Matter & Its States — Science

  1. Grade 6 — States of Matter & the Particle Model
  2. Grade 6 — Physical Properties & Physical Changes
  3. Grade 6 — Mixtures, Solutions & Separating Them
  4. Grade 8 — States of Matter & Phase Changes
  5. Grade 8 — Physical vs. Chemical Properties
  6. Grade 11 — Kinetic Molecular Theory & Phase Changes
  7. Grade 11 — The Gas Laws
  8. Grade 11 — The Ideal Gas Law

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Problem 1

A sealed syringe holds 4.0 L of gas at 1.5 atm. The plunger is pushed in until the pressure reaches 2.5 atm, with the temperature held constant. What is the new volume of the gas, in litres?
Hints

Temperature is constant and the amount of gas is fixed, so this is a Boyle's Law situation: pressure and volume trade off.

Boyle's Law: P_1V_1 = P_2V_2. Rearrange to solve for V_2.

V_2 = (P_1V_1) ÷ P_2 = (1.5 × 4.0) ÷ 2.5.

Answer and explanation

2.4 L

With temperature and amount of gas constant, use Boyle's Law: P_1V_1 = P_2V_2.

V_2 = (P_1V_1) ÷ P_2 = (1.5 atm × 4.0 L) ÷ 2.5 atm = 6.0 ÷ 2.5.

V_2 = 2.4 L.

Problem 2

A balloon contains 2.0 L of air at 27 °C. It is warmed to 127 °C at constant pressure. Assuming no gas escapes, what is the new volume of the balloon, in litres?
Hints

Gas-law temperatures must be in kelvin: K = °C + 273.

At constant pressure, Charles's Law applies: V_1 ÷ T_1 = V_2 ÷ T_2.

Convert 27 °C → 300 K and 127 °C → 400 K, then V_2 = V_1 × (T_2 ÷ T_1).

Answer and explanation

2.67 L

Convert to kelvin: T_1 = 27 + 273 = 300 K, T_2 = 127 + 273 = 400 K.

Charles's Law: V_2 = V_1 × (T_2 ÷ T_1) = 2.0 × (400 ÷ 300).

V_2 = 2.0 × 1.333… = 2.67 L.

Problem 3

A 600 mL sample of gas is at 250 K and 2.0 atm. It is warmed to 300 K while the pressure drops to 1.5 atm. What is the new volume, in millilitres?
Hints

Nothing is held constant here, so use the Combined Gas Law: (P_1V_1) ÷ T_1 = (P_2V_2) ÷ T_2.

Solve for V_2 = V_1 × (P_1 ÷ P_2) × (T_2 ÷ T_1).

V_2 = 600 × (2.0 ÷ 1.5) × (300 ÷ 250).

Answer and explanation

960 mL

Combined Gas Law, solved for V_2: V_2 = V_1 × (P_1 ÷ P_2) × (T_2 ÷ T_1).

V_2 = 600 × (2.0 ÷ 1.5) × (300 ÷ 250) = 600 × 1.333… × 1.2.

V_2 = 600 × 1.6 = 960 mL.

Problem 4

A solid plastic block has a mass of 120 g and a volume of 150 cm³. Water has a density of 1.00 g/cm³. What is the density of the block, in g/cm³ (and will it float in water)?
Hints

Density = mass ÷ volume.

Compare the block's density with water's (1.00 g/cm³): less dense floats, more dense sinks.

120 ÷ 150 = ?

Answer and explanation

0.8 g/cm³

Density = mass ÷ volume = 120 g ÷ 150 cm³.

= 0.80 g/cm³, which is less than water's 1.00 g/cm³, so the block floats.

Problem 5

A 5.6 L sample of gas is at 1.0 atm and 273 K. Using PV = nRT with R = 0.0821 L·atm/(mol·K), how many moles of gas are present?
Hints

Rearrange the ideal gas law to solve for the number of moles: n = PV ÷ (RT).

Make sure the temperature is in kelvin (it already is here).

n = (1.0 × 5.6) ÷ (0.0821 × 273).

Answer and explanation

0.25 mol

Ideal gas law, solved for moles: n = PV ÷ (RT).

n = (1.0 atm × 5.6 L) ÷ (0.0821 × 273 K) = 5.6 ÷ 22.41.

n = 0.25 mol.

Problem 6

A rigid steel cylinder of gas (fixed volume) is moved from a 20 °C room into a 60 °C oven and allowed to reach the new temperature. Using kinetic molecular theory, which prediction is correct?
  1. The pressure decreases because the warmer molecules spread out and strike the walls less often.
  2. The volume increases because the gas expands to fill more space.
  3. The pressure increases because the molecules move faster and strike the walls harder and more frequently.
  4. Nothing measurable changes because the number of gas molecules stays the same.
Hints

The container is rigid, so the volume cannot change — think about what that forces to change instead.

In kinetic molecular theory, raising the temperature raises the average speed (kinetic energy) of the molecules.

Faster molecules hit the walls harder and more often; with the volume fixed, that shows up as a change in pressure.

Answer and explanation

The pressure increases because the molecules move faster and strike the walls harder and more frequently.

The steel cylinder is rigid, so volume is fixed — rule out any answer that says the volume increases.

Heating raises the molecules' average kinetic energy, so they move faster and collide with the walls harder and more frequently.

With volume held constant, more forceful and more frequent collisions mean higher pressure, so the pressure increases (choice C).

Problem 7

A student needs to separate a dry mixture of sand, iron filings, and table salt into three pure samples. Which procedure works?
  1. Dissolve the whole mixture in water, then filter once to separate all three components.
  2. Use a magnet to pull out the iron filings; stir the rest into water and filter to collect the sand; then evaporate the water to recover the salt.
  3. Heat the mixture until the salt melts, then pour the liquid off to leave the sand and iron behind.
  4. Use a magnet to pull out the salt, then filter the sand away from the iron filings.
Hints

Each step should exploit ONE physical property: magnetism, solubility, or particle size (filtering).

Only one of the three is magnetic, and only one dissolves in water.

Sand is insoluble and stays behind on the filter; dissolved salt passes through and is left when the water evaporates.

Answer and explanation

Use a magnet to pull out the iron filings; stir the rest into water and filter to collect the sand; then evaporate the water to recover the salt.

Iron is magnetic while salt and sand are not, so a magnet removes the iron filings first.

Salt dissolves in water but sand does not, so filtering the salt-water and sand slurry catches the sand on the filter.

Evaporating the filtrate leaves pure salt behind — three properties, three clean separations, so choice B works.

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