The problems are on the ideal gas law worksheet. Try them on your own first, then compare with the answers below.
Quick answer: An ideal gas is a collection of tiny, randomly moving particles whose average kinetic energy is proportional to the kelvin temperature. These worked answers correct the 20 problems of the AP Physics 2 worksheet on ideal gas law, from Kelvin readings to design an experiment on conduction: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit in bold. Compare your reasoning as well as your final number, because AP scoring rewards the physics argument.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 9
Problem 1 Kelvin readings
Use T(K) = T(°C) + 273.1. 12 + 273 = 285 K.
2. 400 − 273 = 127 °C.
3. Absolute zero is 0 K, so −273 °C.
Problem 2 Counting molecules
1. N = nNA = 0.25 × 6.02 × 10²³ = 1.5 × 10²³ molecules.2. m = nM = 0.25 × 28 = 7.0 g.
Problem 3 A bicycle pump
1. 4.0 cm² = 4.0 × 10⁻⁴ m².2. p = F/A = 120 / 4.0 × 10⁻⁴ = 3.0 × 10⁵ Pa.
3. In a smaller volume each particle reaches the walls more often, so there are more collisions per second and a larger average force per unit area. Higher pressure comes from more frequent collisions.
Problem 4 Energy of one particle
1. 27 + 273 = 300 K.2. K = (3/2)kBT = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J.
3. 327 °C = 600 K, so K = 1.24 × 10⁻²⁰ J. The kelvin temperature doubled, so the average energy doubled (ratio 2.00).
Problem 5 Heat through a window
1. 4.0 mm = 4.0 × 10⁻³ m.2. H = kAΔT/L = 0.80 × 1.5 × 12 / 4.0 × 10⁻³ = 3600 W (3.6 kW).
3. H is inversely proportional to L, so the rate halves to 1.8 kW.
Problem 6 Pressure from a particle count
1. 1.5 L = 1.5 × 10⁻³ m³.2. p = (3.0 × 10²² × 1.38 × 10⁻²³ × 310)/(1.5 × 10⁻³) = 8.6 × 10⁴ Pa.
3. This is 85% of atmospheric pressure, so the flask is slightly below atmospheric pressure.
Problem 7 Weather balloon
1. Fixed n gives p1V1/T1 = p2V2/T2.2. V2 = V1(p1/p2)(T2/T1) = 2.0 × (1.01 × 10⁵/2.6 × 10⁴) × (223/288) = 6.0 m³.
3. The pressure falls by a factor of about 3.9 while the temperature falls only by a factor of 1.3, so the pressure effect dominates and the gas expands.
Problem 8 Speed of nitrogen
1. vrms = √(3RT/M) = √(3 × 8.31 × 293/0.028) = 511 m/s.2. m = 0.028/6.02 × 10²³ = 4.65 × 10⁻²⁶ kg.
3. √(3 × 1.38 × 10⁻²³ × 293/4.65 × 10⁻²⁶) = 511 m/s, which agrees with part 1 within rounding.
Problem 9 Doubling the speed
1. At 1000 K: √(3 × 8.31 × 1000/0.028) = 944 m/s. At 250 K: 472 m/s.2. 944/472 = 2.00.
3. vrms ∝ √T, so doubling the speed needs four times the kelvin temperature.
Problem 10 Graph of p against 1/V
1. A straight line through the origin; the slope equals nRT, which is pV = 60.0 J.2. n = pV/(RT) = 60.0/(8.31 × 295) = 0.0245 mol.
3. p = 60.0/1.20 × 10⁻³ = 5.0 × 10⁴ Pa.
Problem 11 Heat through a brick wall
1. ΔT = 21 − 5 = 16 K.2. H = 0.72 × 12 × 16/0.22 = 628 W.
3. Q = HΔt = 628.4 × 3600 = 2.3 × 10⁶ J.
Problem 12 Pan bases compared
1. Aluminium: 205 × 0.030 × 4.0/0.0030 = 8200 W. Steel: 16 × 0.030 × 4.0/0.0030 = 640 W.2. Ratio = 205/16 = 12.8.
3. Copper conducts heat sideways quickly, spreading the heat from the burner across the base so there are no hot spots.
Problem 13 Sealed can on a stove
1. Heating raises the kelvin temperature, so the particles’ average kinetic energy and their speeds increase.2. Faster particles strike the walls more often and with larger impulses, so the average force per unit area grows. With V and n fixed, p ∝ T.
3. The pressure difference across the wall grows until the wall can no longer resist it, although the number of particles is unchanged.
Problem 14 Helium against air
1. From pV = nRT with n = m/M: pV = (m/M)RT, so ρ = m/V = pM/(RT).2. Helium: 0.166 kg/m³. Air: 1.20 kg/m³.
3. Net lift = (ρair − ρHe) × 1.0 × 9.81 = (1.203 − 0.166) × 9.81 = 10.2 N.
Problem 15 Spot the error
1. T was not converted to kelvin (25 °C is 298 K), and V was not converted to m³ (3.0 L = 3.0 × 10⁻³ m³).2. p = (0.15 × 8.31 × 298)/(3.0 × 10⁻³) = 1.2 × 10⁵ Pa.
3. 10.4 Pa is about one ten-thousandth of atmospheric pressure, which is absurd for a closed container with 0.15 mol of gas.
Problem 16 A heated jar lid
1. p2 = p1T2/T1 = 1.00 × 10⁵ × 453/293 = 1.55 × 10⁵ Pa.2. F = (p2 − p0)A = (5.46 × 10⁴) × 5.0 × 10⁻³ = 273 N.
3. Since 273 N exceeds 200 N, the lid is pushed off (or the seal fails).
Problem 17 Bubble from a lake bed
1. p = p0 + ρgh = 1.01 × 10⁵ + 1000 × 9.81 × 12 = 2.19 × 10⁵ Pa.2. V2 = V1(p1/p2)(T2/T1) = 1.0 × (2.19 × 10⁵/1.01 × 10⁵)(295/280) = 2.3 cm³.
3. The pressure ratio is 2.17 but the temperature ratio is only 1.05, so the pressure change dominates.
Problem 18 Two layers in series
1. Wood: 0.020/(0.15 × 2.0) = 0.0667 K/W. Foam: 0.040/(0.030 × 2.0) = 0.667 K/W.2. The temperature differences add: ΔTtotal = H(RW + RF), so H = 30/0.7333 = 41 W.
3. ΔT across wood = HRW = 40.91 × 0.06667 = 2.73 K, so the interface is at 25 − 2.73 = 22.3 °C. Almost the entire drop occurs across the foam.
Problem 19 Heavier means faster?
1. Both have K = (3/2)kBT = 6.21 × 10⁻²¹ J, so the energies are equal.2. Ratio = √(MN₂/MO₂) = √(28/32) = 0.935.
3. At the same temperature all ideal gas particles have the same average translational kinetic energy. Because K = ½mv², the heavier oxygen molecules must move more slowly, about 6.5% slower than nitrogen. The claim fails on both counts.
Problem 20 Design an experiment on conduction
1. Independent: rod length L. Dependent: rate of heat flow H, found from the temperature rise of a water bath per second. Controlled: rod material, cross-sectional area, and the temperatures at the two ends.2. Use rods of identical material and diameter in several lengths. Keep one end in a beaker of boiling water and the other in a known mass of cold water with a thermometer, insulate the sides, and record the temperature rise over a fixed time for each length, then repeat each reading.
3. Plot H against 1/L. Because H = kAΔT/L, the graph should be a straight line through the origin with slope kAΔT.
