Kvant Physics Problem 366

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Problem

A spherical flask with a capacity of $V = 1~\text{л}$ was evacuated and sealed. A monomolecular layer of air remained on the walls of the flask. Estimate the pressure in the flask when it is heated to $300^\circ~\text{C}$, given that at this temperature the walls of the flask are completely degassed.

A. V. Mitrofanov

Setup and Assumptions

A sealed spherical flask has volume $V = 1~\text{L} = 1.0 \times 10^{-3}\text{m}^3$. The inner wall initially carries a monomolecular layer of air, which provides the only gas in the system. After heating to $T = 300^\circ\text{C} = 573\text{K}$, all adsorbed molecules desorb into the volume.

The unknown is the resulting gas pressure $P$ inside the flask after complete degassing.

The gas is treated as ideal. Interactions between molecules are neglected except for adsorption on the wall before heating. The monolayer is assumed to correspond to one molecular layer with surface number density equal to one molecule per characteristic molecular cross-sectional area $a_0$. The value $a_0 = 1.0 \times 10^{-19}~\text{m}^2$ is taken as a standard estimate for a dense molecular layer of air. The gas is assumed spatially uniform after desorption.

Physical Principles

The number of molecules in the monolayer is related to the surface area $S$ of the flask by

$$N = \frac{S}{a_0}.$$

The spherical flask of volume $V$ has radius $R$ determined by

$$V = \frac{4}{3}\pi R^3,$$

and surface area

$$S = 4\pi R^2.$$

The pressure of an ideal gas is given by the equation of state

$$PV = NkT,$$

where $k = 1.38 \times 10^{-23}~\text{J/K}$ is Boltzmann’s constant and $T$ is the absolute temperature.

Derivation

The radius of the flask follows from the volume,

$$R = \left(\frac{3V}{4\pi}\right)^{1/3}.$$

Substituting $V = 1.0 \times 10^{-3}~\text{m}^3$ gives

$$R = \left(\frac{3 \times 10^{-3}}{4\pi}\right)^{1/3}.$$

The numerical value of the fraction is

$$\frac{3 \times 10^{-3}}{4\pi} = 2.39 \times 10^{-4}~\text{m}^3,$$

so

$$R = (2.39 \times 10^{-4})^{1/3}\text{m} = 6.2 \times 10^{-2}\text{m}.$$

The surface area becomes

$$S = 4\pi R^2 = 4\pi (6.2 \times 10^{-2})^2~\text{m}^2.$$

Evaluating the square,

$$(6.2 \times 10^{-2})^2 = 3.84 \times 10^{-3}~\text{m}^2,$$

so

$$S = 4\pi \cdot 3.84 \times 10^{-3}\text{m}^2 = 4.83 \times 10^{-2}\text{m}^2.$$

The number of molecules in the monolayer is

$$N = \frac{S}{a_0} = \frac{4.83 \times 10^{-2}}{1.0 \times 10^{-19}} = 4.83 \times 10^{17}.$$

After heating, all molecules occupy the full volume, so the pressure follows from the ideal gas law,

$$P = \frac{NkT}{V}.$$

Substituting values gives

$$P = \frac{(4.83 \times 10^{17})(1.38 \times 10^{-23}\text{J/K})(573\text{K})}{1.0 \times 10^{-3}~\text{m}^3}.$$

First the product in the numerator,

$$(4.83 \times 10^{17})(1.38 \times 10^{-23}) = 6.67 \times 10^{-6}~\text{J/K}.$$

Multiplying by temperature,

$$(6.67 \times 10^{-6})(573) = 3.82 \times 10^{-3}~\text{J}.$$

Dividing by volume,

$$P = \frac{3.82 \times 10^{-3}}{1.0 \times 10^{-3}}\text{Pa} = 3.82\text{Pa}.$$

Result

The pressure after complete degassing at $300^\circ\text{C}$ is

$$P = \frac{S}{a_0}\frac{kT}{V},$$

with $S = 4.83 \times 10^{-2}\text{m}^2$, $a_0 = 1.0 \times 10^{-19}\text{m}^2$, $T = 573~\text{K}$, $V = 1.0 \times 10^{-3}~\text{m}^3$.

Numerically,

$$\boxed{P \approx 3.8~\text{Pa}}.$$

Sanity Checks

The expression $P = NkT/V$ has dimensions of pressure since $kT$ has units of energy and $N/V$ has units of inverse volume, giving energy per volume which is pascals.

The result is many orders of magnitude below atmospheric pressure $10^5~\text{Pa}$, consistent with the assumption that only a monomolecular layer of gas is present initially.

The pressure scales linearly with surface area and temperature and inversely with volume, so a larger flask or thinner adsorption layer would reduce the result proportionally, while higher temperature increases it proportionally.

The dominant sensitivity lies in the assumed molecular area $a_0$, since a factor of two uncertainty in $a_0$ directly produces a factor of two uncertainty in the final pressure.