The Ideal Gas Law — Practice · Grade 11 Science

Chapter: States of Matter & Gases

What this practice covers

By the end of The Ideal Gas Law you should be able to:

10 questions, multiple choice, numeric and written answer, each with a worked explanation.

Vocabulary used: ideal gas law, ideal gas constant (R), mole (n), STP, molar volume, Avogadro's Law, ideal gas.

Problem 1

In the ideal gas law PV = nRT, what does n represent?
  1. the mass in grams
  2. the number of collisions
  3. the number of moles of gas
  4. the number of neutrons
Answer and explanation

the number of moles of gas

n is the amount of gas measured in moles.

Problem 2

Which value of R should you use when pressure is in atm, volume in liters, and temperature in kelvin?
  1. 8.314 J/(mol·K)
  2. 22.4 L/mol
  3. 6.02 × 10²³
  4. 0.0821 L·atm/(mol·K)
Answer and explanation

0.0821 L·atm/(mol·K)

R = 0.0821 L·atm/(mol·K) is the value that matches atm, liters, and kelvin.

Problem 3

What volume does 3.0 mol of an ideal gas occupy at STP?
Answer and explanation

67.2 L

Volume = moles × 22.4 L/mol = 3.0 × 22.4 = 67.2 L.

Problem 4

How many moles of gas are present in a 11.2 L sample at STP?
Answer and explanation

0.5 mol

Moles = volume ÷ 22.4 L/mol = 11.2 ÷ 22.4 = 0.50 mol.

Problem 5

A 10.0 L container holds 2.0 mol of gas at 300 K. What is the pressure, in atm?
Answer and explanation

4.93 atm

P = nRT/V = (2.0 × 0.0821 × 300) / 10.0 = 49.26 / 10.0 ≈ 4.93 atm.

Problem 6

Explain why 1 mole of oxygen gas and 1 mole of helium gas occupy the same volume at STP, even though oxygen molecules are much heavier.
Answer and explanation

By Avogadro's Law, equal volumes at the same temperature and pressure hold equal numbers of particles, regardless of their mass. The ideal gas law depends only on the number of moles (n), not the type or weight of the gas, so one mole of any ideal gas fills 22.4 L at STP.

Problem 7

Find pressure for 1.0 mol in 10 L at 300 K using the same R. Show the reasoning, not just the conclusion.
Hints

Match the units of pressure and R.

Answer and explanation

P=nRT/V=1.0×8.31×300/10=249.3 kPa, about 250 kPa.

Problem 8

Why do real gases depart from the ideal model especially at high pressure and low temperature?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Finite particle volume and intermolecular attractions become more important under those conditions.

Problem 9

Density connects amount and mass without replacing either quantityDensity connects amount and mass without replacing either quantityn = m/MₘPV = (m/Mₘ)RTρ = m/V = PMₘ/(RT)Mₘ = ρRT/PA density comparison must keep P, T and units compatible.
Original algebraic derivation; it does not assume liquid density follows the same temperature law.
Interpret the supplied visual. If temperature increases at fixed pressure for the same ideal gas, what happens to its density?
Answer and explanation

It decreases inversely with absolute temperature because the amount per volume decreases under the model.

Substituting n=m/Mₘ into the ideal gas law gives ρ=PMₘ/(RT). Gas density therefore depends on pressure and temperature as well as chemical identity. Rearranging to Mₘ=ρRT/P can estimate a molar mass when the sample is sufficiently ideal and its composition is appropriate. A mixture’s result can represent an average rather than one pure species. State the gas conditions and uncertainty before interpreting a density as unique substance identification.

It decreases inversely with absolute temperature because the amount per volume decreases under the model.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

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.

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