IB Chemistry Quiz: Apply Ideal Gases
20 questions · exam conditions
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Apply Ideal GasesQuestion 1 of 20

A rigid 5.00 dm³ container holds 0.100 mol of N₂ gas and 0.400 mol of Ar gas at 298 K. Assuming ideal behavior, what is the total pressure inside the container?

49.5 kPa
248 kPa
19.8 kPa
198 kPa
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IB Chemistry Quiz

IB Chemistry Quiz: Apply Ideal Gases

Practice Apply 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.

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This quiz focuses on Apply 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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Question 1

A rigid 5.00 dm³ container holds 0.100 mol of N₂ gas and 0.400 mol of Ar gas at 298 K. Assuming ideal behavior, what is the total pressure inside the container?

  1. 49.5 kPa
  2. 248 kPa (correct answer)
  3. 19.8 kPa
  4. 198 kPa
Explanation: The total pressure depends on the total number of moles of gas. The total moles are n_total = n(N₂) + n(Ar) = 0.100 mol + 0.400 mol = 0.500 mol. Using the ideal gas law, P = nRT/V. P = (0.500 mol × 8.31 kPa dm³ K⁻¹ mol⁻¹ × 298 K) / 5.00 dm³ ≈ 248 kPa. Distractors A and D are the partial pressures of N₂ and Ar, respectively. Distractor C results from accidentally multiplying the moles instead of adding them.

Question 2

Which statement correctly describes a key assumption of the ideal gas model?

  1. The average kinetic energy of gas particles is directly proportional to the temperature in degrees Celsius.
  2. Collisions between gas particles and with the container walls are inelastic, resulting in energy loss.
  3. Attractive and repulsive forces between gas particles are considered to be negligible. (correct answer)
  4. The volume of the gas particles is significant compared to the total volume of the container.
Explanation: The ideal gas model is based on several assumptions. One key assumption is that there are no intermolecular forces (attractive or repulsive) between the gas particles. This is stated in choice C. Choice A is incorrect because kinetic energy is proportional to the absolute temperature in Kelvin, not Celsius. Choice B is incorrect because collisions are assumed to be perfectly elastic. Choice D is incorrect because the volume of the particles themselves is assumed to be negligible compared to the container volume.

Question 3

A 1.50 g sample of a gaseous oxide of chlorine occupies 0.579 dm³ at 298 K and 95.0 kPa. What is the molecular formula of the oxide?

  1. Cl₂O
  2. ClO₂ (correct answer)
  3. Cl₂O₅
  4. Cl₂O₇
Explanation: First, calculate the number of moles (n) using the ideal gas law, n = PV/RT. n = (95.0 kPa × 0.579 dm³) / (8.31 kPa dm³ K⁻¹ mol⁻¹ × 298 K) ≈ 0.0222 mol. Next, calculate the molar mass (M) using M = m/n. M = 1.50 g / 0.0222 mol ≈ 67.5 g mol⁻¹. Finally, compare this experimental molar mass to the molar masses of the options: Cl₂O (86.9 g mol⁻¹), ClO₂ (67.5 g mol⁻¹), Cl₂O₅ (150.9 g mol⁻¹), Cl₂O₇ (182.9 g mol⁻¹). The calculated molar mass matches that of ClO₂.

Question 4

Which of the following gases would show the greatest deviation from ideal behavior when all are subjected to the same conditions of high pressure and low temperature?

  1. H₂
  2. CH₄
  3. Ne
  4. NH₃ (correct answer)
Explanation: Deviation from ideal behavior is most significant for gases with strong intermolecular forces (IMFs) and large molecular volumes. Ammonia (NH₃) is a polar molecule capable of hydrogen bonding, which is a much stronger type of IMF than the London dispersion forces present in the nonpolar H₂, CH₄, and Ne. These strong attractive forces in NH₃ cause the most significant deviation from the ideal gas model's assumption of no IMFs.

Question 5

A 0.586 g sample of a gaseous compound occupies a volume of 250 cm³ at 100 °C and 1.01 × 10⁵ Pa. What is the molar mass of the compound in g mol⁻¹?

  1. 19.3 g mol⁻¹
  2. 53.3 g mol⁻¹
  3. 7.20 g mol⁻¹
  4. 72.0 g mol⁻¹ (correct answer)
Explanation: First, convert all units to be consistent with the gas constant R (8.31 J K⁻¹ mol⁻¹). V = 250 cm³ = 2.50 × 10⁻⁴ m³. T = 100 + 273 = 373 K. P = 1.01 × 10⁵ Pa. Rearrange the ideal gas law PV = nRT to solve for moles: n = PV/RT. n = (1.01 × 10⁵ Pa × 2.50 × 10⁻⁴ m³) / (8.31 J K⁻¹ mol⁻¹ × 373 K) ≈ 0.00814 mol. Then, molar mass M = m/n = 0.586 g / 0.00814 mol ≈ 72.0 g mol⁻¹. Distractor A results from using T in Celsius. Distractor B results from incorrectly applying the molar volume at STP. Distractor C results from a decimal place error in the volume conversion.

Question 6

20 cm³ of a gaseous hydrocarbon was completely combusted with 100 cm³ of oxygen. The final volume of the gaseous mixture was 70 cm³ after cooling to the initial temperature. After treatment with aqueous potassium hydroxide, the volume decreased to 30 cm³. All volumes were measured at the same temperature and pressure. What is the molecular formula of the hydrocarbon?

  1. C₂H₄
  2. C₃H₈
  3. C₃H₆
  4. C₂H₆ (correct answer)
Explanation: According to Avogadro's law, volume ratios are equal to mole ratios for gases at constant T and P. The volume decrease upon adding KOH (which absorbs acidic CO₂) is the volume of CO₂ produced: V(CO₂) = 70 cm³ - 30 cm³ = 40 cm³. The remaining 30 cm³ is excess O₂. The volume of O₂ reacted is the initial volume minus the excess: V(O₂) = 100 cm³ - 30 cm³ = 70 cm³. The volume ratio of hydrocarbon : O₂ : CO₂ is 20:70:40, which simplifies to 1:3.5:2. From the general combustion equation CₓHᵧ + (x+y/4)O₂ → xCO₂ + (y/2)H₂O, we can deduce that x=2 and x+y/4=3.5. Substituting x=2 gives 2+y/4=3.5, so y/4=1.5, and y=6. The formula is C₂H₆.

Question 7

A fixed mass of an ideal gas exerts a pressure P when its volume is V and its absolute temperature is T. What will the pressure be if the volume is changed to V/2 and the absolute temperature is changed to 2T?

  1. P/4
  2. P
  3. 2P
  4. 4P (correct answer)
Explanation: Using the combined gas law, (P₁V₁)/T₁ = (P₂V₂)/T₂. We are given P₁=P, V₁=V, T₁=T, V₂=V/2, T₂=2T. We need to find P₂. (P × V) / T = (P₂ × (V/2)) / (2T). The V and T terms cancel out: P = (P₂/2)/2 = P₂/4. Rearranging gives P₂ = 4P. Alternatively, conceptually: halving the volume doubles the pressure (inverse relationship), and doubling the absolute temperature also doubles the pressure (direct relationship). The total effect is a multiplication by 2 × 2 = 4.

Question 8

Real gases deviate from the ideal gas law because its assumptions are not perfectly true. One assumption is that gas particles have negligible volume. Under which condition does this specific assumption cause the most significant deviation from ideal behavior?

  1. High pressure, as the volume of the particles becomes a significant fraction of the container volume. (correct answer)
  2. Low pressure, as the volume of the container is maximized.
  3. Low temperature, as this allows intermolecular forces to become dominant.
  4. High temperature, as the effective volume of particles increases with kinetic energy.
Explanation: The assumption of negligible particle volume holds well when the container volume is large compared to the total volume of the particles. At high pressure, the gas is compressed, and the container volume is small. In this situation, the volume occupied by the particles themselves is no longer negligible in comparison to the container's volume. This causes a significant deviation from ideal behavior. Low temperature primarily increases the effect of intermolecular forces (choice C). High temperature and low pressure (choice B) are the conditions where gases behave most ideally.

Question 9

A gas occupies a volume of 2.50 dm³ at 27 °C in a cylinder with a frictionless piston. The gas is heated until its volume is 4.00 dm³. Assuming the pressure remains constant, what is the final temperature of the gas in degrees Celsius?

  1. 43.2 °C
  2. 480 °C
  3. 207 °C (correct answer)
  4. 16.9 °C
Explanation: Since pressure and amount of gas are constant, Charles's Law (V₁/T₁ = V₂/T₂) applies. The temperature must be in Kelvin. T₁ = 27 °C + 273 = 300 K. Rearranging the formula to find T₂: T₂ = (V₂ × T₁) / V₁ = (4.00 dm³ × 300 K) / 2.50 dm³ = 480 K. The question asks for the answer in degrees Celsius, so convert back: T₂ = 480 K - 273 = 207 °C. Distractor A results from performing the calculation using Celsius temperatures. Distractor B is the correct temperature in Kelvin, but the question asks for Celsius. Distractor D results from an incorrect setup of the proportion.

Question 10

A rigid 10.0 dm³ container holds an ideal gas at 200 kPa. A valve is opened, and some gas escapes until the pressure drops to 150 kPa. The temperature remains constant throughout the process. What percentage of the original gas molecules escaped?

  1. 75%
  2. 50%
  3. 33%
  4. 25% (correct answer)
Explanation: From the ideal gas law, PV = nRT. Since the volume (V), gas constant (R), and temperature (T) are all constant, the pressure (P) is directly proportional to the number of moles (n). The fraction of gas remaining is equal to the fraction of the pressure remaining: n_final / n_initial = P_final / P_initial = 150 kPa / 200 kPa = 0.75. This means 75% of the gas remains in the container. The percentage that escaped is 100% - 75% = 25%.

Question 11

A flexible balloon contains a fixed mass of an ideal gas at room temperature. The balloon is then submerged in a container of liquid nitrogen (-196 °C). Which row correctly describes the changes to the gas inside the balloon?

  1. Average kinetic energy decreases; Volume decreases (correct answer)
  2. Average kinetic energy decreases; Volume remains constant
  3. Average kinetic energy remains constant; Volume decreases
  4. Average kinetic energy increases; Volume decreases
Explanation: The temperature of the gas decreases significantly when submerged in liquid nitrogen. Since the average kinetic energy of gas particles is directly proportional to the absolute temperature, the average kinetic energy must decrease. The balloon is flexible, which means the internal pressure of the gas will tend to equalize with the constant external atmospheric pressure. According to Charles's Law (V ∝ T at constant P and n), a decrease in temperature will lead to a decrease in volume. Therefore, both average kinetic energy and volume decrease.

Question 12

10.0 g of pure calcium carbonate is completely decomposed by heating. The carbon dioxide gas produced is collected at 50 °C and 95.0 kPa. What is the approximate volume of gas collected? (Mr(CaCO₃) = 100.09)

  1. 2.27 dm³
  2. 0.44 dm³
  3. 2.83 dm³ (correct answer)
  4. 2.61 dm³
Explanation: First, calculate the moles of CaCO₃: n = m / M = 10.0 g / 100.09 g mol⁻¹ ≈ 0.0999 mol. The stoichiometry of the reaction CaCO₃(s) → CaO(s) + CO₂(g) is 1:1, so n(CO₂) ≈ 0.0999 mol. The temperature must be in Kelvin: T = 50 + 273.15 = 323.15 K. Using the ideal gas law, V = nRT/P: V = (0.0999 mol × 8.31 kPa dm³ K⁻¹ mol⁻¹ × 323.15 K) / 95.0 kPa ≈ 2.83 dm³. Distractor A results from incorrectly using the molar volume at STP (0.0999 mol × 22.7 dm³ mol⁻¹). Distractor B results from using the temperature in Celsius instead of Kelvin. Distractor D results from using standard temperature (298 K) instead of the given temperature.

Question 13

The measured pressure of a real gas is often lower than the pressure predicted by the ideal gas law for the same conditions of volume, temperature, and amount. What is the primary reason for this negative deviation in pressure?

  1. The volume of the gas molecules reduces the effective volume of the container.
  2. The average speed of real gas molecules is lower than that of ideal gas molecules at the same temperature.
  3. Attractive forces between molecules reduce the force of their collisions with the container walls. (correct answer)
  4. Collisions between real gas molecules are inelastic, leading to a continuous loss of system energy.
Explanation: Pressure is caused by the collisions of gas molecules with the walls of the container. The ideal gas law assumes there are no intermolecular forces. In a real gas, molecules exert attractive forces on each other. A molecule about to strike a wall is pulled back by its neighbors, which lessens the force of its impact. This reduction in collision force results in a lower measured pressure compared to the ideal prediction. Choice A describes the effect of molecular volume, which tends to increase the pressure relative to the ideal model. Choice B is incorrect as temperature defines the average kinetic energy. Choice D is a factor in real gas behavior but C is the direct cause of the negative pressure deviation.

Question 14

A sample of methane gas occupies 5.6 L at STP. When this sample is transferred to a 3.2 L container and heated to 127°C, what pressure must be applied to maintain the gas in the new container?

  1. 2.04 atm because the pressure increases due to both volume decrease and temperature increase
  2. 1.75 atm because the temperature effect partially compensates for volume reduction
  3. 2.57 atm because the combined gas law requires proportional pressure adjustment (correct answer)
  4. 1.46 atm because methane deviates from ideal behavior at higher temperatures
Explanation: Using combined gas law: P2=P1×V1V2×T2T1=1.00×5.63.2×400273=2.57 atmP_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 1.00 \times \frac{5.6}{3.2} \times \frac{400}{273} = 2.57 \text{ atm}. Choice A uses incorrect temperature conversion. Choice B underestimates the combined effect. Choice D incorrectly invokes non-ideal behavior for this moderate pressure.

Question 15

A weather balloon contains 50.0 L of helium at ground level (1.00 atm, 20°C). As it rises to an altitude where the pressure is 0.30 atm and temperature is -40°C, what volume will the balloon occupy if it remains flexible?

  1. 127 L because the pressure decrease dominates over temperature decrease (correct answer)
  2. 95.2 L because both pressure and temperature changes reduce the volume expansion
  3. 166 L because the combined gas law requires proportional volume adjustment
  4. 78.4 L because the temperature decrease partially compensates for pressure drop
Explanation: Using combined gas law: V2=V1×P1P2×T2T1=50.0×1.000.30×233293=50.0×3.33×0.795=132 LV_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 50.0 \times \frac{1.00}{0.30} \times \frac{233}{293} = 50.0 \times 3.33 \times 0.795 = 132 \text{ L}. The closest answer is 127 L (choice A). Choice B underestimates the pressure effect. Choice C overestimates the combined effect. Choice D incorrectly emphasizes temperature compensation.

Question 16

At 300 K, a gas mixture in a 10.0 L container exerts a pressure of 2.4 atm. If 25% of the gas molecules are removed and the temperature is simultaneously increased to 450 K, what is the final pressure?

  1. 1.8 atm because the removal of molecules decreases pressure despite heating
  2. 2.7 atm because the temperature increase dominates over molecular removal (correct answer)
  3. 3.6 atm because both temperature and concentration changes increase pressure
  4. 1.2 atm because molecular removal has greater impact than temperature increase
Explanation: Initial: P1=2.4P_1 = 2.4 atm, n1=nn_1 = n, T1=300T_1 = 300 K. Final: n2=0.75nn_2 = 0.75n, T2=450T_2 = 450 K. Using P1n1T1=P2n2T2\frac{P_1}{n_1 T_1} = \frac{P_2}{n_2 T_2}: P2=P1×n2n1×T2T1=2.4×0.75×450300=2.4×0.75×1.5=2.7P_2 = P_1 \times \frac{n_2}{n_1} \times \frac{T_2}{T_1} = 2.4 \times 0.75 \times \frac{450}{300} = 2.4 \times 0.75 \times 1.5 = 2.7 atm. Choice A underestimates temperature effect. Choice C ignores molecular removal. Choice D overestimates removal impact.

Question 17

A sealed container initially holds 3.0 L of an ideal gas at 2.0 atm and 273 K. The container expands to 4.5 L while the pressure decreases to 1.2 atm. Which statement best explains the observed changes?

  1. The temperature decreased to 245 K because the gas underwent isothermal expansion followed by isobaric cooling
  2. The temperature increased to 327 K because both volume expansion and pressure reduction require energy input
  3. The temperature decreased to 245 K because the combined effect of volume increase and pressure decrease results in cooling (correct answer)
  4. The temperature remained constant at 273 K because the pressure and volume changes exactly compensated
Explanation: Using the combined gas law: P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}, so T2=P2V2T1P1V1=1.2×4.5×2732.0×3.0=245 KT_2 = \frac{P_2V_2T_1}{P_1V_1} = \frac{1.2 \times 4.5 \times 273}{2.0 \times 3.0} = 245 \text{ K}. Choice A gives correct temperature but wrong mechanism. Choice B incorrectly predicts heating. Choice D incorrectly assumes compensation.

Question 18

The density of an unknown gas is measured to be 1.66 g dm⁻³ at a temperature of 50 °C and a pressure of 1.01 × 10⁵ Pa. What is the likely identity of the gas? (Aᵣ: C=12.01, O=16.00, S=32.07, Ar=39.95)

  1. Carbon dioxide (CO₂) (correct answer)
  2. Sulfur dioxide (SO₂)
  3. Argon (Ar)
  4. Oxygen (O₂)
Explanation: The ideal gas equation can be rearranged in terms of density (ρ) as M = ρRT/P. First, ensure units are consistent. T = 50 + 273 = 323 K. P = 1.01 × 10⁵ Pa = 101 kPa. ρ = 1.66 g dm⁻³. R = 8.31 kPa dm³ K⁻¹ mol⁻¹. M = (1.66 g dm⁻³ × 8.31 kPa dm³ K⁻¹ mol⁻¹ × 323 K) / 101 kPa ≈ 44.1 g mol⁻¹. Now, compare this molar mass to the options: O₂ (32.00 g mol⁻¹), SO₂ (64.07 g mol⁻¹), Ar (39.95 g mol⁻¹), CO₂ (44.01 g mol⁻¹). The calculated molar mass corresponds to carbon dioxide.

Question 19

A weather balloon contains 2500 dm³ of helium at 293 K and 101 kPa. What volume, in dm³, will it occupy at an altitude where the temperature is 253 K and the pressure is 25.0 kPa?

  1. 8730 dm³ (correct answer)
  2. 11700 dm³
  3. 536 dm³
  4. 10100 dm³
Explanation: This problem uses the combined gas law, (P₁V₁)/T₁ = (P₂V₂)/T₂. Rearranging to solve for V₂ gives V₂ = (P₁V₁T₂)/(T₁P₂). Plugging in the values: V₂ = (101 kPa × 2500 dm³ × 253 K) / (293 K × 25.0 kPa) ≈ 8730 dm³. Distractor B results from inverting the temperature ratio (T₁/T₂). Distractor C results from inverting the pressure ratio (P₂/P₁). Distractor D results from ignoring the temperature change.

Question 20

For which of the following gases, under the same non-ideal conditions, would the value of the expression PV/nRT be closest to 1?

  1. CO₂
  2. HCl
  3. He (correct answer)
  4. CH₄
Explanation: The expression PV/nRT equals 1 for a perfectly ideal gas. A real gas will have a value closest to 1 if it behaves most ideally. Ideal behavior is favored by weak intermolecular forces and small particle size. Helium (He) is a small atom with the weakest London dispersion forces among the choices. CH₄ is nonpolar but larger. HCl is polar and has dipole-dipole forces. CO₂ is nonpolar but has a significantly higher molar mass than He, resulting in stronger dispersion forces. Therefore, He is the most ideal gas in the list.