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This deck focuses on Deviation From Ideal Gas Law, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.
Study Deviation From Ideal Gas Law in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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Identify a gas that behaves nearly ideally under ordinary conditions.
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Helium. Small, light atoms with weak intermolecular forces.
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This deck focuses on Deviation From Ideal Gas Law, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: Helium. Small, light atoms with weak intermolecular forces.
Answer: An ideal gas perfectly follows the Ideal Gas Law. A theoretical gas that obeys all gas law assumptions perfectly.
Answer: V2a. For 1 mole, n=1, so the correction becomes V2a.
Answer: (P+V2an2)(V−nb)=nRT. Modified ideal gas law accounting for real gas behavior.
Answer: Strong intermolecular forces. Large 'a' means strong attractive forces between molecules.
Answer: R=0.0821mol×KL×atm. The standard value used in gas law calculations.
Answer: Corrects for volume and pressure deviations. Adds corrections for molecular size and intermolecular forces.
Answer: An ideal gas perfectly follows the Ideal Gas Law. A theoretical gas that obeys all gas law assumptions perfectly.
Answer: Increased intermolecular attractions. Lower temperatures enhance attractive forces between gas molecules.
Answer: Correcting Ideal Gas Law for real gases. Modifies ideal gas law to account for real gas behavior.
Answer: Gas particles exert no intermolecular forces. Assumes no attractive or repulsive forces between particles.
Answer: Low temperature. Reduces kinetic energy, allowing intermolecular forces to dominate.
Answer: PV=nRT. The fundamental relationship between pressure, volume, moles, and temperature.
Answer: Correcting Ideal Gas Law for real gases. Modifies ideal gas law to account for real gas behavior.
Answer: 'R' is the universal gas constant. A proportionality constant linking pressure, volume, moles, and temperature.
Answer: Deviation=2020.8. Pressure correction term: V2a for 1 mole of gas.
Answer: Gases deviate more from ideal behavior. Higher pressure compresses gas, making particle volume more significant.
Answer: Intermolecular attraction forces. Parameter representing strength of intermolecular attractive forces.
Answer: Decreases observed pressure. Attractive forces reduce pressure exerted on container walls.
Answer: V2an2. Accounts for attractive forces reducing observed pressure.
Answer: Gas particles exert no intermolecular forces. Assumes no attractive or repulsive forces between particles.
Answer: Larger size increases deviation. Larger molecules occupy more space, increasing deviation from ideality.
Answer: High pressure. Forces particles closer, making volume and intermolecular forces significant.
Answer: P=VnRT=2.463atm. Using P=VnRT with given values yields 2.463 atm.
Answer: At very high pressures. Extreme pressure makes particle volume and forces significant.
Answer: V2a. For 1 mole, n=1, so the correction becomes V2a.
Answer: Decreases observed pressure. Attractive forces reduce pressure exerted on container walls.
Answer: Volume occupied by gas particles. Total volume excluded due to finite molecular size.
Answer: Assumes no volume and no interactions between particles. Ignores real effects: molecular size and intermolecular forces.
Answer: High pressure. Forces particles closer, making volume and intermolecular forces significant.
Answer: n=RTPV=0.914mol. Using n=RTPV with given values yields 0.914 mol.
Answer: Gases deviate more from ideal behavior. Higher pressure compresses gas, making particle volume more significant.
Answer: P′=3+1020.5=3.005atm. Adding pressure correction: P+V2an2 for n=1.
Answer: Volume of gas particles. The finite size of gas molecules reduces available volume.
Answer: PV=nRT. The fundamental relationship between pressure, volume, moles, and temperature.
Answer: T=nRPV=121.95K. Rearranging ideal gas law: T=nRPV gives 121.95 K.
Answer: V2an2. Accounts for attractive forces reducing observed pressure.
Answer: P=VnRT=2.463atm. Using P=VnRT with given values yields 2.463 atm.
Answer: Boyle's Law. States that PV is constant at fixed temperature.
Answer: Equals 1. Compressibility factor equals 1 for perfect ideal behavior.
Answer: Gas particles have no volume. Assumes particles are point masses with negligible size.
Answer: Non-zero volume and intermolecular forces. Real gases have finite size and experience intermolecular attractions.
Answer: Pressure correction for intermolecular forces. Accounts for reduced pressure due to intermolecular attractions.
Answer: V−nb. Available volume after subtracting space occupied by gas particles.
Answer: Gases deviate more from ideal behavior. Higher pressure compresses gas, making particle volume more significant.
Answer: Charles's Law. States that V/T is constant at fixed pressure.
Answer: Intermolecular forces become significant. Reduced kinetic energy allows attractive forces to affect behavior.
Answer: Charles's Law. States that V/T is constant at fixed pressure.
Answer: Above it, gases behave more ideally. Higher temperatures overcome intermolecular forces, approaching ideal behavior.
Answer: V=PnRT=22.414L. Standard molar volume calculation at STP conditions.
Answer: Finite volume of gas particles. Molecular volume becomes significant fraction of container volume.