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Lesson 16 · 55 min

Vapours, humidity and changes of phase

Explain saturation, evaporation and boiling, then interpret dew-point and wet/dry-bulb information.

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Vapour is a gas phase, not visible mist

Water vapour consists of individual gaseous water molecules and is normally invisible. The visible cloud above cooling steam contains tiny condensed droplets, not visible water molecules. A liquid can lose molecules from its surface even below its boiling temperature. Molecules have a distribution of energies, and some near the surface can escape the attractions holding them in the liquid. Evaporation and condensation are opposing molecular processes; neither requires every to have the same .

In a closed space containing liquid at constant temperature, escaping molecules initially increase the vapour population. As more molecules return to the liquid, condensation becomes more frequent. Saturation is dynamic : evaporation and condensation occur at equal average rates. Molecules still cross the boundary in both directions. An unsaturated vapour has partial below the saturation value at that temperature. Sustained supersaturation is a more specialised situation and is not the ordinary model used in these calculations.

Key definition

Sustained supersaturation

A more specialised situation and is not the ordinary equilibrium model used in these calculations.

Exam keywords:specialisedsituationordinaryequilibrium

Saturated vapour depends on temperature

The saturated vapour pₛ(T) of a pure substance rises with temperature. More energetic molecules can enter the gas phase, so a larger vapour is needed for . While both liquid and vapour remain present at fixed temperature, changing the available volume can cause evaporation or condensation until the is restored. It is therefore not appropriate to apply Boyle's law mechanically to saturated vapour while assuming its mass remains fixed. The quantity in the vapour phase can change.

Distinguish the water-vapour partial pᵥ from the total of moist air. Total includes other gases as well. Relative humidity compares the water contribution with its own saturation at the air temperature, not with the entire atmospheric . Statements that air has a fixed container-like capacity are only shorthand: the underlying relation concerns phase and temperature. A warm parcel and a cool parcel with the same water-vapour amount can have different relative humidities.

Evaporation and boiling are different processes

Evaporation occurs at a liquid surface over a range of temperatures. Increasing temperature, exposed area or removal of humid boundary air can increase its rate under otherwise comparable conditions. Escaping molecules tend to remove , so evaporation can cool the remaining liquid if heat is not supplied quickly enough. Wind does not create water molecules; it can carry away the vapour-rich air next to the surface. High surrounding humidity reduces the net tendency for additional evaporation.

Boiling involves vapour bubbles forming and surviving within the liquid, rather than only molecules leaving its surface. For a simple pure-liquid model, boiling occurs when saturation vapour matches the local external on a bubble. Lower external lowers the boiling temperature; higher raises it. The familiar100°C value for pure water is tied to approximately one standard atmosphere, not every location or vessel. This is a conceptual comparison, not a procedure for operating a pressurised container.

Relative humidity is a temperature-specific ratio

Relative humidity when consistent units are used. If actual water-vapour is2.4kPa and the supplied saturation at the same temperature is3.2kPa, RH is75%. It does not that75%of the air's molecules are water. If the same vapour partial is maintained while temperature rises, pₛ normally rises and RH falls. In a rigid sealed container pᵥ itself can also change with temperature, so follow the conditions given rather than silently assuming every variable constant.

Key equation
RH=100pᵥpₛ(T)

Absolute humidity is commonly expressed as water-vapour mass per volume, for example g/m³. It differs from the dimensionless relative-humidity ratio. A mixing ratio compares water-vapour mass with dry-air mass and is another quantity again. Read the definition provided in a problem instead of substituting one humidity measure for another. -based and -based RH formulas are equivalent under the appropriate same-temperature ideal-gas comparison, but that does not make all moisture quantities numerically identical.

Dew point and a clearly rounded classroom table

The dew point is the temperature to which the air must be cooled, under the stated approximately constant- moisture conditions, before it reaches saturation. The starting temperature and dew point are equal for saturated air in this model. A high RH can result from cooling without adding vapour. Below the dew point, some vapour condenses, so the assumption of unchanged vapour amount no longer holds indefinitely. Dew forms on sufficiently cooled surfaces; fog consists of suspended near-ground droplets, and clouds contain droplets or ice particles aloft.

Consider a supplied rounded classroom reference: pₛ at20°C is2.34kPa and at21°C is2.49kPa. For pᵥ=2.40kPa, the dew point lies between those temperatures. Simple linear interpolation. These are rounded teaching values and a short-interval approximation, not a precision meteorological table. Real saturation- dependence is nonlinear. Label an interpolated result approximate and retain the supplied table's units and temperature range.

Key equation
gives20+2.40−2.342.49−2.34≈20.4°C

Wet and dry bulbs

A dry-bulb thermometer measures the ordinary air temperature with appropriate shielding. A ventilated wet bulb has a wetted surface; evaporation removes until a lower steady reading is approached under ordinary unsaturated conditions. The wet-bulb depression is dry-bulb minus wet-bulb temperature. At saturation the readings approach equality in the ideal arrangement. A larger depression at comparable and dry-bulb temperature generally indicates lower humidity because more evaporative cooling is possible.

The depression alone is not a universal percentage scale. Use a psychrometric table or an equation calibrated for the instrument, ventilation and . For a stated simplified model. With Td=25°C, Tw=20°C, pₛ(Tw)=2.34kPa and, pᵥ=2.01kPa. If the supplied. The is provided for this classroom calculation, not asserted to fit every hygrometer. A dry wick, contamination, poor ventilation or solar heating can distort readings.

Key equation
pᵥ=pₛ(Tw)−γ(Td−Tw)
γ=0.066kPa°C
pₛ(25°C)=3.17kPa,RH≈63.4%

Interpreting the barometer-tube principle

The syllabus's barometer-tube method can be understood from a supplied diagram without constructing mercury apparatus. An ideal evacuated barometer has. If a saturated vapour occupies the space above the column at the same temperature and other gas effects are negligible. Subtraction gives. The lowered column is explained by vapour above it. Any stated liquid heads or trapped-air effects need separate treatment rather than being silently ignored.

Key equation
pₐ=ρgh₀
pₐ=pᵥ+ρgh
pᵥ=ρg(h₀−h)

Self-check: why can a wet bulb cool without the liquid boiling? Surface evaporation can remove below boiling temperature. Why is visible mist not identical to water vapour? It contains condensed particles. Why can RH rise overnight without added moisture? Cooling can lower saturation until the existing vapour is closer to . In calculations, state what is held constant, use the correct temperature's saturation value and distinguish an relation from the instrumental conditions used to estimate it.

Topic revision bank

Key rules & principles

  • Saturation is dynamic , not absence of molecular motion.
  • Water vapour is invisible; mist contains droplets.
  • Boiling depends on external and saturation vapour .
  • RH compares actual and saturation vapour pressures at the same temperature.
  • Dew point and wet-bulb temperature are distinct quantities.
  • Wet/dry readings require an appropriate table or calibrated model.

Worked examples

Actual vapour pressure is2.4kPa; saturation pressure at the same air temperature is3.2kPa. Find RH.
  1. Use RH=100pᵥ/pₛ.
  2. The units already match.
  3. 100×2.4/3.2=75.

Answer: 75%; not75%water molecules in the air.

A rounded classroom reference gives saturation pressures2.34kPa at20°C and2.49kPa at21°C. Estimate dew point for vapour pressure2.40kPa.
  1. The target lies0.06kPa above the20°C value.
  2. The full1°C interval changes the table value by0.15kPa.
  3. Interpolate20+0.06/0.15=20.4°C.
  4. State that short-interval linear interpolation of rounded data is approximate.

Answer: Approximately20.4°C using the supplied classroom table.

Use the stated model pᵥ=pₛ(Tw)−γ(Td−Tw), with Td25°C,Tw20°C,γ0.066kPa/°C,pₛ(Tw)2.34kPa and pₛ(Td)3.17kPa.
  1. Wet-bulb depression=5°C.
  2. pᵥ=2.34−0.066×5=2.01kPa.
  3. RH=100×2.01/3.17≈63.4%.

Answer: Vapour pressure2.01kPa; RH≈63.4%for this supplied model.

Related videos and resources

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Dew point and Humidity

Khan Academy

Named creator video explaining humidity, relative humidity and dew point. It does not establish all saturated-vapour or phase-change objectives; publisher availability and captions need checking.

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