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Apply Ideal Gases - IB Chemistry

  1. For which of the following gases, under the same non-ideal conditions, would the value of the expression PV/nRT be closest to 1?
    Answer: He 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.
  2. 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?
    Answer: 4P 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.
  3. 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?
    Answer: C₂H₆ 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.
  4. 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?
    Answer: 127 L because the pressure decrease dominates over temperature decrease 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.
  5. 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)
    Answer: Carbon dioxide (CO₂) 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.
  6. 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?
    Answer: Attractive forces between molecules reduce the force of their collisions with the container walls. 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.
  7. 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?
    Answer: NH₃ 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.
  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?
    Answer: High pressure, as the volume of the particles becomes a significant fraction of the container volume. 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.
  9. Which of the following gas samples has the greatest mass when measured at standard temperature and pressure (STP)?
    Answer: 10.0 dm³ of Hydrogen Chloride (HCl) According to Avogadro's law, equal volumes of any ideal gas at the same temperature and pressure contain the same number of moles. Since all four samples have a volume of 10.0 dm³ at STP, they all contain an equal number of moles. Mass is calculated as mass = moles × molar mass. Since the number of moles is the same for all samples, the sample with the greatest mass will be the one with the highest molar mass (Mᵣ). Comparing the molar masses: Ne (20.18), C₂H₄ (28.06), N₂ (28.02), and HCl (36.46). Hydrogen chloride has the highest molar mass, and therefore the greatest mass.
  10. 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?
    Answer: 8730 dm³ 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.
  11. 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?
    Answer: 207 °C 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.
  12. The behavior of SO₂ gas deviates more from the ideal gas model than CH₄ gas at the same temperature and pressure. Which is the best explanation for this observation?
    Answer: SO₂ is a polar molecule with stronger intermolecular forces than the nonpolar CH₄ molecule. Deviation from ideal behavior is primarily caused by intermolecular forces (IMFs) and finite molecular volume. SO₂ is a polar molecule (bent geometry), so it exhibits dipole-dipole forces in addition to London dispersion forces. CH₄ is nonpolar (tetrahedral geometry) and only has London dispersion forces. The IMFs in SO₂ are significantly stronger than in CH₄, causing it to deviate more from ideality. Choice A is incorrect; at the same temperature, all gases have the same average kinetic energy. Choice C refers to intramolecular bonds, which do not determine gas behavior. Choice D is factually incorrect; SO₂ is a larger molecule than CH₄.
  13. 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?
    Answer: 248 kPa 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.
  14. What is the density of argon gas (Ar) in g dm⁻³ at 25 °C and 150 kPa? (Aᵣ(Ar) = 39.95)
    Answer: 2.42 g dm⁻³ The ideal gas law can be rearranged to solve for density (ρ = m/V). Starting with PV = (m/M)RT, we get ρ = PM/RT. Ensure units are consistent. T = 25 + 273.15 = 298.15 K. P = 150 kPa. M = 39.95 g mol⁻¹. R = 8.31 kPa dm³ K⁻¹ mol⁻¹. ρ = (150 kPa × 39.95 g mol⁻¹) / (8.31 kPa dm³ K⁻¹ mol⁻¹ × 298.15 K) ≈ 2.42 g dm⁻³. Distractor A results from inverting the formula to ρ = RT/PM. Distractor C results from using the temperature in Celsius (25 °C). Distractor D results from a pressure unit error, likely using 150 Pa instead of 150 kPa.
  15. Which of the following gas samples contains the greatest number of atoms? (Aᵣ: H=1.01, C=12.01, O=16.00, Cl=35.45)
    Answer: 8.03 g of CH₄ First, calculate the moles of molecules for each sample. n=m/M. A: 1.01 g / 2.02 g mol⁻¹ = 0.5 mol H₂. B: 8.03 g / 16.05 g mol⁻¹ = 0.5 mol CH₄. C: 22.01 g / 44.01 g mol⁻¹ = 0.5 mol CO₂. D: 35.45 g / 70.90 g mol⁻¹ = 0.5 mol Cl₂. All samples contain 0.5 moles of molecules. The question asks for the total number of atoms. A: 0.5 mol × 2 atoms/molecule = 1.0 mol atoms. B: 0.5 mol × 5 atoms/molecule = 2.5 mol atoms. C: 0.5 mol × 3 atoms/molecule = 1.5 mol atoms. D: 0.5 mol × 2 atoms/molecule = 1.0 mol atoms. Therefore, the sample of CH₄ contains the greatest number of atoms.
  16. Which statement correctly describes a key assumption of the ideal gas model?
    Answer: Attractive and repulsive forces between gas particles are considered to be negligible. 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.
  17. 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?
    Answer: 72.0 g mol⁻¹ 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.
  18. 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?
    Answer: ClO₂ 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.
  19. 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?
    Answer: 25% 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%.
  20. 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?
    Answer: 2.7 atm because the temperature increase dominates over molecular removal 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.

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