343 exam-style questions on The particulate nature of matter with answers and explanations. Six real samples below — the full set is free with an account.
In the kinetic theory of gases, the absolute temperature of an ideal gas is proportional to the average...
Answer: kinetic energy of the molecules
Average translational KE = (3/2)kT, so temperature (in kelvin) is proportional to the average kinetic energy of the molecules.
Which statement correctly distinguishes an isothermal from an adiabatic process for an ideal gas?
Answer: Isothermal: ΔU = 0 (Q = W); Adiabatic: Q = 0 (ΔU = −W)
Using Q = ΔU + W, where W is the work done by the gas: isothermal keeps T constant, so ΔU = 0 and Q = W; adiabatic exchanges no heat (Q = 0), so ΔU = −W. Swapping the two conditions gives the most tempting wrong option.
A fixed mass of ideal gas is held at constant volume while its absolute (kelvin) temperature is doubled. Its pressure will
Answer: double
At constant volume p ∝ T (Gay-Lussac's law), so doubling the kelvin temperature doubles the pressure.
Two 6.0 Ω resistors are connected in parallel. Their combined resistance is
Answer: 3.0 Ω
1/R = 1/6.0 + 1/6.0 = 2/6.0, so R = 3.0 Ω. Two equal resistors in parallel give half the value of one.
The radiation emitted by a black body has maximum intensity at a wavelength of 580 nm. Using Wien's displacement law (constant = 2.9 × 10⁻³ m K), the surface temperature of the body is approximately:
Answer: 5000 K
T = 2.9 × 10⁻³ / (580 × 10⁻⁹) = 5000 K. Power-of-ten slips give 500 K or 50 000 K; hotter bodies peak at shorter wavelengths.
The absolute temperature of a black body is doubled while its surface area stays constant. The total power it radiates increases by a factor of:
Answer: 16
The Stefan–Boltzmann law gives P = σAT⁴, so doubling T multiplies the power by 2⁴ = 16. Choosing 2 or 4 comes from forgetting the fourth-power dependence.
That was 6 of 343 The particulate nature of matter questions.
Practise all 343 freeA student writes: “In each cycle a heat engine takes in 600 J from the hot reservoir and rejects 240 J to the cold reservoir, so its efficiency is 240/600 = 40 %.”
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