Gases and Solutions MCQs – 40 Practice Questions with Answers and Explanations

Master the gas laws, the ideal gas equation and solution calculations with 40 free practice MCQs. The questions start with GCSE-level basics and build up to A-Level, IB and AP stretch topics such as partial pressures and real gases. Every answer comes with a full worked explanation, so you can see exactly where the marks come from – and where the unit traps are hiding.

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Section 1: The Gas Laws

Q1. A gas occupies a volume of 300 cm3 at a pressure of 200 kPa. It is allowed to expand to a pressure of 100 kPa at constant temperature. What is its new volume?

  • (A) 150 cm3
  • (B) 600 cm3
  • (C) 900 cm3
  • (D) 60 cm3
Show Answer

Answer: B – Boyle’s law: P1V1 = P2V2, so V2 = (200 × 300) / 100 = 600 cm3. Halving the pressure doubles the volume.

Q2. A sample of gas has a volume of 250 cm3 at 27 °C. It is heated to 127 °C at constant pressure. What is the new volume?

  • (A) 1176 cm3
  • (B) 298 cm3
  • (C) 333 cm3
  • (D) 375 cm3
Show Answer

Answer: C – Charles’s law uses kelvin: 27 °C = 300 K and 127 °C = 400 K. V2 = 250 × (400 / 300) = 333 cm3. Using the Celsius values directly (option A) is the classic trap.

Q3. A sealed can contains gas at a pressure of 100 kPa at 27 °C. The can is heated to 327 °C. What is the new pressure, assuming the volume stays constant?

  • (A) 200 kPa
  • (B) 50 kPa
  • (C) 110 kPa
  • (D) 1210 kPa
Show Answer

Answer: A – Pressure law: P / T is constant. The kelvin temperature doubles (300 K to 600 K), so the pressure doubles: 100 × (600 / 300) = 200 kPa.

Q4. Why does the pressure of a fixed mass of gas increase when it is heated in a container of constant volume?

  • (A) The particles expand and take up more space
  • (B) More gas particles are produced by the heat
  • (C) The particles move more slowly and press together
  • (D) The particles move faster and hit the walls more often and with greater force
Show Answer

Answer: D – Heating gives the particles more kinetic energy. They collide with the container walls more frequently and harder, and pressure is the force of these collisions per unit area.

Q5. At constant temperature, the volume of a gas is halved. What happens to its pressure?

  • (A) It halves
  • (B) It doubles
  • (C) It falls to one quarter
  • (D) It stays the same
Show Answer

Answer: B – Boyle’s law says pressure and volume are inversely proportional at constant temperature, so halving the volume doubles the pressure.

Q6. A gas occupies 500 cm3 at 300 K and 100 kPa. What volume will it occupy at 200 kPa and 600 K?

  • (A) 125 cm3
  • (B) 250 cm3
  • (C) 500 cm3
  • (D) 1000 cm3
Show Answer

Answer: C – Combined gas law: V2 = P1V1T2 / (T1P2) = (100 × 500 × 600) / (300 × 200) = 500 cm3. The doubling of pressure and doubling of temperature cancel out.

Q7. A student heats a gas from 25 °C to 50 °C at constant pressure and predicts that its volume will double. Which statement correctly judges this prediction?

  • (A) It is wrong: the kelvin temperature only rises from 298 K to 323 K, so the volume increases by only about 8%
  • (B) It is correct: the temperature doubled, so the volume doubles
  • (C) It is wrong: the volume will halve instead
  • (D) It is correct, but only because the pressure also doubles
Show Answer

Answer: A – Charles’s law uses the kelvin scale. Going from 298 K to 323 K is a ratio of 323 / 298 = 1.08, so the volume grows by roughly 8%, nowhere near double.

Q8. Samples of helium, oxygen and carbon dioxide are all at the same temperature. Which statement about the average kinetic energy of their particles is correct?

  • (A) It is greatest for carbon dioxide
  • (B) It is greatest for helium
  • (C) It is greatest for oxygen
  • (D) It is the same for all three gases
Show Answer

Answer: D – At a given temperature, the average kinetic energy per particle is the same for every gas. Lighter particles simply move faster on average to have the same kinetic energy.

Q9. A sealed aerosol can has an internal pressure of 150 kPa at 300 K. At what temperature would the pressure reach 200 kPa?

  • (A) 225 K
  • (B) 350 K
  • (C) 400 K
  • (D) 600 K
Show Answer

Answer: C – Pressure law: T2 = T1 × (P2 / P1) = 300 × (200 / 150) = 400 K. Option A comes from inverting the pressure ratio.

Q10. A balloon warms up in the Sun and expands, while the pressure inside stays almost equal to atmospheric pressure. What is the best kinetic theory explanation?

  • (A) The gas particles themselves get bigger when heated
  • (B) The particles move faster and push the balloon walls outwards until the pressures balance again
  • (C) Extra gas particles enter the balloon from outside
  • (D) The particles are attracted to the warmer balloon walls
Show Answer

Answer: B – Faster particles hit the walls harder, pushing them out. The balloon expands until the internal pressure falls back to match the outside pressure.

Section 2: The Ideal Gas Equation

Q11. What volume is occupied by 1.0 mol of an ideal gas at 100 kPa and 300 K? (R = 8.31 J K−1 mol−1)

  • (A) 2.49 dm3
  • (B) 0.249 dm3
  • (C) 24.9 dm3
  • (D) 249 dm3
Show Answer

Answer: C – V = nRT / P = (1.0 × 8.31 × 300) / 100 000 = 0.0249 m3 = 24.9 dm3. Remember 1 m3 = 1000 dm3.

Q12. A flask of volume 5.0 dm3 contains 0.50 mol of gas at 298 K. What is the pressure in the flask?

  • (A) 248 kPa
  • (B) 2.48 kPa
  • (C) 24.8 kPa
  • (D) 248 Pa
Show Answer

Answer: A – Convert the volume first: 5.0 dm3 = 5.0 × 10−3 m3. P = nRT / V = (0.50 × 8.31 × 298) / (5.0 × 10−3) = 248 000 Pa = 248 kPa. Option D is what you get if you forget the dm3 to m3 conversion.

Q13. A 0.88 g sample of a gas occupies 500 cm3 at 100 kPa and 300 K. What is the molar mass (Mr) of the gas?

  • (A) 22
  • (B) 44
  • (C) 58
  • (D) 88
Show Answer

Answer: B – n = PV / RT = (100 000 × 5.0 × 10−4) / (8.31 × 300) = 50 / 2493 = 0.020 mol. Mr = mass / moles = 0.88 / 0.020 = 44 (this could be CO2 or propane).

Q14. At room temperature and pressure (RTP), one mole of any gas occupies 24 dm3. What volume does 0.25 mol of a gas occupy at RTP?

  • (A) 5.6 dm3
  • (B) 96 dm3
  • (C) 0.96 dm3
  • (D) 6.0 dm3
Show Answer

Answer: D – Volume = 0.25 × 24 = 6.0 dm3. Option A uses 22.4 dm3, which is the molar volume at STP, not RTP.

Q15. A volume of 150 cm3 is equal to which volume in m3?

  • (A) 1.5 × 10−2 m3
  • (B) 1.5 × 10−4 m3
  • (C) 1.5 × 10−6 m3
  • (D) 0.15 m3
Show Answer

Answer: B – 1 m3 = 106 cm3, so 150 cm3 = 150 / 106 = 1.5 × 10−4 m3. This conversion is the most common source of errors in PV = nRT questions.

Q16. What is the density of methane (Mr = 16) at RTP, where the molar volume is 24 dm3?

  • (A) 0.67 g dm−3
  • (B) 1.5 g dm−3
  • (C) 0.067 g dm−3
  • (D) 384 g dm−3
Show Answer

Answer: A – One mole of methane has a mass of 16 g and occupies 24 dm3 at RTP, so density = 16 / 24 = 0.67 g dm−3. Option B is the inverted division.

Q17. How many moles of gas are in a 250 cm3 sample at 200 kPa and 300 K?

  • (A) 20 mol
  • (B) 0.20 mol
  • (C) 0.020 mol
  • (D) 2.0 mol
Show Answer

Answer: C – V = 250 cm3 = 2.5 × 10−4 m3. n = PV / RT = (200 000 × 2.5 × 10−4) / (8.31 × 300) = 50 / 2493 = 0.020 mol. Option A is the result of leaving the volume in cm3.

Q18. For PV = nRT to work correctly with R = 8.31 J K−1 mol−1, which units must be used for pressure and volume?

  • (A) kPa and dm3
  • (B) Pa and dm3
  • (C) atm and m3
  • (D) Pa and m3
Show Answer

Answer: D – The value R = 8.31 is in SI units, so pressure must be in pascals (Pa) and volume in cubic metres (m3). Using kPa or dm3 without converting gives answers out by factors of 1000.

Q19. What volume is occupied by 2.0 mol of gas at 300 kPa and 400 K?

  • (A) 11.1 dm3
  • (B) 22.2 dm3
  • (C) 0.0222 dm3
  • (D) 44.4 dm3
Show Answer

Answer: B – V = nRT / P = (2.0 × 8.31 × 400) / 300 000 = 0.0222 m3. Converting to dm3 (× 1000) gives 22.2 dm3. Option C forgets that final conversion.

Q20. A 2.5 dm3 flask contains 0.10 mol of gas at a pressure of 100 kPa. What is the temperature of the gas?

  • (A) 301 K
  • (B) 30 K
  • (C) 250 K
  • (D) 28 K
Show Answer

Answer: A – T = PV / nR = (100 000 × 2.5 × 10−3) / (0.10 × 8.31) = 250 / 0.831 = 301 K. Note 28 is the same temperature in °C – the equation always needs kelvin.


Section 3: Real Gases & Partial Pressures (AP / A-Level stretch)

Q21. A gas mixture has a total pressure of 300 kPa. The partial pressure of nitrogen is 180 kPa and the partial pressure of oxygen is 90 kPa. What is the partial pressure of the argon in the mixture?

  • (A) 90 kPa
  • (B) 60 kPa
  • (C) 30 kPa
  • (D) 120 kPa
Show Answer

Answer: C – Dalton’s law: the partial pressures add up to the total pressure. p(Ar) = 300 − 180 − 90 = 30 kPa.

Q22. A mixture contains 3 mol of nitrogen and 1 mol of oxygen at a total pressure of 200 kPa. What is the partial pressure of the nitrogen?

  • (A) 150 kPa
  • (B) 50 kPa
  • (C) 100 kPa
  • (D) 200 kPa
Show Answer

Answer: A – Mole fraction of N2 = 3 / 4 = 0.75. Partial pressure = mole fraction × total pressure = 0.75 × 200 = 150 kPa.

Q23. A gas mixture contains 2 mol of helium, 3 mol of neon and 5 mol of argon. What is the mole fraction of argon?

  • (A) 0.20
  • (B) 0.30
  • (C) 0.25
  • (D) 0.50
Show Answer

Answer: D – Total moles = 2 + 3 + 5 = 10 mol. Mole fraction of Ar = 5 / 10 = 0.50.

Q24. Under which conditions does a real gas behave most like an ideal gas?

  • (A) High pressure and low temperature
  • (B) Low pressure and high temperature
  • (C) Low pressure and low temperature
  • (D) High pressure and high temperature
Show Answer

Answer: B – At low pressure the molecules are far apart, so their own volume and the forces between them matter least. At high temperature they move so fast that attractions have little effect.

Q25. A real gas deviates from ideal behaviour at very high pressure mainly because…

  • (A) the attractions between molecules become stronger
  • (B) the particles slow down
  • (C) the volume of the molecules themselves is no longer negligible compared with the volume of the container
  • (D) collisions between particles become inelastic
Show Answer

Answer: C – The ideal gas model assumes particles have zero volume. When a gas is compressed, the particles’ own volume is a significant fraction of the total, so the free space is less than the measured volume and the pressure is higher than predicted.

Q26. A real gas deviates from ideal behaviour at very low temperature mainly because…

  • (A) the forces of attraction between the slow-moving molecules become significant
  • (B) the molecules swell as they cool down
  • (C) the molecules stop moving completely
  • (D) the container contracts and squeezes the gas
Show Answer

Answer: A – At low temperature the particles move slowly, so intermolecular attractions can pull them towards each other (the gas is close to condensing). This lowers the pressure below the ideal value.

Q27. At the same temperature, how many times faster does hydrogen (Mr = 2) effuse than oxygen (Mr = 32)?

  • (A) 2 times
  • (B) 16 times
  • (C) 8 times
  • (D) 4 times
Show Answer

Answer: D – Graham’s law: rate ratio = sqrt(Mr of O2 / Mr of H2) = sqrt(32 / 2) = sqrt(16) = 4. Hydrogen effuses 4 times faster.

Q28. Methane (Mr = 16) and sulfur dioxide (Mr = 64) are at the same temperature. Which statement about their diffusion rates is correct?

  • (A) Sulfur dioxide diffuses twice as fast as methane
  • (B) Methane diffuses four times as fast as sulfur dioxide
  • (C) They diffuse at the same rate
  • (D) Methane diffuses twice as fast as sulfur dioxide
Show Answer

Answer: D – Rate ratio = sqrt(64 / 16) = sqrt(4) = 2. The lighter gas (methane) diffuses faster, by a factor of 2, not 4 – remember the square root in Graham’s law.

Q29. At moderate pressures, the measured pressure of a real gas is slightly lower than the ideal gas equation predicts. Why?

  • (A) The molecules occupy space, leaving less room to move
  • (B) The molecules lose energy every time they collide
  • (C) Attractions between molecules pull them away from the walls, reducing the force of collisions with the walls
  • (D) The molecules repel each other strongly at all distances
Show Answer

Answer: C – Intermolecular attractions slow a molecule as it approaches the wall, so it hits the wall less hard. This is the effect corrected by the “a” term in the van der Waals equation (the “b” term corrects for molecular volume).

Q30. A flask contains 0.40 mol of nitrogen and 0.10 mol of oxygen at a total pressure of 150 kPa. What is the partial pressure of the oxygen?

  • (A) 30 kPa
  • (B) 120 kPa
  • (C) 15 kPa
  • (D) 37.5 kPa
Show Answer

Answer: A – Mole fraction of O2 = 0.10 / 0.50 = 0.20. Partial pressure = 0.20 × 150 = 30 kPa.

Section 4: Solutions & Concentration

Q31. A solution is made by dissolving 0.50 mol of sodium chloride in water and making the volume up to 250 cm3. What is its concentration?

  • (A) 0.0020 mol dm−3
  • (B) 0.20 mol dm−3
  • (C) 0.50 mol dm−3
  • (D) 2.0 mol dm−3
Show Answer

Answer: D – Concentration = moles / volume in dm3 = 0.50 / 0.250 = 2.0 mol dm−3. Option A comes from dividing by 250 cm3 without converting to dm3.

Q32. How many moles of hydrochloric acid are there in 500 cm3 of a 0.20 mol dm−3 solution?

  • (A) 100 mol
  • (B) 0.10 mol
  • (C) 0.40 mol
  • (D) 10 mol
Show Answer

Answer: B – Moles = concentration × volume in dm3 = 0.20 × 0.500 = 0.10 mol. Option A is the trap of using 500 cm3 directly.

Q33. What mass of sodium hydroxide (Mr = 40) is needed to make 500 cm3 of a 0.10 mol dm−3 solution?

  • (A) 0.20 g
  • (B) 20 g
  • (C) 2.0 g
  • (D) 4.0 g
Show Answer

Answer: C – Moles needed = 0.10 × 0.500 = 0.050 mol. Mass = 0.050 × 40 = 2.0 g.

Q34. A 25.0 cm3 portion of a 2.0 mol dm−3 acid is diluted with water to a total volume of 250 cm3. What is the new concentration?

  • (A) 0.20 mol dm−3
  • (B) 0.020 mol dm−3
  • (C) 20 mol dm−3
  • (D) 0.50 mol dm−3
Show Answer

Answer: A – M1V1 = M2V2, so M2 = (2.0 × 25.0) / 250 = 0.20 mol dm−3. The solution was diluted by a factor of 10, so the concentration falls by a factor of 10.

Q35. What volume of a 1.0 mol dm−3 stock solution is needed to prepare 500 cm3 of a 0.10 mol dm−3 solution?

  • (A) 5.0 cm3
  • (B) 50 cm3
  • (C) 100 cm3
  • (D) 500 cm3
Show Answer

Answer: B – Moles needed = 0.10 × 0.500 = 0.050 mol. Volume of stock = 0.050 / 1.0 = 0.050 dm3 = 50 cm3, which is then made up to 500 cm3 with water.

Q36. A water sample contains 5 mg of a dissolved substance per kilogram of solution. What is this concentration in parts per million (ppm)?

  • (A) 5 ppm
  • (B) 0.005 ppm
  • (C) 5000 ppm
  • (D) 50 ppm
Show Answer

Answer: A – One ppm is one part of solute in a million parts of solution, and 1 mg per kg is exactly 1 ppm. So 5 mg per kg = 5 ppm.

Q37. Which statement about the solubility of gases in water is correct?

  • (A) Gases become more soluble as the temperature rises
  • (B) Temperature has no effect on gas solubility
  • (C) Gases become less soluble as the temperature rises
  • (D) Gases become more soluble as the pressure falls
Show Answer

Answer: C – Gas solubility decreases as temperature increases, which is why warm water holds less dissolved oxygen for fish. (Gas solubility increases with pressure; most solid solutes are the opposite on temperature.)

Q38. What is a saturated solution?

  • (A) A solution containing the maximum mass of solute that can dissolve at that temperature, in equilibrium with undissolved solute
  • (B) A solution that can never dissolve any more solute, even on heating
  • (C) A solution containing equal masses of solute and solvent
  • (D) Another name for a concentrated solution
Show Answer

Answer: A – Saturation is temperature-dependent: the solution holds the maximum dissolved solute and is in equilibrium with any solid present. A concentrated solution is not necessarily saturated.

Q39. A student is preparing a standard solution of sodium carbonate in a volumetric flask. Which step is essential for an accurate concentration?

  • (A) Measuring the water with a measuring cylinder instead of a balance
  • (B) Rinsing the beaker and transferring all the washings into the volumetric flask
  • (C) Adding the solid directly into the flask when it is already full of water
  • (D) Filling the flask to the mark before the solid has dissolved
Show Answer

Answer: B – Every trace of the weighed solute must reach the flask. The correct method is: weigh accurately, dissolve in a beaker with a little water, transfer with all washings, then make up to the mark and invert to mix.

Q40. In a titration, 25.0 cm3 of sodium hydroxide solution is exactly neutralised by 20.0 cm3 of 0.100 mol dm−3 hydrochloric acid (they react 1:1). What is the concentration of the sodium hydroxide?

  • (A) 0.100 mol dm−3
  • (B) 0.125 mol dm−3
  • (C) 0.080 mol dm−3
  • (D) 0.0080 mol dm−3
Show Answer

Answer: C – Moles of HCl = 0.100 × (20.0 / 1000) = 0.00200 mol, so moles of NaOH = 0.00200 mol (1:1 ratio). Concentration = 0.00200 / 0.0250 = 0.080 mol dm−3.

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