Automobile Emission Control Questions and Answers

Answer: C

Correct answer: 0.12 kg

Multiply the carbon mass by its working capacity:

\[ 1.5 \times 0.08 = 0.12\ \mathrm{kg} \]

Working capacity is deliberately less than the total adsorption capacity, since the canister is never purged completely dry. The figure must cover the vapour generated over a hot day's parking plus refuelling losses where the system handles them - and once it saturates, further vapour escapes to atmosphere.

Answer: D

The decile emits \( 0.10 \times 5 = 0.5 \) units against the rest's \( 0.90 \times 1 = 0.9 \), so its share is \( 0.5/1.4 = 0.357 \). A small population of badly maintained vehicles carrying so large a share of the total is exactly why inspection and maintenance programmes are worth running.

Answer: A

Correct answer: 15 per cent of the test

Express the pre-light-off period as a fraction of the cycle:

\[ \frac{90}{600} \times 100 = 15\% \]

Fifteen per cent of the duration accounts for a far larger share of the emissions, since the catalyst is converting almost nothing during it. That disproportion is what drives close-coupled converters, low thermal mass substrates and electrically heated catalysts - all measures to shorten those ninety seconds.

Answer: B

A window of one per cent admits values from 0.99 to 1.01, so only the stoichiometric reading qualifies. It is a remarkably narrow band, and holding the engine within it is precisely what the closed loop fuelling system exists to do.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the reason for the first. The three-way catalyst must oxidise carbon monoxide and hydrocarbons while simultaneously reducing nitrogen oxides, and those requirements conflict: oxidation wants oxygen, reduction wants none.

Only within a narrow window around stoichiometric does the exhaust suit both. Holding the mixture there needs a lambda sensor and closed loop control, oscillating rapidly about the target - which is why the catalyst and the sensor arrived together.

Answer: D

Correct answer: Statement I is incorrect but Statement II is correct

The second statement gives the mechanism correctly, and it refutes the first. Thermal formation of nitrogen oxides needs high temperature and available oxygen together.

A deeply rich mixture is hot but oxygen-starved, so formation falls. The peak occurs slightly lean of stoichiometric, where the flame is still hot and oxygen is in surplus. That is why the emission curves for nitrogen oxides and for carbon monoxide peak on opposite sides of stoichiometric.

Answer: A

Both hold and the second is the reason: reduction of nitrogen oxides is impossible in an oxygen-rich exhaust. Separate devices are therefore needed, an oxidation catalyst for the one duty and a reduction system for the other.

Answer: B

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the mechanism. Recirculated exhaust is inert: it takes up heat as the charge burns without releasing any, so the peak flame temperature falls.

Since thermal formation of nitrogen oxides rises steeply with temperature, a modest reduction cuts the emission substantially. The cost is combustion quality - too much recirculation makes the burn slow and unstable - which is why the rate is scheduled against load and speed rather than applied continuously.

Answer: C

Correct answer: Statement I is incorrect but Statement II is correct

The second statement is true and it is exactly why the first fails. Until light-off the catalyst converts almost nothing, so the first minute or two passes essentially untreated exhaust.

That fixed quantity is emitted whatever the journey length, so it cannot be ignored - it simply becomes a smaller share of a longer trip while remaining the same absolute amount. Averaged over a fleet of short urban journeys it dominates, which is why cold start is a regulated test condition.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the care needed. The canister holds adsorbed fuel vapour, and purging it admits an uncontrolled quantity of fuel to the intake.

An unscheduled purge therefore drives the mixture rich, upsetting the closed loop control and the catalyst's operating window. The controller purges when conditions allow - warm engine, steady load - and trims the injection to compensate, learning the vapour concentration as it goes.