Adiabatic Processes and Lapse Rates Questions and Answers

Answer: A

A transmissometer projects a beam over a baseline of a few tens of metres and measures how much reaches the receiver, from which the extinction coefficient and hence the visibility follow. Forward scatter sensors are now more common because they need no long baseline. Both infer visibility from optical extinction rather than measuring what an observer would see.

Answer: B

A forward scatter sensor samples a volume of a few cubic centimetres and assumes it is representative of the whole path, which fails in patchy fog. Human observers integrate over a much longer path and judge the visibility of actual objects. The discrepancy is a recognised issue at airports where fog is often highly variable across the site.

Answer: C

Supercooled droplets freeze on contact with objects below zero, building rime that loads power lines, ices aircraft and coats road surfaces. The visibility reduction is accompanied by this additional hazard. The latent heat released does warm the surface slightly but nothing like enough to prevent the accretion, so the last option misstates a real but minor effect.

Answer: D

Pure water droplets can remain liquid down to about minus 40 degrees, below which homogeneous nucleation freezes them regardless of any nucleus being present. Between 0 and minus 40 freezing depends on the availability of ice nuclei, which are scarce at the warmer end of the range. This is why mixed phase cloud is so common in mid-latitudes.

Answer: A

Mineral dust, particularly clay minerals, and biological particles such as certain bacteria and pollen are the most effective ice nuclei, and they are far scarcer than the nuclei on which droplets form. Sea salt and sulphate are excellent condensation nuclei but poor ice nuclei. The scarcity of ice nuclei is what allows supercooled water to persist at all.

Answer: A

The correct plot is P.

As the parcel rises its total pressure falls, and since no water is added or removed its mixing ratio stays fixed, so its vapour pressure falls in proportion to the total pressure: slowly, by roughly 10% per kilometre. The saturation vapour pressure depends only on temperature, and the parcel cools at the dry adiabatic rate of about 9.8 °C per km, so it falls much faster. With no nuclei for condensation, the two curves simply cross and the parcel becomes supersaturated above the crossing point. Plot P shows exactly this: a gently decreasing solid line crossed by a steeply decreasing dashed line.

In Q the vapour pressure falls much faster than the saturation value, in R the saturation vapour pressure increases with height, and in S the vapour pressure stays constant although the pressure decreases.

Answer: C

The process is Adiabatic heating or cooling.

In an adiabatic process no heat is exchanged with the surroundings. Rising air expands as pressure falls and cools; sinking air is compressed and warms. Dry air changes temperature at about 9.8°C per kilometre in this way, which is how rising air reaches its dew point and forms clouds.

Advection is the horizontal transport of heat or moisture by the wind. Deflation is wind erosion of loose material, and sublimation is the change of ice directly to vapour.

Answer: C

The dry adiabatic lapse rate is 9.8 K/km.

A rising parcel of unsaturated air expands and cools adiabatically. The rate is set by gravity and the specific heat of air at constant pressure:

\[ \begin{aligned} \Gamma_{d} &= \frac{g}{c_{p}} = \frac{9.81}{1004} \\ &\approx 9.8\ \mathrm{K/km} \end{aligned} \]

6.5 K/km is the average environmental lapse rate of the standard atmosphere. Around 5 to 6 K/km is typical of the saturated (moist) adiabatic lapse rate at low levels, which is smaller because condensation releases latent heat. 1.6 K/km is far too small.

Answer: B

The parcel is at -4 degree Celsius at 4 km.

Below the lifting condensation level the parcel is unsaturated and cools at the dry adiabatic lapse rate, about 10 degrees Celsius per km. Above the LCL it is saturated and cools at the given moist rate of 7 degrees per km.

\[ \begin{aligned} T_{2\,km} &= 30 - 10 \times 2 = 10\ ^{\circ}C \\ T_{4\,km} &= 10 - 7 \times 2 = -4\ ^{\circ}C \end{aligned} \]

Using the moist rate all the way up gives \(30 - 28 = 2\) degrees, which is not offered; using the dry rate all the way gives -10. None of -6, 0 or 6 degrees follows from applying the two rates to their own layers, and the warmer answers ignore the faster dry cooling below the cloud base.

Answer: C

The rate is about 10°C/1000 meters.

Unsaturated air that rises expands and cools adiabatically, and air that sinks is compressed and warms, at the dry adiabatic lapse rate:

\[ \begin{aligned} \Gamma_{d} &= \frac{g}{c_{p}} = \frac{9.81}{1004} \\ &\approx 9.8\ \text{K/km} \end{aligned} \]

That is close to 10°C per 1,000 m. Once air is saturated, latent heat released by condensation slows the cooling to the saturated (moist) adiabatic rate, about 5-6°C per 1,000 m. The other options, 20, 25 and 30°C per 1,000 m, are far steeper than any adiabatic rate.