Progressive Thermal physics practice questions

Use Clevolab for progressive practice in thermal physics. This page shows what the topic covers, what skills the current set targets, and a few real examples from the reviewed question bank.

Physics Progressive Thermal physics

About this topic

Clevolab treats thermal physics as repeated practice with explanation, not just answer checking. This page is designed to make the topic legible before you open the app.

The current Progressive set covers internal energy, gases, temperature, heat transfer, and thermal reasoning. 155 reviewed questions currently published for this page.

What you can practise

  • Internal energy
  • Specific heat capacity
  • Ideal gases
  • Thermal processes
  • Microscopic and macroscopic links

Real sample questions from the current set

These examples come from the reviewed questions currently stored for this topic. They are here so the page shows the actual flavour of Clevolab, not just a summary.

Level 1

Sample question

As temperature increases, the average molecular speed in a gas

  • AIncreasesCorrect answer
  • BDecreasesAnswer option
  • CStays the sameAnswer option
  • DBecomes zeroAnswer option

Why this answer is right

Hotter gases have faster-moving molecules. For an ideal gas, $v_\mathrm{rms} = \sqrt{3k_\mathrm{B}T/m}$, which rises with $T$.

Kinetic theory relates speed to temperature. For molecular mass $m$, $$v_{\mathrm{rms}} = \sqrt{\frac{3k_{\mathrm{B}}T}{m}}.$$ Since $v_{\mathrm{rms}}$ scales with $\sqrt{T}$, increasing temperature increases typical molecular speeds.

Level 3

Sample question

In a gas moving with uniform bulk speed $u$, how does the temperature measured in the lab compare with that in the comoving frame (nonrelativistic)?

  • AIt increases with $u$Answer option
  • BIt decreases with $u$Answer option
  • CIt is the sameCorrect answer
  • DIt becomes zeroAnswer option

Why this answer is right

Temperature reflects random kinetic energy, not net flow. A uniform drift adds the same velocity to all molecules, leaving the random component and thus the temperature unchanged in nonrelativistic physics.

Decompose molecular velocity into bulk and random parts: $$\mathbf{v}=\mathbf{u}+\mathbf{c}.$$ Temperature is linked to the mean of $c^2$, not $u^2$: $$\tfrac{3}{2}kT=\tfrac{1}{2}m\langle c^2\rangle.$$ A uniform drift changes $\langle v^2\rangle$ but leaves $\langle c^2\rangle$ unchanged. Thus the temperature is the same in both frames.

Level 5

Sample question

In the Clausius approximation $\ln p_{\mathrm{sat}}=\text{const}-\dfrac{L}{R}\,\dfrac{1}{T}$, what is the slope of a plot of $\ln p_{\mathrm{sat}}$ versus $1/T$?

  • A$+\dfrac{L}{T}$Answer option
  • B$+\dfrac{R}{L}$Answer option
  • C$-\dfrac{R}{L}$Answer option
  • D$-\dfrac{L}{R}$Correct answer

Why this answer is right

Comparing $\ln p_{\mathrm{sat}}=\text{const}-\dfrac{L}{R}\,\dfrac{1}{T}$ with a line in $1/T$, the slope is $-L/R$. The negative sign reflects decreasing vapor pressure with cooling.

The Clausius–Clapeyron integration under $\Delta v\approx RT/p$ yields $$\ln p=\text{const}-\frac{L}{R}\frac{1}{T}.$$ Treating $x=1/T$ as the horizontal axis gives a line $$\ln p=\text{const}-\frac{L}{R}x,$$ so the slope is $$-\frac{L}{R}.$$ This linearization is accurate when $L$ is weakly $T$-dependent and vapor behaves ideally.

How this page fits into Clevolab

Clevolab is broader than any one exam mode. GCSE and A-level pages are useful entry points, while the wider project is about sharpening understanding through repeated topic practice.

Related topics