States of Matter & Gases — Chapter quiz · Physical Science: Chemistry, Grade 11

Relate particle motion, attractions, gas models, and phase-change energy.

Problem 1

Convert 127°C to kelvin.
Answer and explanation

400 K

K = °C + 273 = 127 + 273 = 400 K.

Problem 2

According to the kinetic molecular theory, the average kinetic energy of gas particles is directly proportional to the…
  1. mass of each particle
  2. volume
  3. Kelvin temperature
  4. pressure
Answer and explanation

Kelvin temperature

Average kinetic energy is directly proportional to the absolute (Kelvin) temperature.

Problem 3

Which phase change is endothermic (absorbs energy)?
  1. condensation
  2. deposition
  3. melting
  4. freezing
Answer and explanation

melting

Melting absorbs energy to loosen the solid arrangement; freezing, condensation, and deposition all release energy.

Problem 4

A gas occupies 8.0 L at 1.0 atm. At constant temperature, what is its volume at 2.0 atm?
Answer and explanation

4 L

Boyle's Law: V₂ = P₁V₁/P₂ = (1.0 × 8.0)/2.0 = 4.0 L.

Problem 5

A gas occupies 3.0 L at 300 K. At constant pressure, what is its volume at 400 K?
Answer and explanation

4 L

Charles's Law: V₂ = V₁ × (T₂/T₁) = 3.0 × (400/300) = 4.0 L.

Problem 6

For a gas in a rigid sealed container, if the Kelvin temperature doubles, the pressure will…
  1. be cut in half
  2. stay the same
  3. become zero
  4. double
Answer and explanation

double

Gay-Lussac's Law: at constant volume, pressure is directly proportional to Kelvin temperature, so doubling T doubles P.

Problem 7

Explain why the pressure inside a sealed aerosol can rises when it is heated. Use both particle motion and a gas law in your answer.
Answer and explanation

Heating raises the average kinetic energy of the gas particles, so they move faster and strike the walls harder and more often, increasing the pressure. Because the can is rigid (constant volume), Gay-Lussac's Law applies: P/T is constant, so pressure is directly proportional to the Kelvin temperature. As T rises, P rises — which is why an overheated can may burst.

Problem 8

What volume does 1.0 mol of an ideal gas occupy at STP?
Answer and explanation

22.4 L

The molar volume of any ideal gas at STP is 22.4 L.

Problem 9

What volume does 4.0 mol of gas occupy at STP?
Answer and explanation

89.6 L

Volume = moles × 22.4 = 4.0 × 22.4 = 89.6 L.

Problem 10

A 5.0 L container holds 0.50 mol of gas at 300 K. What is the pressure, in atm? (R = 0.0821 L·atm/mol·K)
Answer and explanation

2.46 atm

P = nRT/V = (0.50 × 0.0821 × 300)/5.0 = 12.315/5.0 ≈ 2.46 atm.

Problem 11

For molecules of similar size, which intermolecular force is the strongest?
  1. hydrogen bonding
  2. London dispersion forces
  3. dipole-dipole forces
  4. they are all equal in strength
Answer and explanation

hydrogen bonding

For comparably sized molecules, the usual order is hydrogen bonding > dipole-dipole > London dispersion.

Problem 12

Which substance can form hydrogen bonds between its molecules?
  1. F₂
  2. HF
  3. CH₄
  4. CO₂
Answer and explanation

HF

HF has hydrogen bonded to fluorine, one of the three atoms (N, O, F) required for hydrogen bonding.

Problem 13

Explain the following claim and supply a valid example or counterexample: Average kinetic energy is not the same as total internal energy.
Answer and explanation

Energy changes particle arrangements/interactions during phase coexistence rather than mainly raising the kinetic-temperature scale. Average kinetic energy is not the same as total internal energy.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 14

Independent chapter check: Two equal-mass ideal-gas particles are compared through samples at 250 K and 500 K. Give the ratio of mean translational energy and characteristic speed, and explain why Celsius values cannot replace kelvin in these ratios.
Answer and explanation

The energy ratio is 2 and the characteristic speed ratio √2≈1.41. Kelvin is an absolute scale proportional to this mean energy; Celsius has an offset and its numeric ratios do not express the same physical relation.

The energy ratio is 2 and the characteristic speed ratio √2≈1.41. Kelvin is an absolute scale proportional to this mean energy; Celsius has an offset and its numeric ratios do not express the same physical relation.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 15

Explain the following claim and supply a valid example or counterexample: Use absolute pressure and kelvin; Celsius doubling is not absolute-temperature doubling.
Answer and explanation

V₂=2(450/300)=3L; the larger absolute temperature gives proportionally larger volume. Use absolute pressure and kelvin; Celsius doubling is not absolute-temperature doubling.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 16

Independent chapter check: A fixed ideal-gas amount occupies 3.00 L at 100 kPa absolute and 300 K. Find its volume at 150 kPa and 350 K, and name one model limitation.
Answer and explanation

V₂=3.00(100/150)(350/300)=2.33 L. Assume the same gas amount and ideal behavior; leakage, reaction or condensation would require another account.

V₂=3.00(100/150)(350/300)=2.33 L. Assume the same gas amount and ideal behavior; leakage, reaction or condensation would require another account.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 17

Explain the following claim and supply a valid example or counterexample: 22.4L/mol assumes about 273 K and 1atm; it is not universal at room temperature.
Answer and explanation

V=nRT/P=24.63L in the ideal model. Match R to atmosphere and liter units. 22.4L/mol assumes about 273 K and 1atm; it is not universal at room temperature.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 18

Independent chapter check: A gas has density 1.20 g/L at 300 K and 1.00 atm. Using R=0.0821 L·atm/(mol·K), estimate molar mass and explain why identity remains uncertain.
Answer and explanation

Mₘ=ρRT/P=1.20×0.0821×300/1.00≈29.6 g/mol. The estimate assumes ideal behavior and suitable composition; several substances or mixtures can have a similar average molar mass.

Mₘ=ρRT/P=1.20×0.0821×300/1.00≈29.6 g/mol. The estimate assumes ideal behavior and suitable composition; several substances or mixtures can have a similar average molar mass.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 19

Explain the following claim and supply a valid example or counterexample: Hydrogen bonding is not simply every bond containing H; ordinary boiling does not break covalent molecules apart.
Answer and explanation

All particles have dispersion attractions; larger polarizable electron clouds can produce substantial attraction. Hydrogen bonding is not simply every bond containing H; ordinary boiling does not break covalent molecules apart.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 20

Independent chapter check: A student says a nonpolar molecule has no attractions and boiling always breaks covalent bonds. Correct both claims and identify the scale of the relevant interaction.
Answer and explanation

Nonpolar molecules still experience dispersion attractions between molecules. Ordinary vaporization of a stable molecular substance mainly changes intermolecular arrangement/interactions, while breaking its internal covalent bonds is a chemical change requiring a different account.

Nonpolar molecules still experience dispersion attractions between molecules. Ordinary vaporization of a stable molecular substance mainly changes intermolecular arrangement/interactions, while breaking its internal covalent bonds is a chemical change requiring a different account.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 21

Vaporize 2 g water at its boiling point using Lv=2260 J/g.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

4520 J

2×2260.

Problem 22

What assumption lets heating time stand for added energy?
Answer and explanation

Approximately constant power with known or negligible heat loss.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 23

Independent chapter check: For 20.0 g ice already at 0°C, calculate energy to melt it and warm the liquid to 10.0°C. Use Lf=334 J/g and c=4.18 J/(g·°C), neglecting apparatus/losses.
Answer and explanation

Melting: 20.0×334=6680 J. Warming: 20.0×4.18×10.0=836 J. Total=7516 J≈7.52 kJ. The sample receives positive heat under the stated boundary.

Melting: 20.0×334=6680 J. Warming: 20.0×4.18×10.0=836 J. Total=7516 J≈7.52 kJ. The sample receives positive heat under the stated boundary.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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