A Level Physics 15.2 Equation of State MCQs – 30 Practice Questions with Answers (Cambridge 9702)

This page covers Section 15.2, Equation of State, from the Cambridge International AS & A Level Physics (9702) syllabus – an A Level (A2) topic in the Ideal Gases unit. Practise the ideal gas equation pV = nRT, the combined gas equation, and using the equation of state to find amounts, molar masses and densities, then check each worked answer as you go.

Formula recap (see the A Level Physics Formula Sheet for the full list):

  • pV = nRT, where n is the number of moles and R = 8.31 J mol−1 K−1
  • pV = NkT, where N is the total number of molecules and k = 1.38 × 10−23 J K−1
  • k = R / NA and N = nNA, with NA = 6.02 × 1023 mol−1
  • For a fixed mass of gas: p1V1 / T1 = p2V2 / T2
  • ρ = pM / (RT) for the density of a gas of molar mass M
  • T must always be in kelvin: T / K = θ / °C + 273

The Ideal Gas Equation

Q1. Which equation correctly relates the pressure p, volume V, amount n (in moles) and thermodynamic temperature T of an ideal gas?

  • (A) pV = nRT
  • (B) pV = nR / T
  • (C) p / V = nRT
  • (D) pV = RT / n
Show Answer

Answer: A – The equation of state of an ideal gas is pV = nRT, with p in Pa, V in m3 and T in kelvin.

Q2. A fixed cylinder contains 2.0 mol of an ideal gas in a volume of 0.050 m3 at a temperature of 300 K. What is the pressure of the gas?

  • (A) 5.0 × 104 Pa
  • (B) 1.0 × 105 Pa
  • (C) 2.0 × 105 Pa
  • (D) 1.0 × 104 Pa
Show Answer

Answer: B – p = nRT / V = (2.0 × 8.31 × 300) / 0.050 = 4986 / 0.050 = 99 720 Pa ≈ 1.0 × 105 Pa.

Q3. What volume is occupied by 1.0 mol of an ideal gas at a pressure of 1.0 × 105 Pa and a temperature of 273 K?

  • (A) 2.27 m3
  • (B) 0.227 m3
  • (C) 0.0227 m3
  • (D) 2.27 × 10−4 m3
Show Answer

Answer: C – V = nRT / p = (1.0 × 8.31 × 273) / (1.0 × 105) = 2269 / 1.0 × 105 = 0.0227 m3 (about 23 litres, the molar volume at this temperature and pressure).

Q4. An ideal gas of amount 2.0 mol is held at a pressure of 2.0 × 105 Pa in a container of volume 0.025 m3. The temperature of the gas is approximately

  • (A) 150 K
  • (B) 600 K
  • (C) 75 K
  • (D) 300 K
Show Answer

Answer: D – T = pV / (nR) = (2.0 × 105 × 0.025) / (2.0 × 8.31) = 5000 / 16.62 = 301 K ≈ 300 K.

Q5. A container of volume 0.020 m3 holds 0.50 mol of an ideal gas at a temperature of 27 °C. What is the pressure of the gas?

  • (A) 6.2 × 104 Pa
  • (B) 5.6 × 103 Pa
  • (C) 6.9 × 104 Pa
  • (D) 1.2 × 105 Pa
Show Answer

Answer: A – First convert the temperature: T = 27 + 273 = 300 K. Then p = nRT / V = (0.50 × 8.31 × 300) / 0.020 = 1246.5 / 0.020 = 62 325 Pa ≈ 6.2 × 104 Pa. Using 27 K instead of 300 K gives the much smaller (wrong) value in option B.

Q6. A sample of 0.020 mol of an ideal gas occupies a volume of 500 cm3 at 300 K. What is the pressure of the gas?

  • (A) 1.0 × 10−1 Pa
  • (B) 1.0 × 105 Pa
  • (C) 1.0 × 102 Pa
  • (D) 5.0 × 104 Pa
Show Answer

Answer: B – Convert the volume: 500 cm3 = 500 × 10−6 m3 = 5.0 × 10−4 m3. Then p = nRT / V = (0.020 × 8.31 × 300) / (5.0 × 10−4) = 49.86 / 5.0 × 10−4 = 99 720 Pa ≈ 1.0 × 105 Pa. Option A comes from forgetting the cm3 to m3 conversion.

Q7. A container of volume 0.0125 m3 contains 3.01 × 1023 molecules of an ideal gas at 300 K. Using pV = NkT, the pressure of the gas is approximately

  • (A) 1.0 × 104 Pa
  • (B) 5.0 × 104 Pa
  • (C) 1.0 × 105 Pa
  • (D) 2.0 × 105 Pa
Show Answer

Answer: C – p = NkT / V = (3.01 × 1023 × 1.38 × 10−23 × 300) / 0.0125 = 1246 / 0.0125 = 99 700 Pa ≈ 1.0 × 105 Pa. Note that 3.01 × 1023 molecules is 0.50 mol, so nRT gives the same result.

Q8. The Boltzmann constant k is given by k = R / NA. Its value and unit are

  • (A) 8.31 J mol−1 K−1
  • (B) 5.0 × 1046 J K−1
  • (C) 3.3 × 10−24 J K−1
  • (D) 1.38 × 10−23 J K−1
Show Answer

Answer: D – k = R / NA = 8.31 / (6.02 × 1023) = 1.38 × 10−23 J K−1. It is the gas constant per molecule rather than per mole.

Q9. At the level of the equation of state, an ideal gas is one whose molecules

  • (A) have a large volume and attract each other strongly
  • (B) behave as point particles and exert no forces on each other except during collisions
  • (C) attract each other strongly at all separations
  • (D) lose energy in every collision with the container walls
Show Answer

Answer: B – The ideal gas model assumes molecules of negligible volume (point particles) with no intermolecular forces except during collisions, which are elastic. The full kinetic theory assumptions are developed in Section 15.3.

Q10. A gas bottle of volume 25 litres contains 1.0 mol of an ideal gas at 300 K. What is the pressure of the gas?

  • (A) 1.0 × 102 Pa
  • (B) 1.0 × 103 Pa
  • (C) 1.0 × 105 Pa
  • (D) 2.5 × 105 Pa
Show Answer

Answer: C – Convert the volume: 25 litres = 25 × 10−3 m3 = 0.025 m3. Then p = nRT / V = (1.0 × 8.31 × 300) / 0.025 = 2493 / 0.025 = 99 720 Pa ≈ 1.0 × 105 Pa. Using V = 25 m3 gives the incorrect option A.

Gas Changes & the Combined Equation

Q11. For a fixed mass of an ideal gas, which combination of p, V and T remains constant whatever change the gas undergoes?

  • (A) pV / T
  • (B) pT / V
  • (C) pVT
  • (D) V / (pT)
Show Answer

Answer: A – From pV = nRT with n fixed, pV / T = nR = constant, so p1V1 / T1 = p2V2 / T2. Boyle’s law, Charles’s law and the pressure law are each special cases of this combined equation.

Q12. A fixed mass of gas at constant temperature has a pressure of 1.0 × 105 Pa and a volume of 0.60 m3. It is compressed to a volume of 0.20 m3. Its new pressure is

  • (A) 3.3 × 104 Pa
  • (B) 3.0 × 105 Pa
  • (C) 1.0 × 105 Pa
  • (D) 9.0 × 105 Pa
Show Answer

Answer: B – Boyle’s law (constant T): p2 = p1V1 / V2 = (1.0 × 105 × 0.60) / 0.20 = 3.0 × 105 Pa. Reducing the volume to one third triples the pressure.

Q13. A fixed mass of gas at constant temperature has a volume of 0.040 m3 at a pressure of 1.5 × 105 Pa. When the pressure is increased to 3.0 × 105 Pa, the volume becomes

  • (A) 0.080 m3
  • (B) 0.010 m3
  • (C) 0.020 m3
  • (D) 0.060 m3
Show Answer

Answer: C – V2 = p1V1 / p2 = (1.5 × 105 × 0.040) / (3.0 × 105) = 0.020 m3. Doubling the pressure halves the volume.

Q14. Gas trapped in a sealed, rigid container has a pressure of 1.2 × 105 Pa at 300 K. The container is heated until the gas temperature is 600 K. The new pressure is

  • (A) 6.0 × 104 Pa
  • (B) 1.2 × 105 Pa
  • (C) 4.8 × 105 Pa
  • (D) 2.4 × 105 Pa
Show Answer

Answer: D – Pressure law (constant V): p2 = p1T2 / T1 = 1.2 × 105 × 600 / 300 = 2.4 × 105 Pa. The thermodynamic temperature doubles, so the pressure doubles.

Q15. A sealed can contains gas at a pressure of 3.0 × 105 Pa at 27 °C. The can is heated to 127 °C at constant volume. The pressure of the gas is now

  • (A) 4.0 × 105 Pa
  • (B) 1.4 × 106 Pa
  • (C) 3.0 × 105 Pa
  • (D) 2.3 × 105 Pa
Show Answer

Answer: A – Convert to kelvin: T1 = 27 + 273 = 300 K and T2 = 127 + 273 = 400 K. Then p2 = p1T2 / T1 = 3.0 × 105 × 400 / 300 = 4.0 × 105 Pa. Option B comes from using the Celsius temperatures directly (127 / 27), which is wrong.

Q16. A fixed mass of gas at constant pressure occupies 0.30 m3 at 300 K. It is heated to 450 K. Its new volume is

  • (A) 0.20 m3
  • (B) 0.45 m3
  • (C) 0.30 m3
  • (D) 0.60 m3
Show Answer

Answer: B – Charles’s law (constant p): V2 = V1T2 / T1 = 0.30 × 450 / 300 = 0.45 m3.

Q17. A fixed mass of gas has p1 = 1.0 × 105 Pa, V1 = 0.50 m3 and T1 = 300 K. It is changed to V2 = 0.25 m3 and T2 = 450 K. The new pressure p2 is

  • (A) 7.5 × 104 Pa
  • (B) 1.5 × 105 Pa
  • (C) 6.0 × 105 Pa
  • (D) 3.0 × 105 Pa
Show Answer

Answer: D – p2 = p1V1T2 / (T1V2) = (1.0 × 105 × 0.50 × 450) / (300 × 0.25) = 1.0 × 105 × 2 × 1.5 = 3.0 × 105 Pa. Halving the volume doubles p and raising T by a factor 1.5 multiplies p by 1.5 again.

Q18. A fixed mass of gas initially has p1 = 2.0 × 105 Pa, V1 = 0.040 m3 and T1 = 300 K. After a change, p2 = 1.0 × 105 Pa and V2 = 0.060 m3. The final temperature T2 is

  • (A) 225 K
  • (B) 400 K
  • (C) 300 K
  • (D) 150 K
Show Answer

Answer: A – T2 = p2V2T1 / (p1V1) = (1.0 × 105 × 0.060 × 300) / (2.0 × 105 × 0.040) = 1.8 × 106 / 8.0 × 103 = 225 K.

Q19. For a fixed mass of gas held at constant volume, a graph of pressure p against thermodynamic temperature T is

  • (A) a straight line that does not pass through the origin
  • (B) a curve of decreasing gradient
  • (C) a straight line passing through the origin
  • (D) a horizontal line
Show Answer

Answer: C – At constant V, p = (nR / V) T, so p is directly proportional to T and the graph is a straight line through the origin. (A graph of p against Celsius temperature is also straight but cuts the temperature axis at −273 °C, the position of absolute zero.)

Q20. A fixed mass of ideal gas is changed so that its pressure is doubled and its volume is halved. The temperature of the gas

  • (A) doubles
  • (B) halves
  • (C) quadruples
  • (D) stays the same
Show Answer

Answer: D – T = pV / (nR). The product pV is unchanged, since (2p)(V / 2) = pV, so with n fixed the temperature is unchanged.


Using the Equation of State

Q21. A vessel of volume 8.31 × 10−3 m3 contains an ideal gas at a pressure of 3.0 × 105 Pa and a temperature of 300 K. The amount of gas in the vessel is

  • (A) 0.10 mol
  • (B) 1.0 mol
  • (C) 10 mol
  • (D) 3.0 mol
Show Answer

Answer: B – n = pV / (RT) = (3.0 × 105 × 8.31 × 10−3) / (8.31 × 300) = 2493 / 2493 = 1.0 mol.

Q22. How many molecules are there in 0.50 mol of an ideal gas?

  • (A) 3.01 × 1023
  • (B) 6.02 × 1023
  • (C) 1.20 × 1024
  • (D) 3.01 × 1022
Show Answer

Answer: A – N = nNA = 0.50 × 6.02 × 1023 = 3.01 × 1023 molecules.

Q23. A sample of ideal gas at a pressure of 1.0 × 105 Pa occupies a volume of 4.14 × 10−3 m3 at 300 K. Using pV = NkT, the number of molecules in the sample is

  • (A) 1.0 × 1020
  • (B) 6.0 × 1023
  • (C) 1.0 × 1023
  • (D) 2.4 × 1022
Show Answer

Answer: C – N = pV / (kT) = (1.0 × 105 × 4.14 × 10−3) / (1.38 × 10−23 × 300) = 414 / (4.14 × 10−21) = 1.0 × 1023 molecules.

Q24. A 44 g sample of a gas occupies a volume of 0.0227 m3 at a pressure of 1.0 × 105 Pa and a temperature of 273 K. The molar mass of the gas is approximately

  • (A) 11 g mol−1
  • (B) 44 g mol−1
  • (C) 88 g mol−1
  • (D) 4.4 g mol−1
Show Answer

Answer: B – n = pV / (RT) = (1.0 × 105 × 0.0227) / (8.31 × 273) = 2270 / 2269 ≈ 1.0 mol. Molar mass M = m / n = 44 g / 1.0 mol = 44 g mol−1 (equivalently, M = mRT / (pV)).

Q25. A gas has molar mass 0.029 kg mol−1. At a pressure of 1.0 × 105 Pa and a temperature of 300 K, its density is approximately

  • (A) 0.12 kg m−3
  • (B) 12 kg m−3
  • (C) 0.86 kg m−3
  • (D) 1.2 kg m−3
Show Answer

Answer: D – From pV = nRT with n = m / M, the density is ρ = m / V = pM / (RT) = (1.0 × 105 × 0.029) / (8.31 × 300) = 2900 / 2493 = 1.16 kg m−3 ≈ 1.2 kg m−3 (close to the density of air).

Q26. The pressure of a fixed mass of ideal gas is halved while its temperature is kept constant. The density of the gas

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

Answer: C – ρ = pM / (RT). With M and T unchanged, ρ is proportional to p, so halving the pressure halves the density (the gas expands to twice the volume while its mass stays the same).

Q27. Container A, of volume 0.020 m3, holds gas at 1.0 × 105 Pa. Container B, of volume 0.010 m3, holds gas at 3.0 × 105 Pa. Both are at the same temperature. A tap between them is opened and the gases mix at constant temperature. The final pressure is approximately

  • (A) 1.7 × 105 Pa
  • (B) 2.0 × 105 Pa
  • (C) 4.0 × 105 Pa
  • (D) 1.3 × 105 Pa
Show Answer

Answer: A – At constant T the total value of pV is conserved: pfinal (0.020 + 0.010) = (1.0 × 105 × 0.020) + (3.0 × 105 × 0.010) = 2000 + 3000 = 5000. So pfinal = 5000 / 0.030 = 1.67 × 105 Pa ≈ 1.7 × 105 Pa. It is not the simple average of the two pressures, because the volumes differ.

Q28. A real gas behaves least like an ideal gas when it is at

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

Answer: B – At high pressure the molecules are close together, so their own volume is a significant fraction of the container volume; at low temperature they move slowly, so the attractive forces between them have a noticeable effect. Both ideal-gas assumptions then fail.

Q29. Which statement best explains why a real gas deviates from ideal behaviour at high pressure?

  • (A) The molecules move faster, so they hit the walls more often
  • (B) The molecules stop colliding with each other
  • (C) The temperature can no longer be measured accurately
  • (D) The molecules are close together, so their own volume and the attractions between them can no longer be ignored
Show Answer

Answer: D – The ideal model treats molecules as point particles with no intermolecular forces. At high pressure the separation between molecules is small, the finite volume of the molecules matters, and the attractive forces between them are no longer negligible, so pV differs from nRT.

Q30. A steel cylinder of volume 0.050 m3 is filled with oxygen (molar mass 32 g mol−1) at a pressure of 1.0 × 107 Pa and a temperature of 300 K. The mass of oxygen in the cylinder is approximately

  • (A) 6.4 kg
  • (B) 0.64 kg
  • (C) 64 kg
  • (D) 3.2 kg
Show Answer

Answer: A – n = pV / (RT) = (1.0 × 107 × 0.050) / (8.31 × 300) = 5.0 × 105 / 2493 ≈ 200 mol. Mass = nM = 200 × 0.032 kg = 6.4 kg.

Keep practising: Go back to the Physics MCQs hub for more topic-by-topic practice, and check the A Level Physics Formula Sheet whenever you need the equation of state or any other 9702 formula.

These are original practice questions written for revision; they are not Cambridge past-paper questions, and this page is not affiliated with Cambridge Assessment International Education.