At the same temperature, hydrogen molecules (M = 2 g/mol) and oxygen molecules (M = 32 g/mol) are compared. Which statement correctly describes their relationship?
AOxygen molecules have more kinetic energy because they are heavier
BHydrogen molecules have more kinetic energy because they move faster
CBoth have the same average kinetic energy; hydrogen moves faster because the same energy means a higher speed for lighter molecules
DHydrogen molecules have less kinetic energy because they contribute less mass to the gas
At the same temperature T, every molecule has the same average kinetic energy ⟨KE⟩ = (3/2)kT regardless of mass. Since KE = ½mv², equal energy with smaller mass means higher speed: v_rms = √(3kT/m). Hydrogen (m ∝ 2) has v_rms ≈ 1930 m/s; oxygen (m ∝ 32) has v_rms ≈ 484 m/s — 4 times slower (√(32/2) = 4). Options A and B both fail because they assume heavier mass means more energy, which is wrong — temperature sets energy, and mass sets how fast that energy translates to speed.
Question 2 Multiple Choice
The temperature of a gas sample doubles from 300 K to 600 K. By what factor does v_rms change?
AIt doubles
BIt increases by a factor of √2 ≈ 1.41
CIt increases by a factor of 4
DIt stays the same — temperature only affects average kinetic energy, not speed
Since v_rms = √(3kT/m), v_rms scales as √T. If T doubles, v_rms increases by √2 ≈ 1.41. It does not double (that would require T to quadruple). Option D is wrong: kinetic energy and speed are directly related through KE = ½mv², so changing KE necessarily changes v_rms.
Question 3 True / False
At the same temperature, a sample of oxygen gas and a sample of hydrogen gas have the same average translational kinetic energy per molecule.
TTrue
FFalse
Answer: True
This is a direct consequence of the kinetic theory result ⟨KE⟩ = (3/2)kT. Temperature, not mass, determines average kinetic energy per molecule. This is why heavier molecules move more slowly at the same temperature — the same energy is distributed into a larger mass, requiring a lower speed to maintain KE = ½mv².
Question 4 True / False
The rms speed v_rms of gas molecules equals their average speed ⟨|v|⟩.
TTrue
FFalse
Answer: False
v_rms = √⟨v²⟩ and ⟨|v|⟩ are different quantities. For a Maxwell-Boltzmann distribution, v_rms is about 8% larger than ⟨|v|⟩ (v_rms = √(3kT/m), ⟨|v|⟩ = √(8kT/πm)). The rms speed is also distinct from the most probable speed (the peak of the distribution), which is the smallest of the three. The rms speed is the relevant quantity for connecting to kinetic energy because ⟨KE⟩ = ½m⟨v²⟩ = ½mv_rms².
Question 5 Short Answer
Why do we define and use the rms speed v_rms = √⟨v²⟩ instead of simply using the average speed ⟨|v|⟩ to characterize molecular motion?
Think about your answer, then reveal below.
Model answer: The rms speed connects directly to kinetic energy: ⟨KE⟩ = ½m⟨v²⟩ = ½mv_rms². Using v_rms makes the energy formula exact and algebraically clean. The average velocity vector ⟨v⟩ is actually zero (molecules move equally in all directions), so that is useless as a speed measure. The average speed ⟨|v|⟩ is nonzero but requires integrating over the Maxwell-Boltzmann distribution and does not connect as neatly to thermodynamic quantities like temperature.
This is a conceptual choice driven by the goal of connecting molecular motion to thermodynamics. The 'square' in rms naturally pairs with the v² in kinetic energy, making the correspondence between temperature and molecular speed clean and direct. Using ⟨|v|⟩ would introduce factors of √(π/8) into the temperature-energy relationship, obscuring the physics without any compensating benefit.