)}=\boxed{2.41\ \text{mol/m}^3}.$$ Because concentration is directly proportional to partial pressure at fixed $T$, the interface value is exactly $1/\phi=2.5\times$ the top value.

Quantity	Result
Vapour concentration at top, $c_A(z{=}0)$	0.966 mol/m³
Vapour concentration at interface, $c_A(z{=}L)$	2.41 mol/m³

Check

The printed sketch draws a tapered vessel (top 1 ft, bottom 2 ft) while the text calls the tank “cylindrical, 2 ft diameter.” This contradiction only affects the flux/rate of evaporation (which depends on area and column length); the two boundary concentrations asked here depend solely on $T$, $P$, $\phi$ and $p^\text{sat}$, so both answers are unaffected. The density and molar masses given are red herrings for this part.

Concept & Theory

Topic. Boundary concentrations for evaporation of a vapour into a stagnant gas.

Key relations. Ideal-gas concentration $c_A=p_A/RT$; relative humidity $\phi=p_A/p^\text{sat}$; saturation at a gas–liquid interface, $p_{A,i}=p^\text{sat}(T)$.

Why this works. Thermodynamic equilibrium at the interface pins the vapour there to its saturation pressure at the surface temperature, while the ambient relative humidity fixes the far boundary. Concentration follows from the ideal-gas equation of state because the humid air is dilute and near atmospheric pressure. These two endpoint concentrations are precisely the driving-force limits for the subsequent Stefan-diffusion flux (not requested here).

Common pitfall. Multiplying by the molar mass to get a “concentration” in kg/m³ — the question asks for a molar concentration, so stop at mol/m³. A second trap is forgetting that only the top carries the 40% factor; the interface is always saturated.

Source. Geankoplis, Transport Processes and Separation Process Principles (4th ed.), §6.2 (molecular diffusion of gases); Welty et al., §26 (mass-transfer boundary conditions).

Question 1 of 7: Water-vapour concentration above an open tank

Question 1: Water-vapour concentration above an open tank (Part A — equal value)