Composition and Structure of the Atmosphere Questions and Answers

Answer: A

Chapman set out the four reactions of formation and destruction that determine the natural ozone concentration in a pure oxygen atmosphere. The scheme predicts a layer in roughly the observed place but overestimates the amount, because catalytic cycles involving nitrogen, hydrogen and halogen species destroy ozone faster than Chapman chemistry alone allows. The Brewer-Dobson circulation handles transport rather than chemistry. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: B

Pure oxygen chemistry gives roughly twice the observed ozone amount, and the discrepancy is made up by catalytic destruction cycles involving nitrogen, hydrogen and halogen radicals. Identifying those cycles was a major achievement of atmospheric chemistry in the 1970s. The altitude of the predicted layer is approximately right, which is why the scheme remains the starting point for any discussion. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: C

Molina and Rowland showed that chlorofluorocarbons, being inert in the troposphere, reach the stratosphere intact and are then photolysed to release chlorine atoms that catalytically destroy ozone. A single chlorine atom can destroy many thousands of ozone molecules. Volcanic sulphate does provide surfaces for heterogeneous chemistry, which makes that a genuine contributory factor rather than a wholly wrong answer. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: D

The Montreal Protocol of 1987 set binding schedules for phasing out the production of ozone depleting substances and has been strengthened several times since. It is generally regarded as the most successful international environmental agreement, with stratospheric chlorine now declining. The Vienna Convention was the framework it sat within, so that option is close but does not name the instrument that set the limits. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: A

Temperatures below about 195 kelvin allow polar stratospheric clouds to form, and their surfaces convert reservoir chlorine species into forms that photolyse readily when sunlight returns. The isolation of the vortex prevents replenishment. The rapid loss then occurs in spring, which is why the hole appears in September and October rather than in midwinter darkness. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: B

The chemistry requires sunlight to photolyse the activated chlorine, so although the clouds form in midwinter darkness the ozone loss occurs rapidly once the sun returns in September and October. By December the vortex breaks down and ozone rich air is mixed in. The sequence of cold, then activation, then sunlight, then loss is worth following through in order.

Answer: C

Pressure at 5.5 kilometres is close to 500 hectopascals, half the surface value, and under hydrostatic balance the pressure is proportional to the mass above, so exactly half the atmosphere lies below. This gives the 500 hectopascal surface its special status in synoptic analysis. The result follows in one line from hydrostatic balance and is worth deriving rather than memorising. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: D

With a mean temperature near 250 kelvin the scale height \( RT/g \) works out at about 8 kilometres, so pressure falls by a factor of about 2.7 over that distance. The value depends on temperature and therefore varies with latitude and season. The boundary layer depth and tropopause height happen to be of comparable magnitude, which makes them plausible distractors. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: A

The environmental lapse rate averages 6.5 kelvin per kilometre in the standard atmosphere, which lies between the dry adiabatic value of 9.8 and typical saturated values of 4 to 7. That placement makes the troposphere conditionally unstable, stable to dry ascent but unstable to saturated ascent. Confusing the environmental lapse rate with the dry adiabatic rate is the commonest error in this area. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

The International Standard Atmosphere defines a mean sea level pressure of 1013.25 hectopascals, a temperature of 288.15 kelvin and a lapse rate of 6.5 kelvin per kilometre to 11 kilometres. It is a reference for aircraft altimetry and instrument calibration rather than a description of any actual day. Humidity is specified as zero, so it says nothing about water vapour. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.