GASEOUS STATE / GAS LAWS
Questions 26–53 — Questions with Answers
26. State Boyle’s law.
Answer: At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume.
Thus, PV = constant, or P₁ V₁ = P₂ V₂ .
27. Name two items that can serve as a model for Gay Lussac’s law and explain.
Answer: A sealed aerosol can and a pressure cooker can serve as models. At constant volume, the pressure
of a fixed mass of gas is directly proportional to its absolute temperature: P/T = constant. Heating increases the
pressure.
28. Give the mathematical expression that relates gas volume and temperature.
Answer: Charles’ law: At constant pressure, V ∝ T (in kelvin), hence V/T = constant or V₁ /T₁ = V₂ /T₂ .
29. What are ideal gases? In what way do real gases differ from ideal gases?
Answer: An ideal gas is a hypothetical gas whose molecules have negligible volume and no intermolecular
forces and which obeys PV = nRT exactly. Real gases have finite molecular volume and intermolecular
attractions/repulsions, so they deviate from ideal behaviour, especially at high pressure and low temperature.
30. Can a Van der Waals gas with a = 0 be liquefied? Explain.
Answer: No. If a = 0, intermolecular attractive forces are absent. Since liquefaction requires intermolecular
attraction, such a gas cannot be liquefied by pressure alone.
31. Suppose there is a tiny sticky area on the wall of a container of gas. Molecules on this area stick there
permanently. Is the pressure greater or less than on the ordinary walls?
Answer: The pressure is less. Molecules striking the sticky area do not rebound, so they transfer less
momentum to the wall. Since pressure is due to momentum transfer, the pressure there is reduced.
32. Explain the following observations: (a) Aerated water bottles are kept under water during summer (b)
Liquid ammonia bottle is cooled before opening the seal (c) Automobile tyre is inflated to slightly lesser
pressure in summer (d) Weather balloon becomes larger as it ascends to higher altitude.
Answer: (a) Cooling keeps the gas pressure lower and reduces the tendency of the bottle to burst.
(b) Cooling lowers the vapour pressure of ammonia, making the bottle safer to open.
(c) Air pressure inside a tyre increases when temperature rises; therefore it is inflated to a slightly lower
pressure in summer.
(d) Atmospheric pressure decreases with altitude, so the gas inside the balloon expands and its volume
increases.
33. Give suitable explanation for the following facts about gases: (a) Gases don’t settle at the bottom of a
container (b) Gases diffuse through all the space available to them.
Answer: Gas molecules are in continuous random motion. Hence they do not remain at the bottom but spread
throughout the container. Their continuous motion and large intermolecular spaces allow gases to diffuse and
occupy all available space.
34. Suggest why there is no hydrogen (H₂ ) in our atmosphere. Why does the moon have no atmosphere?
Answer: Hydrogen molecules are very light and have high average speed. At the temperature of the upper
atmosphere, a fraction of H₂ molecules can reach escape velocity and escape into space. The Moon has much
weaker gravity and cannot retain atmospheric gases; therefore it has essentially no atmosphere.
35. Determine whether a gas approaches ideal behaviour or deviates from ideal behaviour: (a) it is
compressed to a smaller volume at constant temperature (b) the temperature is raised while keeping the
volume constant (c) more gas is introduced into the same volume and at the same temperature.
Answer: (a) Deviates from ideal behaviour, because high pressure/small volume makes molecular volume and
intermolecular forces significant.
Gas Laws • Questions 26–53
(b) Approaches ideal behaviour as temperature increases, because intermolecular attractions become less
significant.
(c) Deviates from ideal behaviour because increasing the amount of gas in the same volume increases
pressure and makes intermolecular effects significant.
36. Which of the following gases would you expect to deviate from ideal behaviour under conditions of low
temperature: F₂ , Cl₂ or Br₂ ? Explain.
Answer: Br₂ . At low temperature, intermolecular attractions become important. Br₂ has the largest molar
mass and strongest dispersion forces among F₂ , Cl₂ and Br₂ , so it shows the greatest deviation from ideal
behaviour.
37. Distinguish between diffusion and effusion.
Answer: Diffusion is the spontaneous mixing of gas molecules throughout a space due to random molecular
motion. Effusion is the escape of gas molecules through a very small hole into a vacuum or low-pressure
region. Graham’s law applies to both processes.
38. Aerosol cans carry clear warning of heating of the can.
Answer: An aerosol can is nearly constant in volume. On heating, the temperature of the gas increases and, by
Gay-Lussac’s law, its pressure increases. Excessive pressure can cause the can to rupture or explode.
39. Would it be easier to drink water with a straw on the top of Mount Everest?
Answer: No. Atmospheric pressure is much lower at Mount Everest, so there is less external pressure to push
water up the straw. Hence drinking through a straw is more difficult.
40. Write the Van der Waals equation for a real gas. Explain the correction terms for pressure and volume.
Answer: Van der Waals equation: (P + an²/V²)(V − nb) = nRT. The pressure correction, an²/V², accounts for
intermolecular attraction because the observed pressure is lower than ideal. The volume correction, nb,
accounts for the finite volume occupied by gas molecules, so the free volume is V − nb.
41. Derive the values of critical constants in terms of Van der Waals constants.
Answer: For one mole: P = RT/(V−b) − a/V². At the critical point, (∂P/∂V)T = 0 and (∂²P/∂V²)T = 0. Solving
these conditions gives the critical constants: Vc = 3b; Pc = a/(27b²); Tc = 8a/(27Rb).
42. Why do astronauts have to wear protective suits when they are on the surface of the Moon?
Answer: The Moon has essentially no atmosphere and therefore no normal atmospheric pressure or
breathable oxygen. Astronaut suits provide pressure, oxygen and protection from the extreme thermal
environment and radiation.
43. When ammonia combines with HCl, NH₄ Cl is formed as white dense fumes. Why do more fumes
appear near HCl?
Answer: NH₃ + HCl → NH₄ Cl. NH₃ has a lower molar mass than HCl and therefore diffuses faster. The
gases meet closer to the HCl end, so more NH₄ Cl fumes are observed near HCl.
44. A sample of gas at 15°C and 1 atm has a volume of 2.58 dm³. When the temperature is raised to 38°C at
1 atm does the volume of the gas increase? If so, calculate the final volume.
Answer: Yes. At constant pressure, Charles’ law applies: V₂ = V₁ T₂ /T₁ . T₁ = 288.15 K, T₂ = 311.15 K. V₂
= 2.58 × 311.15/288.15 ≈ 2.79 dm³.
45. A sample of gas has a volume of 8.5 dm³ at an unknown temperature. When the sample is submerged
in ice water at 0°C, its volume gets reduced to 6.37 dm³. What is its initial temperature?
Answer: At constant pressure, V₁ /T₁ = V₂ /T₂ . T₁ = V₁ T₂ /V₂ = 8.5 × 273.15/6.37 ≈ 364.5 K = 91.3°C.
46. Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen in a vessel of volume 37.6
dm³ at 298 K, and sample B is in a vessel of volume 16.5 dm³ at 298 K. Calculate the number of moles in
sample B.
Answer: At the same temperature and pressure, V ∝ n. Therefore n₂ = n₁ V₂ /V₁ = 1.5 × 16.5/37.6 ≈ 0.658
mol.
Gas Laws • Questions 26–53
47. Sulphur hexafluoride is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of
the gas in a steel vessel of volume 5.43 dm³ at 69.5°C, assuming ideal gas behaviour.
Answer: Using PV = nRT: T = 342.65 K. P = nRT/V = (1.82)(0.082057)(342.65)/5.43 ≈ 9.42 atm.
48. Argon is an inert gas used in light bulbs to retard vaporization of the tungsten filament. A certain light
bulb containing argon at 1.2 atm and 18°C is heated to 85°C at constant volume. Calculate its final
pressure in atm.
Answer: At constant volume, P₁ /T₁ = P₂ /T₂ . P₂ = 1.2 × (358.15/291.15) ≈ 1.48 atm.
49. A small bubble rises from the bottom of a lake where the temperature and pressure are 6°C and 4 atm,
to the water surface, where the temperature is 25°C and pressure is 1 atm. Calculate the final volume (in
mL) of the bubble, if its initial volume is 1.5 mL.
Answer: Combined gas law: P₁ V₁ /T₁ = P₂ V₂ /T₂ . V₂ = P₁ V₁ T₂ /(P₂ T₁ ) = 4 × 1.5 × 298.15/279.15 ≈ 6.41
mL.
50. Hydrochloric acid is treated with a metal to produce hydrogen gas. Suppose a student collects a
volume of 154.4 × 10⁻ ³ dm³ of a gas at a pressure of 742 mm Hg and a temperature of 298 K. What mass of
hydrogen gas (in mg) did the student collect?
Answer: Using PV = nRT: P = 742/760 atm, V = 0.1544 L, T = 298 K. n = PV/RT ≈ 0.006165 mol. Mass of H₂
= n × 2.016 ≈ 0.01243 g = 12.43 mg.
51. It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N₂ gas to effuse at
the same temperature and pressure. What is the molar mass of the unknown gas?
Answer: By Graham’s law, r ∝ 1/√M and, for equal quantities, t ∝ √M. Thus t_unknown/t_N₂ =
√(M_unknown/M_N₂ ). M_unknown = 28.02 × (192/84)² ≈ 146.4 g mol⁻ ¹.
52. A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO₂ at 300 K. The total pressure in the tank
is 9.21 atm. Calculate the partial pressure (in atm) of each gas in the mixture.
Answer: n(O₂ ) = 52.5/32.00 = 1.641 mol. n(CO₂ ) = 65.1/44.01 = 1.479 mol. Total moles ≈ 3.120 mol. By
Dalton’s law, Pᵢ = xᵢP_total. P(O₂ ) ≈ (1.641/3.120)(9.21) = 4.85 atm; P(CO₂ ) ≈ (1.479/3.120)(9.21) = 4.36 atm.
53. A combustible gas is stored in a metal tank at a pressure of 2.98 atm at 25°C. The tank can withstand a
maximum pressure of 12 atm after which it will explode. The building in which the tank has been stored
catches fire. Now predict whether the tank will blow up first or start melting. (Melting point of the metal =
1100 K.)
Answer: At constant volume, P/T = constant. Temperature when P reaches 12 atm: T₂ = T₁ (P₂ /P₁ ) = 298.15
× 12/2.98 ≈ 1202 K. This is above the metal melting point of 1100 K. Therefore, the metal will start melting
before the gas pressure reaches 12 atm; the tank will start melting first.
Gas Laws • Questions 26–53
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