Kvant Physics Problem 29

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Problem

Find the evaporation rate from a unit surface area of water into a vacuum at temperature $20^\circ\text{ C}$. (The saturated vapor pressure at this temperature is $17{,}5\text{ мм рт. ст.}$) How long will it take for water poured to the brim in a standard tea saucer to evaporate in a room? The evaporation of a small amount of water practically does not change the room’s air humidity, which is 70%.

Setup and Assumptions

A flat water surface at temperature $T = 20^\circ\text{C} = 293,\text{K}$ evaporates into a vacuum. The saturated vapor pressure of water at this temperature is $p_{\text{sat}} = 17.5,\text{mmHg}$. The unknown is the mass evaporation rate per unit area $j$ in $\text{kg},\text{m}^{-2}\text{s}^{-1}$.

The molecular mass of water is $M = 18.0 \times 10^{-3},\text{kg},\text{mol}^{-1}$. The universal gas constant is $R = 8.314,\text{J},\text{mol}^{-1}\text{K}^{-1}$. The pressure conversion is $1,\text{mmHg} = 133.322,\text{Pa}$, giving $p_{\text{sat}} = 2333,\text{Pa}$.

Molecules leaving the liquid are assumed to form a Maxwellian gas at the interface temperature. The evaporation coefficient is taken as unity. Collisions in the vapor phase are neglected, which corresponds to free molecular effusion.

The second part considers water in a tea saucer exposed to air at relative humidity $70%$. The vapor pressure difference driving evaporation is therefore $p_{\text{sat}} - p_{\infty} = 0.30,p_{\text{sat}}$. The water depth in the saucer is taken as $h = 1.0,\text{cm} = 0.010,\text{m}$, and the density of water is $\rho = 1000,\text{kg},\text{m}^{-3}$. Edge effects and cooling of the liquid are neglected.

Physical Principles

The molecular flux of evaporation from a surface into vacuum is given by the Hertz–Knudsen relation derived from kinetic theory,

$$j = p \sqrt{\frac{M}{2\pi R T}},$$

where $p$ is the vapor pressure above the liquid, $M$ is molar mass, $R$ is the gas constant, and $T$ is absolute temperature.

For evaporation into a surrounding gas with partial vapor pressure $p_{\infty}$, the net flux is determined by the pressure difference,

$$j = (p_{\text{sat}} - p_{\infty}) \sqrt{\frac{M}{2\pi R T}}.$$

The time required to evaporate a liquid layer of thickness $h$ is determined from mass conservation,

$$\frac{m}{A} = \rho h,$$

and

$$t = \frac{\rho h}{j}.$$

Derivation

For evaporation into vacuum, the vapor pressure at the interface equals the saturated pressure, so $p = p_{\text{sat}}$. Substituting into the Hertz–Knudsen expression yields

$$j_0 = p_{\text{sat}} \sqrt{\frac{M}{2\pi R T}}.$$

The factor in the square root is computed from the given constants,

$$2\pi R T = 2\pi \cdot 8.314,\text{J mol}^{-1}\text{K}^{-1} \cdot 293,\text{K} = 1.53 \times 10^{4},\text{J mol}^{-1},$$

$$\frac{M}{2\pi R T} = \frac{1.8 \times 10^{-2},\text{kg mol}^{-1}}{1.53 \times 10^{4},\text{J mol}^{-1}} = 1.18 \times 10^{-6},\text{kg J}^{-1}.$$

Taking the square root gives

$$\sqrt{\frac{M}{2\pi R T}} = 1.08 \times 10^{-3},\text{s m}^{-1}.$$

The vacuum evaporation flux becomes

$$j_0 = 2333,\text{Pa} \cdot 1.08 \times 10^{-3},\text{s m}^{-1} = 2.52,\text{kg m}^{-2}\text{s}^{-1}.$$

For evaporation into air with relative humidity $0.70$, the driving pressure is $0.30 p_{\text{sat}}$, so

$$j = 0.30 \cdot j_0 = 0.30 \cdot 2.52,\text{kg m}^{-2}\text{s}^{-1} = 0.756,\text{kg m}^{-2}\text{s}^{-1}.$$

The mass of water per unit area in a layer of thickness $h = 0.010,\text{m}$ is

$$\frac{m}{A} = \rho h = 1000,\text{kg m}^{-3} \cdot 0.010,\text{m} = 10.0,\text{kg m}^{-2}.$$

The evaporation time follows as

$$t = \frac{10.0,\text{kg m}^{-2}}{0.756,\text{kg m}^{-2}\text{s}^{-1}} = 13.2,\text{s}.$$

Result

The evaporation rate into vacuum is

$$j_0 = p_{\text{sat}} \sqrt{\frac{M}{2\pi R T}} = 2.52,\text{kg m}^{-2}\text{s}^{-1}.$$

For air with $70%$ relative humidity, the net evaporation rate is

$$j = 0.756,\text{kg m}^{-2}\text{s}^{-1}.$$

The evaporation time of a $1.0,\text{cm}$ layer of water is

$$t = \frac{\rho h}{j} = 13.2,\text{s} \approx 1.3 \times 10^{1},\text{s}.$$

Sanity Checks

Dimensional analysis of the Hertz–Knudsen expression gives pressure multiplied by the square root of $\text{kg mol}^{-1} / (\text{J mol}^{-1})$, which reduces to $\text{kg m}^{-2}\text{s}^{-1}$ after using $\text{J} = \text{kg m}^2\text{s}^{-2}$. The final units are consistent with a mass flux.

The vacuum flux corresponds to molecular thermal speeds of order $\sqrt{RT/M} \sim 500,\text{m s}^{-1}$, which is consistent with rapid effusion and explains the large numerical value of the flux.

The reduction factor from humidity enters linearly through the vapor pressure difference, so decreasing relative humidity from $100%$ to $70%$ reduces the flux by a factor of $0.30$, matching the structure of the kinetic formula.

The most sensitive point in the calculation is the conversion of $17.5,\text{mmHg}$ to pascals; an error there would scale the final flux and time directly by the same factor.