States of Matter & Gases — Problem set, Grade 11

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

Review the key ideas

Kinetic Molecular Theory & Phase Changes

An ideal-gas kinetic model relates mean translational energy to absolute temperature and pressure to wall momentum transfer. Phase behavior also depends on interparticle interactions and conditions, not temperature alone.

The Gas Laws

Gas-law ratios require fixed quantities to be identified, absolute pressure and kelvin temperature. The combined ideal relation applies to a fixed gas amount within its stated physical domain.

The Ideal Gas Law

PV=nRT connects an ideal gas’s absolute pressure, volume, amount and temperature using compatible units. Molar volume depends on the stated pressure and temperature; 22.4 L/mol is an approximate 273.15 K, 1 atm value.

Intermolecular Forces

Dispersion, permanent-dipole interactions and suitable hydrogen-bonding contexts contribute to molecular behavior. Compare the full structure, size, shape and conditions rather than use one universal force-strength or melting-point ranking.

Heating Curves & Phase-Change Energy

Add the energy for each temperature change and each phase change separately.

Problem 1

State two of the assumptions the kinetic molecular theory makes about an ideal gas.

Answer and explanation

Any two: particles are in constant random motion; particle volume is negligible; there are no attractions between particles; collisions are perfectly elastic; average kinetic energy is proportional to Kelvin temperature.

KMT assumes gas particles move constantly and randomly in straight lines, take up negligible volume themselves, neither attract nor repel one another (elastic collisions), and have an average kinetic energy directly proportional to the Kelvin temperature.

Problem 2

Convert −73°C to kelvin.

Answer and explanation

200 K

K = °C + 273 = −73 + 273 = 200 K.

Problem 3

A gas at 4.0 atm occupies 2.0 L. It is allowed to expand to 8.0 L at constant temperature. Find the new pressure.

Answer and explanation

1.0 atm

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

Problem 4

A balloon holds 2.0 L of gas at 300 K. It is cooled to 150 K at constant pressure. Find the new volume.

Answer and explanation

1.0 L

Charles's Law: V₂ = V₁ × (T₂/T₁) = 2.0 × (150/300) = 1.0 L.

Problem 5

A rigid steel tank reads 3.0 atm at 250 K. It is heated to 500 K. Find the new pressure.

Answer and explanation

6.0 atm

Gay-Lussac's Law (constant volume): P₂ = P₁ × (T₂/T₁) = 3.0 × (500/250) = 6.0 atm.

Problem 6

A gas occupies 5.0 L at 2.0 atm and 400 K. It is changed to 1.0 atm and 200 K. Find the new volume.

Answer and explanation

5.0 L

Combined gas law: V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 5.0 × (2.0/1.0) × (200/400) = 5.0 × 2 × 0.5 = 5.0 L.

Problem 7

How many moles of gas are in a 2.0 L container at 1.5 atm and 300 K? (R = 0.0821 L·atm/mol·K)

Answer and explanation

0.122 mol

n = PV/RT = (1.5 × 2.0)/(0.0821 × 300) = 3.0/24.63 = 0.122 mol.

Problem 8

What volume does 2.5 mol of nitrogen gas occupy at STP?

Answer and explanation

56.0 L

Volume = moles × molar volume = 2.5 × 22.4 = 56.0 L.

Problem 9

A 3.0 L tank holds 5.0 mol of gas at 350 K. Find the pressure. (R = 0.0821 L·atm/mol·K)

Answer and explanation

47.9 atm

P = nRT/V = (5.0 × 0.0821 × 350)/3.0 = 143.675/3.0 ≈ 47.9 atm.

Problem 10

A gas has a density of 1.25 g/L at STP. Find its molar mass, and suggest what the gas might be.

Answer and explanation

28.0 g/mol — likely N₂ (or CO)

At STP, molar mass = density × molar volume = 1.25 g/L × 22.4 L/mol = 28.0 g/mol. A molar mass of 28 g/mol matches nitrogen gas, N₂ (also carbon monoxide, CO).

Problem 11

Br₂ is a liquid at room temperature while Cl₂ is a gas, even though both molecules are nonpolar. Explain why.

Answer and explanation

Br₂ has stronger London dispersion forces, giving it a higher boiling point.

Both molecules have only London dispersion forces. Br₂ is larger with more electrons and a more polarizable electron cloud, so its dispersion forces are stronger. Stronger forces raise the boiling point, making Br₂ a liquid (bp ≈ 59°C) while smaller Cl₂ remains a gas (bp ≈ −34°C).

Problem 12

Rank CH₄, PH₃, and NH₃ by increasing boiling point and explain your ranking.

Answer and explanation

CH₄ < PH₃ < NH₃

CH₄ is nonpolar, so it has only weak London dispersion forces (lowest boiling point). PH₃ is slightly polar with dipole-dipole plus dispersion forces, but phosphorus is not electronegative enough for hydrogen bonding. NH₃ has H bonded to N, so it hydrogen-bonds — the strongest force — giving it the highest boiling point. (Real values: CH₄ ≈ −161°C, PH₃ ≈ −88°C, NH₃ ≈ −33°C.)

Problem 13

Why can melting absorb energy without warming?

Answer and explanation

Energy changes particle arrangements/interactions during phase coexistence rather than mainly raising the kinetic-temperature scale.

Response criteria

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

Problem 14

Chapter review: 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.

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

A2L gas at 300 K warms to 450 K at constant P and n. Find volume.

Answer and explanation

V₂=2(450/300)=3L; the larger absolute temperature gives proportionally larger volume.

Response criteria

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

Problem 16

Chapter review: 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.

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

Find V for 1 mol at 300 K and 1atm using R=.0821.

Answer and explanation

V=nRT/P=24.63L in the ideal model. Match R to atmosphere and liter units.

Response criteria

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

Problem 18

Chapter review: 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.

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

Why can nonpolar molecules still condense?

Answer and explanation

All particles have dispersion attractions; larger polarizable electron clouds can produce substantial attraction.

Response criteria

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

Problem 20

Chapter review: 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.

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

Warm 20 g water by 10°C using c=4.18 J/g°C.

Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

836 J

20×4.18×10.

Problem 22

Does boiling break water into H and O atoms?

Answer and explanation

No. Ordinary boiling primarily overcomes intermolecular attractions; the molecules remain H₂O.

Response criteria

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

Problem 23

Chapter review: 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.

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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