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Understand Ideal Gases - IB Chemistry

  1. 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?
    Answer: 3.0 atm 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.
  2. A mixture of 0.50 mol of H₂(g) and 0.50 mol of N₂(g) is contained in a flask at a total pressure of 200 kPa. What is the partial pressure of the hydrogen gas, H₂(g)?
    Answer: 100 kPa The partial pressure of a gas in a mixture is proportional to its mole fraction. The total number of moles is n_total = n(H₂) + n(N₂) = 0.50 + 0.50 = 1.00 mol. The mole fraction of H₂ is X(H₂) = n(H₂) / n_total = 0.50 / 1.00 = 0.50. The partial pressure is the mole fraction times the total pressure: P(H₂) = X(H₂) × P_total = 0.50 × 200 kPa = 100 kPa.
  3. 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?
    Answer: The volume of the gas particles themselves is negligible. 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.
  4. 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?
    Answer: 2:1 ratio of Ne atoms to Ar atoms 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.
  5. 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?
    Answer: 1.48 atm after removing nitrogen gas 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⋅atm/mol⋅K)(300 K)10.0 L=0.49 atmP_{O_2} = \frac{(0.20 \text{ mol})(0.0821 \text{ L}\cdot\text{atm/mol}\cdot\text{K})(300 \text{ K})}{10.0 \text{ L}} = 0.49 \text{ atm} Ar: PAr=(0.40 mol)(0.0821 L⋅atm/mol⋅K)(300 K)10.0 L=0.98 atmP_{Ar} = \frac{(0.40 \text{ mol})(0.0821 \text{ L}\cdot\text{atm/mol}\cdot\text{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.
  6. 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)?
    Answer: 2.01 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.
  7. 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?
    Answer: The average kinetic energy of the particles is quadrupled. 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.
  8. Which change will cause the greatest increase in the pressure of an ideal gas in a sealed container?
    Answer: Halving the volume while doubling the absolute temperature. 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.
  9. An ideal gas undergoes a process where its pressure doubles while its density increases by a factor of 1.5. What happens to the absolute temperature of the gas during this process?
    Answer: Temperature increases by a factor of 1.33 When you encounter gas law problems involving changes to multiple properties, you need to combine the ideal gas law with the relationship between density and molar mass. The key insight is recognizing how density relates to the other gas variables. Start with the ideal gas law: PV=nRTPV = nRT, which can be rearranged as P=nRTVP = \frac{nRT}{V}. Since density ρ=massV=n×MV\rho = \frac{mass}{V} = \frac{n \times M}{V} (where M is molar mass), we can substitute to get: P=ρRTMP = \frac{\rho RT}{M}. For a fixed amount of gas, M is constant, so PρT=constant\frac{P}{\rho T} = constant. This means P1ρ1T1=P2ρ2T2\frac{P_1}{\rho_1 T_1} = \frac{P_2}{\rho_2 T_2}. Solving for the temperature ratio: T2T1=P2P1×ρ1ρ2\frac{T_2}{T_1} = \frac{P_2}{P_1} \times \frac{\rho_1}{\rho_2}. Given that pressure doubles (P2P1=2\frac{P_2}{P_1} = 2) and density increases by 1.5 (ρ2ρ1=1.5\frac{\rho_2}{\rho_1} = 1.5, so ρ1ρ2=11.5=0.67\frac{\rho_1}{\rho_2} = \frac{1}{1.5} = 0.67): T2T1=2×0.67=1.33\frac{T_2}{T_1} = 2 \times 0.67 = 1.33 Therefore, temperature increases by a factor of 1.33, making C correct. Option A (0.75) incorrectly suggests temperature decreases. Option B (3.0) likely comes from adding the factors instead of using the proper relationship. Option D (2.0) ignores the density change entirely, considering only pressure. Remember: when multiple gas properties change simultaneously, always establish the mathematical relationship between all variables before substituting values. Density problems often require connecting mass, volume, and the ideal gas law.
  10. 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)
    Answer: 12.4 dm³ 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.
  11. Which statement best explains why increasing the temperature of a gas in a rigid container increases its pressure?
    Answer: The molecules collide with the container walls more frequently and with greater momentum. 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).
  12. 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)
    Answer: The pressure in the neon container is greater than in the oxygen container. 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.
  13. 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?
    Answer: 132 L of helium at the higher altitude 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.
  14. 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?
    Answer: The pressure remains the same. 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).
  15. Which of the following gases would have the highest density at 273 K and 101.3 kPa?
    Answer: CO₂ 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.
  16. 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?
    Answer: 41.6 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.
  17. Under which conditions will the behavior of sulfur hexafluoride gas, SF₆(g), deviate most from the ideal gas law?
    Answer: 5000 kPa and 200 K 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.
  18. A 2.0 dm³ container holds N₂(g) at a certain temperature and pressure. An identical 2.0 dm³ container holds CO₂(g) at the same temperature and pressure. Which of the following is correct?
    Answer: The number of molecules in both containers is the same. According to Avogadro's law, equal volumes of ideal gases at the same temperature and pressure contain an equal number of molecules. Since both containers have the same volume, temperature, and pressure, they must contain the same number of moles, and therefore the same number of molecules. The molar masses of N₂ (28.02 g mol⁻¹) and CO₂ (44.01 g mol⁻¹) are different, so the masses and densities will be different. The average kinetic energy is the same (same T), but since the masses of the molecules are different, their average speeds will also be different.
  19. 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)
    Answer: The pressure in container X is equal to the pressure in container Y. 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.
  20. For a real gas, the measured pressure is often lower than the pressure predicted by the ideal gas law. What is the primary reason for this negative deviation?
    Answer: Attractive intermolecular forces reduce the force of collisions with the container walls. The ideal gas law assumes there are no intermolecular forces (IMFs) between gas particles. In a real gas, attractive IMFs exist. These forces pull the molecules toward each other, slightly reducing their speed and the force with which they collide with the container walls. This results in a lower measured pressure compared to the ideal prediction. The volume of molecules (A) causes a positive pressure deviation at very high pressures. Collisions are assumed to be elastic (B). Average kinetic energy is defined by the temperature, so it is the same (D).

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