Solve eight changed-condition cases, explain what each inference needs, then compare feedback and revise.
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Combine pressure corrections before inferring inventory
A stipulated wet-gas sample is saturated with vapor of a hypothetical solvent at 300 K. External pressure is 100 kPa; a positive liquid-level correction makes internal wet pressure 4 kPa lower than external. The hypothetical solvent has a supplied vapor pressure of 12 kPa. Volume is 0.500 L and use R=8.00 kPa L mol⁻¹ K⁻¹. Enter the retained dry-gas amount in mmol. Explain the signs of both corrections and one condition needed before equating retained amount with amount produced.
Quantity / representation
Packet value
External / hydrostatic correction
100 kPa / −4 kPa
Stipulated solvent vapor contribution
12 kPa
Volume / temperature / rounded R
0.500 L / 300 K / 8.00 kPa L mol⁻¹ K⁻¹
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Check inventory before imposing saturation pressure
At fixed T and a fixed gas volume, an ideal-vapor packet requires 10 mmol vapor to reach its supplied saturation pressure. A closed initially evacuated vessel receives only 6 mmol of the volatile substance. There is no reaction or adsorption. At equilibrium, enter P/Psat. Explain whether any liquid remains and why increasing total inventory beyond 10 mmol changes the conclusion.
Quantity / representation
Packet value
Saturation vapor capacity
10 mmol at this T and gas volume
Total conserved inventory
6 mmol
Model
Ideal vapor; pure liquid if present; fixed T and gas volume
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Distinguish equal speed distributions from equal kinetic energy
Compare ideal equilibrium gas A (molar mass 20 g/mol, 400 K) and gas B (45 g/mol, 900 K). Which statement is consistent with speed scale proportional to √(T/M) and mean translational energy per molecule proportional to T? Explain what the equal speed distribution does and does not imply about molecular mass.
Quantity / representation
Packet value
A
20 g/mol, 400 K
B
45 g/mol, 900 K
Independent scalings
Speed distribution: T/M; mean energy per molecule: T
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Use another condition to test an apparently ideal result
A teaching equation writes Z=1+size−attraction. At one condition, size=0.25 and attraction=0.25. Size is fixed at fixed molar volume while attraction is proportional to 1/T. Which interpretation and changed-condition prediction follow? Explain why Z alone at one condition cannot identify the two mechanisms.
Quantity / representation
Packet value
Initial corrections
+0.25 and −0.25
Perturbation
Double absolute T; hold molar volume fixed
Scope
Stipulated equation, not a fitted empirical gas model
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Repair a mixed-salt particle and concentration claim
A final 250 mL solution contains 10 mmol CaCl2 and 20 mmol KCl, both fully dissolved and dissociated. Enter chloride concentration in mol/L. In your explanation give all three ionic amounts, verify total charge, and explain why using the sum of separately poured solvent volumes would be an additional assumption.
Quantity / representation
Packet value
Analytical amounts
10 mmol CaCl₂ and 20 mmol KCl
Final solution volume
250 mL
State assumption
Fully dissolved, fully dissociated; no reaction
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Normalize and bound a solvent preference
Two synthetic equilibrium-qualified records for the same solute at one temperature give dissolved mass A=5.0±0.5 g in 50.0 g solvent A and B=4.0±0.4 g in 100.0 g solvent B. Solvent masses are exact. Enter the smallest allowed ratio of normalized solubility A/B, to at least six decimal places if needed. Explain the opposite interval endpoint and why the conclusion changes if record A has no evidence of equilibrium.
Quantity / representation
Packet value
A dissolved / solvent
5.0 ±0.5 g / 50.0 g
B dissolved / solvent
4.0 ±0.4 g / 100.0 g
Qualification
Persistent excess solid, supplied plateau, same temperature; absolute input bounds
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Challenge one separation match with a second representation
A synthetic chromatographic spot could contain A, B or both because A and B co-migrate under the selected condition. A blank-corrected 1 cm optical packet has absorbances 0.170 and 0.070 at two wavelengths. Under a known two-species additive model, let x and y be concentrations in µmol/L and use the equations shown. Enter x for A. Explain y, why one spot did not establish purity, and what an unmodeled third absorber would invalidate.
Quantity / representation
Packet value
Wavelength 1
0.010x + 0.004y = 0.170
Wavelength 2
0.002x + 0.008y = 0.070
Assumptions
Known A/B absorptivities, matched blank and path, additive linear absorbance, no other absorber
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Separate energy scaling from an observable emission
A selected upper-to-lower transition would reduce matter energy by 3 eV, but the packet supplies zero population in that upper level. Compare its corresponding positive photon with a 2 eV photon and choose the supported statement. Explain the radiation/matter signs and why energy compatibility alone does not establish a spectral intensity.
Quantity / representation
Packet value
Selected matter change
−3 eV
Upper-level population
Zero in the supplied preparation
Photon relations
E=hν=hc/λ
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Retain a snapshot for a human review
The report separates current editable explanations from retained first/latest answers. It includes comparison reasoning only for already exposed cases. Preparing a snapshot does not submit it or establish independent review.
Source and representation boundaries
All eight packets use original stipulated values. The College Board CED, Unit 3 supplies the capability framework; no restricted exam questions are reproduced. The teaching models and their sources are explained in the solubility, optical inference and photon-transition lessons.
A reviewer should request an unfamiliar case and separately judge the pressure/species/energy ledger, particle representation, interval reasoning and evidence controls. This review does not replace broader solid/liquid/gas representation, empirical materials comparison or practical investigation. Saved text and numeric agreement alone cannot establish those outcomes.
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