What is the current EPA NAAQS primary 1-hour standard for SO2?
A.75 ppb (1-hour average)
B.140 μg/m³ (24-hour average)
C.35 ppb (annual arithmetic mean)
D.500 μg/m³ (3-hour secondary standard)
Show solution
Answer · A
The EPA revised the primary SO2 NAAQS in 2010, replacing the older 24-hour and annual standards with a 1-hour standard of 75 ppb. This 1-hour standard protects against short-duration SO2 peaks that can trigger respiratory effects. It is not to be exceeded more than once per year on a 3-year design value basis. Choice B (140 μg/m³ 24-hour) was part of the previous (pre-2010) primary standard now revoked. Choice C (35 ppb annual) is the annual standard for NO2, not SO2. Choice D (500 μg/m³ 3-hour) is the secondary (welfare) standard for SO2, protecting vegetation and ecosystems.
Question 2Environmental Science & Management
Two noise sources operate simultaneously: 75 dB(A) and 72 dB(A). What is the combined sound pressure level?
A.73.5 dB(A)
B.74.5 dB(A)
C.76.8 dB(A)
D.147 dB(A)
Show solution
Answer · C
Convert each sound level to relative acoustic energy before adding: Ltotal = 10·log₁₀(10^(L1/10) + 10^(L2/10)).
Compute the two terms: 10^(75/10) = 3.162×10⁷ and 10^(72/10) = 1.585×10⁷.
Add the energies: 3.162×10⁷ + 1.585×10⁷ = 4.747×10⁷.
Convert back to decibels: Ltotal = 10·log₁₀(4.747×10⁷) = 76.8 dB(A).
Averaging or directly adding the decibel values is incorrect because decibels are logarithmic.
Question 3Site Assessment & Remediation
An aquifer has: Kd = 2.5 mL/g (partition coefficient), bulk density ρb = 1.8 g/cm³, porosity n = 0.30. What is the retardation factor R?
A.6
B.11
C.16
D.21
Show solution
Answer · C
Use the retardation factor equation for sorbing contaminants: R = 1 + (ρb·Kd)/n.
Substitute the given values with compatible g, cm³, and mL units: R = 1 + (1.8 × 2.5)/0.30.
Evaluate the ratio: 1.8 × 2.5 = 4.5, and 4.5/0.30 = 15.
Add the advective term: R = 1 + 15 = 16, meaning the contaminant moves about 16 times slower than groundwater.
Question 4Solid & Hazardous Waste
Under RCRA, which document is required to accompany hazardous waste shipments from a generator to a TSDF?
A.Waste stream profile and constituent characterization report (40 CFR Part 261.11)
B.Safety Data Sheet per OSHA 29 CFR 1910.1200, Section 14 transport hazard information
C.Uniform Hazardous Waste Manifest (EPA Form 8700-22)
D.RCRA Part B hazardous waste storage and treatment facility permit (EPA Form 8700-23)
Show solution
Answer · C
The Uniform Hazardous Waste Manifest (EPA Form 8700-22) is required under RCRA 40 CFR Part 262 to accompany all off-site shipments of hazardous waste from a generator to a TSDF. The manifest must be signed by the generator, each transporter, and the receiving TSDF. If the generator doesn’t receive a copy signed by the TSDF within 35 days, they must file an exception report. Choice A (waste stream profile) and Choice B (SDS) are not the required tracking document. Choice D confuses the facility’s operating permit (Part B) with the shipment tracking document.
Question 5Water & Wastewater Engineering
A treatment lagoon (V = 15,000 m³) receives a flow of 500 m³/day. What is the hydraulic detention time?
A.10 days
B.20 days
C.30 days
D.45 days
Show solution
Answer · C
Hydraulic Retention Time (HRT) is calculated as HRT = V/Q, where V is the lagoon volume and Q is the volumetric flow rate. Substituting the given values: HRT = 15,000 m³ ÷ 500 m³/day = 30 days. HRT is a critical design parameter governing the available reaction time for BOD removal, solids settling, and pathogen die-off; selecting the wrong volume-to-flow ratio (e.g., inverting Q/V or misreading units) produces the plausible-looking but incorrect alternatives of 10, 20, or 45 days.
Question 6Water Resources Engineering
A watershed has CN = 75 and receives P = 3.5 inches of rainfall. Using the NRCS runoff equation, what is the runoff depth Q?
A.0.85 in
B.1.10 in
C.1.30 in
D.1.57 in
Show solution
Answer · C
S = 1000/CN − 10 = 1000/75 − 10 = 3.33 in. Ia = 0.2S = 0.667 in. Q = (P−Ia)²/(P−Ia+S) = (3.5−0.667)²/(3.5−0.667+3.33) = (2.833)²/6.163 = 8.026/6.163 = 1.30 in.
Question 7Air Quality Engineering
A baghouse processes 60,000 m³/h of gas. The air-to-cloth ratio is 1.2 m³/(m²·min). What total filter area is required?
A.417 m²
B.625 m²
C.833 m²
D.1,250 m²
Show solution
Answer · C
Convert the gas flow to the same time basis as the air-to-cloth ratio: 60,000 m³/h ÷ 60 min/h = 1,000 m³/min.
Use the definition of air-to-cloth ratio, A/C = Q/A, so A = Q/(A/C).
Substitute the values: A = 1,000 m³/min ÷ 1.2 m³/(m²·min) = 833 m².
A much smaller area would come from mishandling the hour-to-minute conversion, while a larger area such as 1,250 m² would correspond to using a lower filtration velocity.
Question 8Environmental Science & Management
The method detection limit (MDL) for a target analyte is 0.005 mg/L. A sample analysis yields a measured value of 0.003 mg/L. How should this result be reported?
A.Report as 0.003 mg/L (the measured value)
B.Report as '<MDL' or '<0.005 mg/L' (not detected)
C.Report as 0.004 mg/L (average of measured and MDL)
D.The analysis is invalid and must be repeated
Show solution
Answer · B
Results below the MDL cannot be distinguished from instrument noise at the required statistical confidence level. Per EPA guidance, results < MDL are reported as 'not detected' or '<MDL'. The numeric value may be flagged but should not be reported as a confirmed measurement.
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