IB Chemistry Quiz: Understand Ideal Gases
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Understand Ideal GasesQuestion 1 of 20

A gas has a density of 2.50 g dm⁻³ at 300 K and 150 kPa. What is the approximate molar mass of the gas in g mol⁻¹?

20.8
41.6
5.0
12.5
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IB Chemistry Quiz

IB Chemistry Quiz: Understand Ideal Gases

Practice Understand Ideal Gases in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Understand Ideal Gases, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A gas has a density of 2.50 g dm⁻³ at 300 K and 150 kPa. What is the approximate molar mass of the gas in g mol⁻¹?

  1. 20.8
  2. 41.6 (correct answer)
  3. 5.0
  4. 12.5
Explanation: Start with the ideal gas law PV = nRT. Substitute n = m/M and rearrange to get PM = (m/V)RT. Since density d = m/V, the equation becomes PM = dRT. Rearrange to solve for molar mass: M = dRT/P. Use consistent units. d = 2.50 g dm⁻³, R = 8.31 J K⁻¹ mol⁻¹, T = 300 K, P = 150 kPa. Convert units so they cancel: It is easiest to use P in kPa and V in dm³, so R = 8.31 J K⁻¹ mol⁻¹, which is 8.31 kPa dm³ K⁻¹ mol⁻¹. M = (2.50 g dm⁻³ × 8.31 kPa dm³ K⁻¹ mol⁻¹ × 300 K) / 150 kPa = (2.50 × 8.31 × 2) g mol⁻¹ = 41.55 g mol⁻¹. This is approximately 41.6 g mol⁻¹.

Question 2

12.16 g of magnesium reacts completely with excess hydrochloric acid in a flask connected to a gas syringe at 25 °C and 100 kPa. What volume of gas is collected? (Aᵣ(Mg) = 24.31) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

  1. 12.4 dm³ (correct answer)
  2. 11.4 dm³
  3. 6.2 dm³
  4. 24.8 dm³
Explanation: First, calculate the moles of Mg: n(Mg) = 12.16 g / 24.31 g mol⁻¹ = 0.5002 mol. From the stoichiometry, the mole ratio of Mg to H₂ is 1:1, so n(H₂) = 0.5002 mol. The conditions are 25 °C (298 K) and 100 kPa. Use the ideal gas law, V = nRT/P. V = (0.5002 mol × 8.31 J K⁻¹ mol⁻¹ × 298 K) / 100000 Pa = 0.01238 m³ = 12.38 dm³. This is approximately 12.4 dm³. Distractor B uses the STP molar volume (0.5002 x 22.7). Distractor C halves the result. Distractor D doubles the result.

Question 3

A 40.0 cm³ sample of methane, CH₄, is completely combusted in 100.0 cm³ of oxygen at constant temperature and pressure. What is the final volume of the gaseous mixture? (Assume water is a liquid with negligible volume). CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

  1. 40.0 cm³
  2. 60.0 cm³ (correct answer)
  3. 80.0 cm³
  4. 140.0 cm³
Explanation: First, determine the limiting reactant. The stoichiometric ratio is 1 mole CH₄ to 2 moles O₂. Thus, 40.0 cm³ of CH₄ requires 2 × 40.0 = 80.0 cm³ of O₂. Since we have 100.0 cm³ of O₂, oxygen is in excess and CH₄ is the limiting reactant. Volume of O₂ reacted = 80.0 cm³. Volume of O₂ remaining = 100.0 - 80.0 = 20.0 cm³. Volume of CO₂ produced is based on the limiting reactant: 40.0 cm³ CH₄ × (1 cm³ CO₂ / 1 cm³ CH₄) = 40.0 cm³ CO₂. The final volume is the sum of the remaining excess reactant and the gaseous products: V_final = V(O₂ remaining) + V(CO₂ produced) = 20.0 cm³ + 40.0 cm³ = 60.0 cm³.

Question 4

Which change will cause the greatest increase in the pressure of an ideal gas in a sealed container?

  1. Doubling the volume while keeping the temperature constant.
  2. Doubling the absolute temperature while keeping the volume constant.
  3. Halving the volume while doubling the absolute temperature. (correct answer)
  4. Halving the number of moles while doubling the volume.
Explanation: Let initial pressure be P. A) Doubling V halves P. B) Doubling T doubles P. C) Halving V doubles P, and doubling T doubles P again. The combined effect is P becoming 4P. D) Halving n halves P, and doubling V halves P again. The combined effect is P becoming P/4. Therefore, halving the volume while doubling the absolute temperature causes the greatest increase (a four-fold increase) in pressure.

Question 5

The ideal gas equation can be modified for real gases with the van der Waals equation: (P + a(n/V)²)(V - nb) = nRT. The term 'nb' corrects for the finite volume of gas particles. Which gas would be expected to have the largest value for the constant 'b'?

  1. He
  2. Ne
  3. Ar
  4. Kr (correct answer)
Explanation: The van der Waals constant 'b' is a correction for the volume occupied by the gas molecules themselves. A larger 'b' value corresponds to a larger molecular size. He, Ne, Ar, and Kr are all noble gases in Group 18. Atomic radius increases down a group. Therefore, Krypton (Kr) is the largest atom among the choices and would be expected to have the largest value for the constant 'b'.

Question 6

Two identical sealed containers, X and Y, contain ideal gases at the same temperature. Container X contains 4 g of H₂ gas. Container Y contains 32 g of CH₄ gas. Which statement is correct? (Aᵣ(H)=1.01, Aᵣ(C)=12.01)

  1. The pressure in container X is equal to the pressure in container Y. (correct answer)
  2. The pressure in container X is half the pressure in container Y.
  3. The pressure in container X is double the pressure in container Y.
  4. The pressure in container Y is four times the pressure in container X.
Explanation: First, calculate the number of moles in each container. M(H₂) = 2.02 g mol⁻¹. M(CH₄) = 16.05 g mol⁻¹. n(H₂) = 4 g / 2.02 g mol⁻¹ ≈ 2 mol. n(CH₄) = 32 g / 16.05 g mol⁻¹ ≈ 2 mol. Since both containers have the same volume (identical containers), the same temperature, and approximately the same number of moles, according to the ideal gas law (P = nRT/V), the pressure in both containers will be approximately equal.

Question 7

A rigid 10.0 dm³ cylinder contains 2.00 mol of an ideal gas at 298 K. An additional 1.00 mol of the same gas is injected into the cylinder, and the temperature is raised to 398 K. What is the ratio of the final pressure to the initial pressure (P_final / P_initial)?

  1. 2.01 (correct answer)
  2. 1.50
  3. 0.50
  4. 1.12
Explanation: Use the ideal gas law in ratio form: (P₂V₂)/(n₂T₂) = (P₁V₁)/(n₁T₁). Since the volume is constant (rigid cylinder), V₁ = V₂. The ratio P₂/P₁ = (n₂/n₁) * (T₂/T₁). The initial moles n₁ = 2.00 mol and initial temperature T₁ = 298 K. The final moles n₂ = 2.00 + 1.00 = 3.00 mol and final temperature T₂ = 398 K. The ratio is P₂/P₁ = (3.00 mol / 2.00 mol) * (398 K / 298 K) = 1.50 * 1.3356 ≈ 2.003. So the ratio is approximately 2.01.

Question 8

A weather balloon contains 50.0 L of helium at ground level (1.0 atm, 20°C). As it rises to an altitude where the pressure is 0.30 atm and temperature is -40°C, assuming the balloon material is perfectly flexible, what volume does the helium occupy?

  1. 132 L of helium at the higher altitude (correct answer)
  2. 167 L of helium at the higher altitude
  3. 125 L of helium at the higher altitude
  4. 200 L of helium at the higher altitude
Explanation: Use the combined gas law: (P₁V₁)/T₁ = (P₂V₂)/T₂. Convert temperatures to Kelvin: T₁ = 20°C + 273 = 293 K, T₂ = -40°C + 273 = 233 K. Solve for V₂: V₂ = (P₁V₁T₂)/(P₂T₁) = (1.0 atm)(50.0 L)(233 K)/[(0.30 atm)(293 K)] = 11,650/87.9 = 132.5 L ≈ 132 L. Choice B uses 273 K instead of 233 K for T₂. Choice C neglects the temperature change. Choice D uses an incorrect pressure ratio calculation.

Question 9

A mixture of ideal gases contains 0.40 mol N₂, 0.20 mol O₂, and 0.40 mol Ar in a 10.0 L container at 300 K. If only the N₂ is removed from the container while temperature and volume remain constant, what is the new total pressure?

  1. 1.97 atm after removing nitrogen gas
  2. 2.46 atm after removing nitrogen gas
  3. 0.98 atm after removing nitrogen gas
  4. 1.48 atm after removing nitrogen gas (correct answer)
Explanation: When you encounter gas mixture problems involving removal of components, you're working with partial pressures and Dalton's Law. The key insight is that removing one gas doesn't affect the partial pressures of the remaining gases. First, calculate the initial partial pressures using PV=nRTPV = nRT. For the remaining gases after N₂ removal: O₂: PO2=(0.20 mol)(0.0821 L\cdotpatm/mol\cdotpK)(300 K)10.0 L=0.49 atmP_{O_2} = \frac{(0.20 \text{ mol})(0.0821 \text{ L·atm/mol·K})(300 \text{ K})}{10.0 \text{ L}} = 0.49 \text{ atm} Ar: PAr=(0.40 mol)(0.0821 L\cdotpatm/mol\cdotpK)(300 K)10.0 L=0.98 atmP_{Ar} = \frac{(0.40 \text{ mol})(0.0821 \text{ L·atm/mol·K})(300 \text{ K})}{10.0 \text{ L}} = 0.98 \text{ atm} The new total pressure is simply: Ptotal=PO2+PAr=0.49+0.98=1.47 atmP_{total} = P_{O_2} + P_{Ar} = 0.49 + 0.98 = 1.47 \text{ atm} This rounds to 1.48 atm, confirming answer D. Looking at the incorrect options: A (1.97 atm) likely represents a calculation error where someone incorrectly added pressures or used wrong gas constants. B (2.46 atm) appears to be the original total pressure before any gas was removed, suggesting the student ignored the N₂ removal entirely. C (0.98 atm) is just the partial pressure of argon alone, indicating the student forgot to include oxygen in the final calculation. Study tip: In gas removal problems, calculate partial pressures separately for each remaining component, then add them. The removed gas simply disappears from your final sum—it doesn't affect the others' individual contributions.

Question 10

Two identical containers at the same temperature contain different ideal gases. Container X holds Ne gas at 3.0 atm, and container Y holds Ar gas at 1.5 atm. If the containers are connected and equilibrium is established, what is the ratio of Ne atoms to Ar atoms in the final mixture?

  1. The ratio depends on the molecular masses of the gases
  2. 2:1 ratio of Ne atoms to Ar atoms (correct answer)
  3. 1:2 ratio of Ne atoms to Ar atoms
  4. 1:1 ratio of Ne atoms to Ar atoms
Explanation: At constant temperature and volume, pressure is directly proportional to the number of moles (and atoms) of gas. Initially, Container X has 3.0 atm of Ne and Container Y has 1.5 atm of Ar. Since PV = nRT and T, V, and R are the same, the mole ratio equals the pressure ratio. Therefore, nₙₑ:nₐᵣ = 3.0:1.5 = 2:1. When connected, this ratio is preserved. Choice A incorrectly suggests molecular mass affects the ratio. Choice C inverts the correct ratio. Choice D ignores the pressure difference.

Question 11

A rigid container holds 2.0 mol of an ideal gas at 300 K and 4.0 atm pressure. If the temperature is increased to 450 K while maintaining constant volume, and then 1.0 mol of gas is removed at constant temperature, what is the final pressure?

  1. 4.0 atm
  2. 3.0 atm (correct answer)
  3. 6.0 atm
  4. 2.0 atm
Explanation: This requires two steps using Gay-Lussac's Law and then Avogadro's Law. First, heating at constant volume: P₂/T₂ = P₁/T₁, so P₂ = (4.0 atm)(450 K)/(300 K) = 6.0 atm. Then removing gas at constant temperature: P₃/n₃ = P₂/n₂, so P₃ = (6.0 atm)(1.0 mol)/(2.0 mol) = 3.0 atm. Choice A assumes no change occurred. Choice C represents only the heating step without accounting for gas removal. Choice D incorrectly applies the temperature ratio to the final mole ratio.

Question 12

A 1.0 dm³ sealed flask contains an ideal gas at 200 K and pressure P. The flask is heated to 400 K, and a leak allows one-quarter of the gas molecules to escape. What is the new pressure in the flask?

  1. 0.5 P
  2. 1.5 P (correct answer)
  3. 2.0 P
  4. 0.75 P
Explanation: Let the initial state be P₁, V₁, n₁, T₁ and the final state be P₂, V₂, n₂, T₂. We have T₁ = 200 K, P₁ = P, n₁ = n, and V₁ = V. The final state has T₂ = 400 K, n₂ = 0.75n (since one-quarter escaped), and V₂ = V (sealed flask). Using the ideal gas law in ratio form: (P₂V₂)/(n₂T₂) = (P₁V₁)/(n₁T₁). Since V₁ = V₂, we have P₂/P₁ = (n₂/n₁)(T₂/T₁). Substituting the values: P₂/P = (0.75n / n) * (400 K / 200 K) = 0.75 * 2 = 1.5. Therefore, the new pressure P₂ = 1.5 P.

Question 13

A sample of gas is moved from a 2.0 dm³ container to a 4.0 dm³ container. The temperature is also changed from 300 K to 600 K. What is the effect on the pressure of the gas?

  1. The pressure is quartered.
  2. The pressure is halved.
  3. The pressure remains the same. (correct answer)
  4. The pressure is doubled.
Explanation: Use the combined gas law: P₁V₁/T₁ = P₂V₂/T₂. We want to find the ratio P₂/P₁. Rearranging gives P₂/P₁ = (V₁/V₂) * (T₂/T₁). Given V₁ = 2.0 dm³, V₂ = 4.0 dm³, T₁ = 300 K, and T₂ = 600 K. So, P₂/P₁ = (2.0 / 4.0) * (600 / 300) = (0.5) * (2) = 1. The pressure ratio is 1, meaning the final pressure is the same as the initial pressure. The effect of doubling the volume (which halves the pressure) is exactly cancelled by the effect of doubling the temperature (which doubles the pressure).

Question 14

An ideal gas is contained in a flexible balloon. The balloon is submerged in a cold water bath, causing its volume to decrease by half. What must be true about the temperature change?

  1. The absolute temperature (in Kelvin) decreased by half. (correct answer)
  2. The temperature in degrees Celsius decreased by half.
  3. The absolute temperature (in Kelvin) was squared.
  4. The temperature change cannot be determined without knowing the pressure.
Explanation: A flexible balloon maintains a pressure that is approximately equal to the external pressure, which is constant. According to Charles's Law (a consequence of the ideal gas law at constant pressure and amount), volume is directly proportional to the absolute temperature (V ∝ T). If the volume decreases by half (V₂ = 0.5V₁), then the absolute temperature must also decrease by half (T₂ = 0.5T₁).

Question 15

Equal masses of oxygen (O₂) and neon (Ne) are placed in separate, identical containers at the same temperature. Assuming ideal gas behavior, which statement is correct? (Aᵣ(O) = 16.00; Aᵣ(Ne) = 20.18)

  1. The pressure in the oxygen container is greater than in the neon container.
  2. The pressure in the neon container is greater than in the oxygen container. (correct answer)
  3. The pressure in both containers is the same.
  4. The average kinetic energy of neon atoms is greater than that of oxygen molecules.
Explanation: The molar mass of O₂ is 32.00 g mol⁻¹ and Ne is 20.18 g mol⁻¹. For the same mass, there will be more moles of the substance with the lower molar mass. Thus, there are more moles of Ne than O₂. According to the ideal gas law (P = nRT/V), pressure is directly proportional to the number of moles when volume and temperature are constant. Therefore, the pressure in the neon container will be greater.

Question 16

At very high pressures, the measured pressure of a real gas is typically greater than that predicted by the ideal gas law. Which assumption of the ideal gas model is responsible for this deviation?

  1. Gas particles are in constant, random motion.
  2. The volume of the gas particles themselves is negligible. (correct answer)
  3. There are no attractive or repulsive forces between gas particles.
  4. Collisions between gas particles are perfectly elastic.
Explanation: The ideal gas law assumes that the volume of the gas particles is zero compared to the volume of the container. At very high pressures, the particles are forced close together, and their actual volume becomes a significant fraction of the container's volume. The 'free' volume available for the gas to move in is less than the total container volume. This effective reduction in volume leads to more frequent collisions and a higher pressure than predicted by the ideal gas law, which assumes the full container volume is available.

Question 17

Which of the following gases would have the highest density at 273 K and 101.3 kPa?

  1. N₂
  2. O₂
  3. F₂
  4. CO₂ (correct answer)
Explanation: According to the ideal gas equation, density (d) can be expressed as d = PM/RT. Since P, R, and T are constant for all the gases, the density is directly proportional to the molar mass (M). We need to find the gas with the highest molar mass. M(N₂) = 28.02 g mol⁻¹; M(O₂) = 32.00 g mol⁻¹; M(F₂) = 38.00 g mol⁻¹; M(CO₂) = 12.01 + 2(16.00) = 44.01 g mol⁻¹. CO₂ has the highest molar mass, so it will have the highest density.

Question 18

Under which conditions will the behavior of sulfur hexafluoride gas, SF₆(g), deviate most from the ideal gas law?

  1. 100 kPa and 500 K
  2. 5000 kPa and 500 K
  3. 5000 kPa and 200 K (correct answer)
  4. 100 kPa and 200 K
Explanation: Real gases deviate most from ideal behavior at high pressures and low temperatures. At high pressure, the volume of the gas particles becomes significant compared to the container volume. At low temperature, the kinetic energy of particles is lower, making intermolecular forces more significant. Comparing the options, 5000 kPa is the highest pressure and 200 K is the lowest temperature, so this combination causes the greatest deviation.

Question 19

Which statement best explains why increasing the temperature of a gas in a rigid container increases its pressure?

  1. The gas molecules expand and occupy more volume, pushing on the container walls.
  2. The attractive forces between molecules are overcome, allowing them to collide with walls.
  3. The molecules collide with the container walls more frequently and with greater momentum. (correct answer)
  4. The number of gas molecules increases as the temperature rises.
Explanation: Pressure is a result of the force and frequency of collisions of gas particles with the container walls. Increasing the temperature increases the average kinetic energy of the particles. This means they move faster, leading to two effects: they hit the walls more often (increased frequency) and each collision imparts a greater force (due to greater momentum). This combined effect increases the total pressure. Gas molecules do not expand (A), attractive forces are assumed negligible in an ideal gas (B), and the number of molecules is constant (D).

Question 20

The temperature of a fixed volume of an ideal gas is increased from 100 K to 400 K. Which statement accurately describes the change in the gas particles?

  1. The average speed of the particles is quadrupled.
  2. The average kinetic energy of the particles is doubled.
  3. The average kinetic energy of the particles is quadrupled. (correct answer)
  4. The frequency of collisions between particles is doubled.
Explanation: The average kinetic energy (KE_avg) of gas particles is directly proportional to the absolute temperature in Kelvin (KE_avg ∝ T). If the temperature is quadrupled from 100 K to 400 K, the average kinetic energy of the particles is also quadrupled. The average speed is proportional to the square root of the temperature (v_rms ∝ √T), so it would only be doubled. The frequency of collisions would increase, but by a factor related to the speed increase (doubled), not quadrupled.