Thermodynamics of Air: Hydrostatics, Clausius–Clapeyron, Adiabatic Processes, Humidity, Stability and Clouds

This chapter answers the second half of Section XE7.1 Atmospheric Science of XE7 Atmospheric and Oceanic Sciences, taken in the thermodynamic part of its list: “Cloud Types; Laws of Thermodynamics; Gas Laws; Hydrostatic Equation; Clausius Clapeyron Equation; Adiabatic Processes, Humidity in the Atmosphere, Atmospheric Stability.” It teaches the laws of thermodynamics and the gas laws applied to air, the hydrostatic equation and scale height, the Clausius–Clapeyron equation, dry and moist adiabatic processes with potential temperature, the measures of humidity, parcel stability and the lifting condensation level, and the main cloud types. The numericals are scale height, saturation vapour pressure, potential temperature, mixing ratio, lifting condensation level and dew point. Constants used: g = 9.81 m/s², R = 287 J kg⁻¹ K⁻¹ for dry air, cp = 1004 J kg⁻¹ K⁻¹, R_v = 461.5 J kg⁻¹ K⁻¹, L_v = 2.5×10⁶ J/kg, ε = R/R_v = 0.622.

1. Laws of thermodynamics and the gas laws applied to air

Air at atmospheric pressures and temperatures obeys the ideal gas law closely: p = ρRT for dry air, with R = R*/M = 8.314/0.02897 = 287 J kg⁻¹ K⁻¹, and e = ρ_v R_v T for water vapour with R_v = 461.5 J kg⁻¹ K⁻¹. Dalton's law gives total pressure p = p_dry + e, the sum of partial pressures. Water vapour is lighter than air (molar mass 18.0 against 28.97), a fact that matters for buoyancy. The specific heats of dry air are cp = 1004 and cv = 717 J kg⁻¹ K⁻¹, with cp − cv = R and γ = cp/cv = 1.4.

  • Zeroth law: bodies in thermal equilibrium with a third share a temperature. It justifies the thermometer.
  • First law, per unit mass: dq = du + p dα = cp dT − α dp, where α = 1/ρ. Heat added either raises temperature or does work of expansion; at constant pressure dq = cp dT.
  • Second law: entropy ds = dq/T. A reversible adiabatic process has ds = 0 (isentropic); heat flows spontaneously only from warm to cold, which is why radiation and convection carry energy outward.
  • Third law (entropy of a perfect crystal tends to zero at 0 K) is of little use in meteorology.

Air parcel idealisation. A parcel is a small body of air that does not mix with its surroundings and always has the pressure of its surroundings (because pressure adjusts at the speed of sound), but not necessarily their temperature. Rising air expands because pressure falls with height, and expansion does work at the expense of internal energy, so the parcel cools. That is the origin of nearly all tropospheric cooling that makes cloud.

2. The hydrostatic equation, scale height and the hypsometric relation

For air at rest vertically, the weight of a thin layer is balanced by the pressure difference across it: dp/dz = −ρg. The pressure at any level is the weight per unit area of the air above it. Using ρ = p/RT gives dp/p = −g dz/(RT). For an isothermal layer this integrates to p = p₀ exp(−z/H) with the scale height H = RT/g, the height over which pressure falls by a factor e (to 37%).

Worked example. At T = 250 K, H = 287 × 250 / 9.81 = 7314 m ≈ 7.3 km. At 5.5 km, p = 1000 exp(−5500/7314) = 1000 × 0.471 = 471 hPa: about half the surface pressure, as it should be. At 288 K, H = 8.4 km; a colder atmosphere is more compressed, so H is smaller. Every 10 m of height near the surface costs roughly 1.2 hPa, since Δp = ρgΔz = 1.2 × 9.81 × 10 = 118 Pa.

Where temperature varies, use the mean layer temperature: the hypsometric equation gives the thickness between two pressure levels as Δz = (R T̄ / g) ln(p₁/p₂). For 1000 to 500 hPa with T̄ = 255 K, Δz = (287 × 255 / 9.81) × ln 2 = 5171 m. A warmer layer is thicker. That is why thickness maps show where air is warm, and why thermal wind (next chapter) is tied to horizontal temperature gradients.

⚠️ Pressure falls exponentially, not linearly
The 10 m = 1.2 hPa rule holds only near the ground. Aloft ρ is smaller, so the same height change costs less pressure. Also p = p₀ exp(−z/H) assumes an isothermal layer; for a real lapse rate use the layer-mean T̄ (hypsometric form) and never T at the surface alone.

3. Clausius–Clapeyron, humidity and the dew point

The Clausius–Clapeyron equation gives the slope of the equilibrium (saturation) vapour pressure e_s of water against temperature: de_s/dT = L_v e_s/(R_v T²). With L_v treated as constant it integrates to e_s(T) = e_s(T₀) exp[(L_v/R_v)(1/T₀ − 1/T)], with L_v/R_v = 2.5×10⁶/461.5 = 5417 K. Taking e_s(273.15 K) = 6.11 hPa gives about 23.6 hPa at 293.15 K (20 °C; tables give 23.4). The fractional rate is (1/e_s)(de_s/dT) = L_v/(R_v T²) = 0.063 K⁻¹ at 293 K: saturation vapour pressure rises by about 6 to 7% per kelvin, so warm air can hold far more vapour.

Measures of humidity (e = vapour pressure, p = total pressure, ε = 0.622)
QuantityDefinition and formulaNote
Mixing ratio wmass of vapour / mass of dry air; w = ε e/(p − e)conserved in unsaturated rise
Specific humidity qmass of vapour / mass of moist air; q = w/(1 + w) ≈ walso conserved; q < w slightly
Relative humidityRH = e/e_s(T) ≈ w/w_s × 100%depends on T: not conserved
Dew point T_dtemperature at which e_s(T_d) = e at constant p and eT_d ≤ T always; measures moisture content
Virtual temperature T_vT_v = T(1 + 0.608 q); dry air that would have the same density as the moist airT_v > T: moist air is lighter

Worked example. At p = 1000 hPa with e = 10 hPa, w = 0.622 × 10 / (1000 − 10) = 6.28×10⁻³ = 6.28 g/kg. If this air is at 20 °C, e_s = 23.4 hPa and RH = 10/23.4 = 43%. Heating the air to 30 °C at constant w leaves e unchanged but raises e_s to 42.4 hPa, so RH falls to 24%. Cooling it instead to the dew point, where e_s = 10 hPa (about 7 °C), gives RH = 100%.

⚠️ Relative humidity is not the amount of water
RH changes with temperature even when no moisture is added or removed, so a cold morning and a hot afternoon with the same w have very different RH. Use w, q or T_d to compare moisture content. For a quick estimate near the surface, T_d ≈ T − (100 − RH)/5 in °C is acceptable for RH above about 50%; it is an approximation, not a law.

4. Adiabatic processes, lapse rates and potential temperature

For a dry parcel with dq = 0, the first law gives cp dT = α dp. Using the hydrostatic equation dp = −ρg dz and α = 1/ρ, this becomes cp dT = −g dz, so the dry adiabatic lapse rate is Γd = g/cp = 9.81/1004 = 9.77 K/km ≈ 9.8 K/km. It is a property of the parcel (a rising, expanding one), not of the environment, and does not depend on the surrounding temperature. A parcel lifted 2 km from 25 °C reaches 25 − 19.5 = 5.5 °C, if it stays unsaturated.

Potential temperature is the temperature a parcel would have if brought dry-adiabatically to the reference pressure p₀ = 1000 hPa: θ = T (p₀/p)^κ, κ = R/cp = 0.286. Integrating the first law shows that θ is constant in a dry adiabatic process (and ds = cp d ln θ, so constant θ is isentropic flow). Example: air at 280 K and 700 hPa has θ = 280 × (1000/700)^0.286 = 280 × 1.107 = 310 K. A layer in which θ increases with height is stable; θ constant means neutral (well mixed); θ decreasing with height is unstable, since Γ > Γd is equivalent to ∂θ/∂z < 0.

Once the parcel is saturated, condensation releases latent heat and the cooling slows to the moist adiabatic lapse rate Γm. Γm is not a constant: about 3 to 4 K/km in warm moist tropical air near the surface (so much vapour condenses per kilometre), rising towards Γd (about 9 K/km) in cold or high air where little vapour remains. A typical working value is about 6 K/km in the mid-troposphere. Always Γm < Γd. The equivalent potential temperature θ_e, the θ the parcel would reach if all its vapour condensed and the latent heat were used, is conserved in both dry and moist adiabatic ascent.

⚠️ Γ, Γd and Γm are three different things
Γ = −∂T/∂z is the lapse rate of the environment, measured by a sounding, and varies from place to place. Γd and Γm are the rates at which a displaced parcel changes temperature. Stability compares Γ with them. Also, a descending saturated parcel warms at Γm only while it carries liquid water; once that evaporates it follows Γd.

5. Atmospheric stability, the lifting condensation level and cloud types

Parcel method. Displace a parcel upward. If it is then colder (denser) than its surroundings it sinks back, and the atmosphere is stable; if warmer, it keeps rising and the atmosphere is unstable. Comparing the environmental lapse rate Γ with Γd and Γm:

Stability classification
ConditionClassBehaviour
Γ < ΓmAbsolutely stableParcel returns whether dry or saturated
Γ > ΓdAbsolutely unstableParcel keeps rising whether dry or saturated
Γm < Γ < ΓdConditionally unstableStable while unsaturated; unstable once lifted past saturation and the level of free convection

Example: with Γd = 9.8 K/km, Γm = 5 K/km and an environment falling at 7 K/km, the layer is conditionally unstable. The troposphere on average (6.5 K/km) lies in this class, which is why moist convection is so common. A temperature inversion (Γ < 0) is extremely stable and traps pollution and haze. Heating a layer at the base or cooling it at the top steepens Γ and destabilises it.

Lifting condensation level (LCL). A parcel lifted dry cools at 9.8 K/km while its dew point falls only about 1.8 K/km (the vapour pressure drops with p). The gap T − T_d closes at about 8 K/km, so the approximation z_LCL ≈ 125 (T − T_d) metres (T, T_d in °C or K, differences only). For T = 30 °C and T_d = 18 °C, z_LCL ≈ 125 × 12 = 1500 m: the cloud base. Convective (potential) instability is a property of a whole layer lifted bodily: it is unstable if θ_e decreases with height (∂θ_e/∂z < 0), typically a moist layer under a dry one. Lifting it saturates its base first, and cooling by Γm then outpaces the top.

Basic cloud genera (WMO) by approximate height of base
LevelCloudsCharacter
High, above about 6 km in temperate latitudesCirrus, cirrostratus, cirrocumulusIce crystals; thin, wispy; halos from cirrostratus
Middle, about 2 to 6 kmAltocumulus, altostratusWater or mixed; altostratus gives a dim sun
Low, below about 2 kmStratus, stratocumulus, nimbostratus (a thick layer with a low base that often extends into the middle level)Layered, stable air; nimbostratus gives steady rain
Vertical developmentCumulus, cumulonimbusConvective, unstable air; cumulonimbus gives showers, thunder, hail

Cloud formation needs cooling of air to its dew point, usually by lifting (convection, fronts, orography) and sometimes by radiative cooling (fog) or mixing; and condensation nuclei (aerosol particles), without which saturation would need to be several hundred percent. Stable air gives layered stratiform cloud; unstable air gives cumuliform cloud. A cloud droplet is about 10 μm across; a raindrop contains a million of them.

Key takeaways

  • Dry air is an ideal gas with p = ρRT, R = 287 J kg⁻¹ K⁻¹, cp = 1004 J kg⁻¹ K⁻¹; the first law is dq = cp dT − α dp, and a reversible adiabatic process is isentropic. A rising parcel cools because it expands against falling pressure.
  • The hydrostatic equation dp/dz = −ρg gives p = p₀ exp(−z/H) with H = RT/g ≈ 7.3 km at 250 K; the hypsometric equation Δz = (R T̄/g) ln(p₁/p₂) gives layer thickness, larger for warmer layers.
  • Clausius–Clapeyron, de_s/dT = L_v e_s/(R_v T²), makes saturation vapour pressure rise about 6 to 7% per kelvin. Mixing ratio w = ε e/(p − e) and dew point measure moisture; RH does not, and falls when air is heated at fixed w.
  • Γd = g/cp = 9.8 K/km; θ = T(1000/p)^0.286 is conserved in dry adiabatic motion; Γm is smaller than Γd, about 3 to 9 K/km, because condensation releases latent heat; θ_e is conserved in moist ascent too.
  • Stability: Γ < Γm absolutely stable, Γ > Γd absolutely unstable, in between conditionally unstable. The LCL is about 125 (T − T_d) m. A layer with θ_e decreasing with height is convectively unstable.
  • Clouds form when air is cooled to its dew point on nuclei, mainly by lifting; stable air gives layer clouds (stratus, nimbostratus), unstable air gives cumulus and cumulonimbus, and high clouds are ice (cirrus family).

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. For an isothermal atmosphere in hydrostatic balance, the fraction of the surface pressure remaining at a height of one scale height H = RT/g is

    1. one half, since H is defined as the half-pressure height
    2. zero, since all the air lies below one scale height
    3. 1/e² ≈ 0.14, since the exponent is −2 at z = H
    4. 1/e ≈ 0.37, since p = p₀ exp(−z/H)
    Show answer

    Answer: D — 1/e ≈ 0.37, since p = p₀ exp(−z/H)

    Integrating dp/p = −dz/H gives p = p₀ exp(−z/H), so at z = H, p = p₀/e = 0.368 p₀. The half-pressure height is H ln 2 = 0.69 H, which is the tempting confusion. The exponent at z = H is −1.

  2. If the mean temperature of an atmospheric layer rises while the pressures at its bottom and top stay fixed, the thickness between the two pressure surfaces

    1. decreases, since warm air is lighter and the same mass of air is squeezed into a thinner layer
    2. increases in proportion to the mean temperature, from the hypsometric equation
    3. stays fixed, since the same mass of air lies between the two pressure surfaces whatever its temperature
    4. increases only if the humidity of the layer also increases at the same time
    Show answer

    Answer: B — increases in proportion to the mean temperature, from the hypsometric equation

    Δz = (R T̄/g) ln(p₁/p₂) is proportional to T̄. The mass between the two pressure surfaces is fixed (it is (p₁ − p₂)/g per unit area), but warmer air has lower density, so that mass occupies a greater depth. The shrinking idea has the density effect upside down.

  3. The Clausius–Clapeyron equation de_s/dT = L_v e_s/(R_v T²) implies that, near 20 °C, saturation vapour pressure

    1. falls with temperature since L_v decreases with temperature
    2. rises by a fixed 1 hPa for each kelvin of warming
    3. is independent of temperature and depends only on total pressure
    4. rises by about 6 to 7% for each kelvin of warming
    Show answer

    Answer: D — rises by about 6 to 7% for each kelvin of warming

    The fractional rate L_v/(R_v T²) = 2.5×10⁶ / (461.5 × 293²) = 0.063 per kelvin. The change in hPa grows with e_s itself, so it is not fixed. Saturation vapour pressure depends on temperature alone, not on the total pressure of other gases.

  4. Air at 15 °C and 1000 hPa is heated to 25 °C without adding or removing water vapour, the pressure being unchanged. Which humidity measure does NOT change?

    1. Relative humidity, since the vapour pressure is unchanged
    2. Saturation vapour pressure, since it depends only on the amount of vapour
    3. Dew point, which depends only on the vapour content at constant pressure
    4. Both relative humidity and dew point
    Show answer

    Answer: C — Dew point, which depends only on the vapour content at constant pressure

    The dew point is fixed by e (and p), and both are unchanged, so T_d is unchanged. Saturation vapour pressure rises with temperature, so relative humidity e/e_s falls. The statement with both fails because RH changes.

  5. For the same temperature and pressure, moist air is less dense than dry air because

    1. water vapour molecules (molar mass 18) replace molecules of average molar mass 28.97
    2. water vapour molecules repel nitrogen and oxygen molecules and so expand the volume of the air
    3. moist air always contains many liquid droplets and the droplets make the parcel lighter
    4. the total pressure of moist air is always lower than that of dry air at the same temperature
    Show answer

    Answer: A — water vapour molecules (molar mass 18) replace molecules of average molar mass 28.97

    At fixed p and T the number of molecules per unit volume is fixed (ideal gas), so replacing heavier molecules by lighter H₂O lowers the mean molar mass and the density. That is why T_v = T(1 + 0.608 q) exceeds T. The other statements invent forces, droplets or a pressure change that the question excludes.

  6. A dry parcel is lifted adiabatically in an environment whose temperature falls at 5 K/km. Compared with Γd = 9.8 K/km, the rate at which the parcel cools is

    1. 5 K/km, because the parcel adopts the temperature of its surroundings
    2. 7.4 K/km, the average of the two values
    3. zero, since an adiabatic process exchanges no heat and so no temperature change
    4. 9.8 K/km, since Γd depends only on g and cp and not on the environment
    Show answer

    Answer: D — 9.8 K/km, since Γd depends only on g and cp and not on the environment

    The parcel cools only by expansion work, at g/cp = 9.8 K/km, whatever the environment does. It therefore becomes colder than the surroundings (which cool at 5 K/km) and is stable. Zero temperature change confuses adiabatic with isothermal.

  7. The moist adiabatic lapse rate is smaller than the dry adiabatic lapse rate because

    1. saturated air is denser than dry air and tends to sink, and the sinking slows the cooling of the parcel
    2. the effective gravitational acceleration is reduced inside a cloud by the buoyancy of the droplets
    3. condensation releases latent heat that offsets part of the expansion cooling
    4. evaporation of cloud droplets during the ascent absorbs heat from the parcel and slows its cooling
    Show answer

    Answer: C — condensation releases latent heat that offsets part of the expansion cooling

    In a rising saturated parcel vapour condenses and L_v dq_s is released, so dT/dz is −(g/cp)/(1 + L_v dq_s/(cp dT)), smaller in magnitude than g/cp. Evaporation would absorb heat, which is the reverse of what happens on ascent.

  8. An environmental sounding shows temperature falling at 7.5 K/km. Take Γd = 9.8 K/km and Γm = 5 K/km. Which statement is correct?

    1. A dry parcel lifted from the surface is unstable because the lapse rate of 7.5 K/km is greater than Γm
    2. The layer is absolutely unstable because its lapse rate exceeds the moist adiabatic lapse rate
    3. The layer is conditionally unstable: stable for dry lift, unstable for saturated lift past the free-convection level
    4. The layer is absolutely stable because its temperature falls with height more slowly than the dry rate
    Show answer

    Answer: C — The layer is conditionally unstable: stable for dry lift, unstable for saturated lift past the free-convection level

    Γm (5) < Γ (7.5) < Γd (9.8) defines conditional instability. A dry parcel cools at 9.8 K/km, faster than the 7.5 K/km of the surroundings, so it sinks back; a saturated parcel cools at only 5 K/km and eventually becomes warmer than the surroundings. Comparing only with Γm or only with Γd gives the wrong classes.

  9. Air at 10 °C has relative humidity 80%. It is heated to 20 °C at constant pressure and constant mixing ratio. Given saturation vapour pressures of 12.3 hPa at 10 °C and 23.4 hPa at 20 °C, the new relative humidity is closest to

    1. 80%, since no vapour was added
    2. 42%, since e = 0.8 × 12.3 is unchanged and e_s = 23.4
    3. 65%, since RH falls in proportion to the temperature in kelvin
    4. 152%, since RH rises with temperature
    Show answer

    Answer: B — 42%, since e = 0.8 × 12.3 is unchanged and e_s = 23.4

    e = 0.80 × 12.3 = 9.84 hPa remains fixed (constant w and p), so RH = 9.84/23.4 = 42%. Treating RH as conserved ignores that e_s has nearly doubled; RH does not scale with absolute temperature.

  10. Select ALL statements that are correct for a parcel rising adiabatically without mixing.

    1. Its potential temperature θ is constant while it is unsaturated
    2. Its mixing ratio w is constant while it is unsaturated
    3. Its relative humidity increases as it rises and cools
    4. Its temperature stays constant because no heat is exchanged
    Show answer

    Answer: A — Its potential temperature θ is constant while it is unsaturated; B — Its mixing ratio w is constant while it is unsaturated; C — Its relative humidity increases as it rises and cools

    θ and w are both conserved for unsaturated ascent; because e_s falls with temperature faster than e falls with pressure, RH rises until saturation at the LCL. Constant temperature would mean isothermal expansion, not adiabatic.

  11. Select ALL correct statements about clouds and their formation.

    1. Cumulonimbus forms in unstable air and can give thunder, showers and hail
    2. Cirrus clouds are composed mainly of ice crystals
    3. Cloud droplets can form in clean air at saturation, without any nuclei
    4. Nimbostratus is a layer cloud that gives steady, widespread rain
    Show answer

    Answer: A — Cumulonimbus forms in unstable air and can give thunder, showers and hail; B — Cirrus clouds are composed mainly of ice crystals; D — Nimbostratus is a layer cloud that gives steady, widespread rain

    Unstable air gives cumuliform cloud and cumulonimbus, high thin clouds are ice, and nimbostratus is the rain-bearing layer cloud. Without condensation nuclei, homogeneous nucleation needs a supersaturation of several hundred percent, which never occurs in the atmosphere.

  12. Select ALL correct statements about humidity variables.

    1. The dew point can never exceed the air temperature
    2. For small w, specific humidity q = w/(1 + w) is slightly smaller than w
    3. Cooling air at constant pressure and constant w raises its relative humidity
    4. A relative humidity of 100% means the air contains the greatest possible mixing ratio at any temperature
    Show answer

    Answer: A — The dew point can never exceed the air temperature; B — For small w, specific humidity q = w/(1 + w) is slightly smaller than w; C — Cooling air at constant pressure and constant w raises its relative humidity

    T_d ≤ T in the absence of supersaturation, q = w/(1 + w) < w, and cooling lowers e_s with e fixed so RH rises. The 100% relative-humidity statement is false because the maximum mixing ratio depends on temperature: 100% RH at 0 °C holds far less vapour than 100% at 30 °C.

  13. Find the scale height H = RT/g, in km, for an isothermal layer at T = 250 K with R = 287 J kg⁻¹ K⁻¹ and g = 9.81 m/s², to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 7.31

    H = 287 × 250 / 9.81 = 7314 m = 7.31 km.

  14. Air at 280 K and 700 hPa is brought dry-adiabatically to 1000 hPa. Taking κ = R/cp = 0.286, find its potential temperature in kelvin, rounded to the nearest integer.

    Numerical answer — type the value.

    Show answer

    Answer: 310

    θ = T (p₀/p)^κ = 280 × (1000/700)^0.286 = 280 × 1.1074 = 310.1 K, so 310.

  15. Surface air has temperature 30 °C and dew point 18 °C. Using the approximation that the lifting condensation level lies 125 m above the surface for each degree of difference T − T_d, find the cloud-base height in metres.

    Numerical answer — type the value.

    Show answer

    Answer: 1500

    z_LCL ≈ 125 × (30 − 18) = 125 × 12 = 1500 m. Check: the dry parcel cools 9.8 K/km and T_d falls 1.8 K/km, closing 12 K at 8 K/km, i.e. 1.5 km.

  16. Total pressure is 1000 hPa and vapour pressure is 10 hPa. With ε = 0.622, find the mixing ratio w = ε e/(p − e) in g/kg, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 6.28

    w = 0.622 × 10 / (1000 − 10) = 6.22/990 = 6.283×10⁻³ = 6.28 g/kg.

  17. Take e_s(273.15 K) = 6.11 hPa, L_v = 2.5×10⁶ J/kg (constant) and R_v = 461.5 J kg⁻¹ K⁻¹. Using the integrated Clausius–Clapeyron form e_s(T) = e_s(T₀) exp[(L_v/R_v)(1/T₀ − 1/T)], find e_s in hPa at T = 293.15 K, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 23.6

    L_v/R_v = 5417 K. 1/273.15 − 1/293.15 = 2.4976×10⁻⁴. Exponent = 1.353; exp(1.353) = 3.869. e_s = 6.11 × 3.869 = 23.6 hPa.