Show a photo/short clip of a scuba tank (200 bar) and a helium party balloon. Ask: 'If I use PV = nRT to calculate how many moles of O₂ are in a scuba tank, will my answer be too high, too low, or exact? What about for the helium balloon?' Take a quick hand vote. Reveal that PV=nRT overestimates moles of O₂ in a scuba tank by several percent, while the He balloon is almost perfectly ideal. Frame the driving question: what molecular properties and conditions make PV=nRT fail? Targets SP 6 Argumentation — students commit to a claim before instruction.

1. 5m

   The two assumptions that fail

   Content

   The ideal gas law rests on two assumptions from kinetic molecular theory: (1) gas particles have negligible volume compared to the container, and (2) gas particles exert no attractive forces on each other. Both assumptions are excellent at low pressure and high temperature, where particles are far apart and moving fast. They fail at high pressure (particles crowded, own volume matters, and neighbors are close enough to attract) and at low temperature (particles slow, attractions redirect trajectories). A real gas is a gas in which these two effects show up as measurable departures from PV = nRT. The direction of the deviation depends on which effect dominates: intermolecular attractions pull molecules inward and reduce wall collisions, giving lower observed P than ideal; excluded volume forces particles into less free space, giving higher observed P than ideal.

   Delivery

   Anchor everything on the two assumptions — write them as bullets on the slide and refer back to them every time deviation is discussed. Ask students to predict which assumption fails first as you compress a gas, and which fails first as you cool a gas. Head off the misconception that 'attractions always raise pressure' — attractions PULL molecules away from the wall, so they LOWER measured pressure. Emphasize this is about what a pressure gauge would actually read versus what PV=nRT predicts.

2. 5m

   Effect 1 — Intermolecular attractions lower measured P

   Content

   Imagine a molecule racing toward the container wall. In an ideal gas, nothing pulls on it; it hits the wall with its full momentum. In a real gas, the neighboring molecules just behind it exert attractive IMFs that tug it backward at the moment of collision, reducing the force it delivers to the wall. Multiply this over ~10²³ collisions per second and the observed pressure is LOWER than PV = nRT would predict. Consequence: Z = PV/nRT < 1 whenever attractions dominate. This effect is strongest at moderate-to-high pressures (neighbors are close enough to attract) and low temperatures (molecules move slowly enough that the attractive tug significantly changes their trajectory). Gases with stronger IMFs — NH₃, H₂O, CO₂ — show larger dips below Z = 1 than He or H₂.

   Delivery

   Use the visual of a molecule being tugged back by neighbors as it approaches the wall. Ask: 'If the wall gets hit softer, does the gauge read higher or lower than ideal?' — walk them to 'lower.' This is the misconception hotspot: many students think attractions squeeze the gas and raise pressure. Push back: attractions act BETWEEN molecules, not between molecules and the wall — they only ever reduce the wall impact. Connect to IMF strength: rank He < N₂ < CO₂ < NH₃ by expected dip depth.

3. 5m

   Effect 2 — Excluded volume raises measured P

   Content

   Now push the pressure much higher. The molecules themselves take up space — call it the excluded volume, b. The free volume actually available for particles to move in is (V − nb), which is smaller than V. Particles collide with the walls MORE often per unit real free volume than PV = nRT assumes, so the observed pressure ends up HIGHER than ideal. Consequence: Z = PV/nRT > 1 whenever finite-volume effects dominate. This effect always wins eventually — at very high pressure every real gas has Z > 1, and the Z-versus-P curve rises steeply. Larger molecules (Xe, CO₂, larger hydrocarbons) show this rise earlier than small ones (He, H₂).

   Delivery

   Use the crowded-particles visual to make 'excluded volume' concrete — literally point out that in the ideal picture we pretend the box is empty except for point particles, while in the real picture a big fraction of the box is filled by molecules themselves. Ask: 'If less space is really available, do molecules hit the wall more or less often?' — build to 'more often, so P is higher.' Preview that this is why every Z curve eventually turns upward at high P, even for gases that dip first.

4. 5m

   Reading a Z-vs-P plot

   Content

   A compressibility-factor plot shows Z = PV/nRT on the y-axis and pressure on the x-axis, for a fixed temperature. Ideal behavior is the horizontal line Z = 1. A typical real-gas curve starts near Z = 1 at low P, dips below 1 as attractions become significant, reaches a minimum, then climbs above 1 as excluded volume takes over at high P. The depth of the dip reports IMF strength; the steepness of the rise reports molecular size. He and H₂ barely dip and rise gently — nearly ideal at ordinary conditions. N₂ shows a shallow dip. CH₄ dips more. CO₂ and NH₃ show deep dips because of stronger IMFs. Raising T flattens the curve toward Z = 1 (attractions matter less than kinetic energy). Lowering T deepens the dip (attractions matter more; the gas is close to condensing).

   Delivery

   Walk students across a single curve first — 'what's happening in region A vs region B?' — then compare multiple gases on the same axes. Explicitly ask, 'Which gas is most ideal here? How do you know?' Address the misconception that all gases deviate equally — the whole point of the plot is that they don't. Also state clearly: ideality is BEST at low P and high T (top-left of any Z plot), not the reverse.

1. 25m

   Ranking real gases with Z-vs-P data + PhET simulation check

   Structure: 5 min set up and Part 1 individually, 10 min Part 2 in pairs at computers, 10 min Part 3 whole-class share-out and consolidation. Targets SP 1 Models and Representations, SP 4 Model Analysis, and SP 6 Argumentation. Walk around and check that pairs correctly identify Z < 1 as attractions-dominated and Z > 1 as volume-dominated; a common error is calling a Z > 1 gas 'more ideal' because Z is 'closer to 1 on average' — push them to reason from the physics, not the average. Student handout: Part 1 — Predict from molecular properties (5 min, individual) Below are four gases at 300 K. Using what you know about IMFs and molecular size, RANK them from most ideal (1) to least ideal (4) at 300 K and moderate pressure. Justify each ranking with ONE sentence about IMF strength or molecular size. - He (molar mass 4 g/mol, only weak dispersion) - N₂ (28 g/mol, weak dispersion, nonpolar) - CH₄ (16 g/mol, moderate dispersion, nonpolar) - CO₂ (44 g/mol, strong dispersion, quadrupole) Ranking (1 = most ideal): ______, ______, ______, ______ Justification: ______________________________________ Part 2 — Test with real Z data (10 min, pairs) Below are actual compressibility factors Z = PV/nRT for these four gases at 300 K at three pressures. Higher-magnitude departure from Z = 1 means MORE deviation, in either direction. - He: Z(1 atm) = 1.0005, Z(100 atm) = 1.049, Z(500 atm) = 1.245 - N₂: Z(1 atm) = 0.9998, Z(100 atm) = 0.984, Z(500 atm) = 1.086 - CH₄: Z(1 atm) = 0.998, Z(100 atm) = 0.783, Z(500 atm) = 0.869 - CO₂: Z(1 atm) = 0.994, Z(100 atm) = 0.213, Z(500 atm) = 0.677 Answer: 1. At 100 atm, rank the four gases by MAGNITUDE of deviation from ideality. Actual ranking: ______, ______, ______, ______ 2. Which gas has Z > 1 at all three pressures? ______ Which effect (attractions or excluded volume) dominates for this gas? ______ 3. Which gas has Z far BELOW 1 at 100 atm? ______ Which effect dominates here, and what molecular feature of this gas causes it? ______ 4. He at 500 atm has Z = 1.245. Does this mean He has strong IMFs? Explain in one sentence. ______________________________________ 5. Compare your Part 1 prediction to the data. Any surprises? What molecular property matters MOST at 100 atm and 300 K — IMFs or size? ______________________________________ Part 3 — PhET simulation check (7 min, pairs) Open: https://phet.colorado.edu/en/simulations/gas-properties 1. Set the sim to Ideal mode. Add ~200 heavy particles. Record P, V, T. Compute PV/nRT — should be ≈ 1. 2. Now shrink the container to increase pressure until particles fill a large fraction of the box. Recompute PV/nRT. - Did Z go up or down? ______ - Which real-gas effect does the sim capture — excluded volume, attractions, or both? ______ 3. Claim–Evidence–Reasoning (write on your handout): Under what conditions (P, T) is the ideal gas law most reliable, and why at the particle level? Claim: ______________________________________ Evidence (cite Z values from Part 2 or the sim): ______________________________________ Reasoning (connect to IMFs and excluded volume): ______________________________________

   Materials

   • Printed student handout (content below)
   • Computers with internet access
   • Calculators

   Example outputs

   • Part 2 Q1: Actual ranking of magnitude of deviation at 100 atm: CO₂ (|Δ|=0.787) > CH₄ (0.217) > N₂ (0.016) > He (0.049). Note He is #3, not #4, because He is already on the rise. Accept CO₂ > CH₄ > He > N₂ with correct reasoning.
   • Part 2 Q2: He — Z > 1 at every P shown. Excluded volume dominates because He's IMFs are so weak they barely register, so the finite-volume rise wins from the start.
   • Part 2 Q3: CO₂ — Z = 0.213 at 100 atm. Intermolecular attractions dominate; CO₂ has strong dispersion forces (large polarizable electron cloud) and a quadrupole moment, and 300 K is close to CO₂'s critical temperature (~304 K), so attractions strongly reduce wall collisions.
   • Part 2 Q4: No — Z > 1 means excluded volume dominates. He deviates upward because attractions are so weak they never mattered. Deviation magnitude ≠ IMF strength; direction matters.
   • Part 3 CER: Claim — the ideal gas law is most reliable at low pressure and high temperature. Evidence — at 1 atm all four gases have Z within 0.01 of 1; at 100 atm CO₂ has Z = 0.213. Reasoning — at low P, particles are far apart so both excluded volume and IMFs are negligible; at high T, kinetic energy overwhelms IMF tugs.
   • Common wrong answer to head off: 'CH₄ deviates most because it has the most atoms.' Wrong — CO₂ deviates more because its IMFs (dispersion + quadrupole) are stronger AND 300 K is near its condensation regime.
   • Part 1 typical justification: He is most ideal because it is small with only weak dispersion; CO₂ is least ideal because it is largest with strongest IMFs (dispersion + quadrupole).

   No-equipment fallback

   If computers are unavailable, replace Part 3 with an additional data set: give students Z values for CO₂ at 250 K, 300 K, and 400 K at 100 atm (Z ≈ 0.05 [liquid], 0.213, 0.68) and ask them to explain why lowering T deepens the deviation, connecting to slower molecules and stronger relative influence of IMFs.

1. Under which set of conditions will a sample of NH₃(g) behave MOST like an ideal gas? A) 500 K and 0.5 atm B) 500 K and 200 atm C) 250 K and 0.5 atm D) 250 K and 200 atm

   multiple choiceA. Ideal behavior is best at high T (kinetic energy dominates over NH₃'s strong hydrogen-bonding IMFs) and low P (particles far apart, excluded volume negligible). Targets SP 4 Model Analysis.

2. At 300 K and 100 atm, N₂ has Z = 0.98 while H₂ has Z = 1.06. Explain, at the particle level, why N₂'s Z is below 1 but H₂'s Z is above 1 at the same conditions. (2-3 sentences)

   short answerN₂ has stronger dispersion forces than H₂ (larger polarizable electron cloud), so IMFs tug molecules away from the wall enough to lower the measured pressure below ideal, giving Z < 1. H₂ has extremely weak IMFs that are essentially negligible at 300 K, so the finite volume of H₂ molecules (excluded volume) dominates and raises the observed pressure above ideal, giving Z > 1. Targets SP 6 Argumentation and SP 1 Models and Representations.

3. A student measures 5.00 mol of CO₂ in a 1.00 L rigid vessel at 300 K. Using PV = nRT, they predict P_ideal. The actual measured pressure is significantly lower than P_ideal. (a) Calculate P_ideal in atm. (R = 0.0821 L·atm/(mol·K)) (b) Explain why P_measured < P_ideal for CO₂ at these conditions, referencing the two assumptions of the ideal gas law and which one dominates here.

   calculation(a) P_ideal = nRT/V = (5.00)(0.0821)(300)/(1.00) = 123 atm. (b) At 300 K and ~123 atm, CO₂ is near its critical temperature (304 K) and highly compressed. Intermolecular attractions (strong dispersion + quadrupole) pull molecules away from the wall at the instant of collision, so each collision delivers less force than PV = nRT assumes — measured P drops below ideal. The excluded-volume effect is present but is out-competed by the strong attractions at this T, giving Z < 1. Targets SP 5 Mathematical Routines and SP 6 Argumentation.

4. The compressibility-factor curves for He, CH₄, and CO₂ are plotted versus P at 300 K on the same axes. Rank the depth of the minimum (dip) of each curve from deepest to shallowest, and justify your ranking.

   short answerDeepest to shallowest dip: CO₂ > CH₄ > He. CO₂ has the strongest IMFs (largest polarizable electron cloud plus quadrupole) and is closest to its critical temperature at 300 K, so attractions cause the deepest depression in Z. CH₄ has moderate dispersion forces, giving a shallower dip. He has essentially only very weak dispersion between light atoms, so its Z barely dips (and in fact climbs above 1 quickly as excluded volume wins). Targets SP 4 Model Analysis and SP 6 Argumentation.

• ☐Printed student handout with Parts 1–3 (one per student)
• ☐Computers or Chromebooks with internet (one per pair) for PhET Gas Properties simulation
• ☐Calculators
• ☐Whiteboard/projector for share-out