Thirty Questions. Thirty Minutes. Go.

Sections 1 to 11 took kinetic theory apart slowly and worked forty-odd problems through it. Section 14 then rebuilt it for speed: the statements asked word for word, the formula cards with a hook on each, the speed ordering drilled until it is automatic, the degree-of-freedom lookup, the proportionality habits and the two special formats. This section finds out whether any of that survives contact with a clock.

There is no new physics below. There are 30 questions built the way this paper builds them, and one rule that matters more than the rest: you are being tested on pace, not on cleverness. If a question here takes you five lines of algebra, you have misread it.

One syllabus note. Several items below rest on material that sits outside the body text of the rationalised syllabus — the Maxwell speed distribution and the formulas for vˉ\bar{v} and vmpv_{mp}, Brownian motion, Graham's law and the pressure law. NEET has asked about every one of them, so every one is drilled here.

How to attempt this set

Key Point: Blank sheet, pen, timer. Attempt all 30 questions in one unbroken sitting, and do not read a single explanation until your last answer is written. A drill you pause to check is a reading exercise, and reading exercises do not build speed.

The setup What it is
Number of questions 30, single correct option
Marking scheme +4+4 correct, 1-1 incorrect, 00 unattempted
Maximum score 30×4=12030 \times 4 = 120 marks
Minimum possible score 30×(1)=3030 \times (-1) = -30 marks
Suggested time limit 30 minutes (45 Physics questions in about 45 minutes, so roughly a minute each)
Allowed a rough sheet and your memory
Not allowed calculator, formula sheet, or a glance back at Section 14

The constants sheet

Every question that needs a number uses these and no others.

Quantity Value
gas constant, RR 8.3148.314 J/(mol K)
Boltzmann constant, kBk_B 1.38×10231.38 \times 10^{-23} J/K
Avogadro number, NAN_A 6.022×10236.022 \times 10^{23} per mole
absolute temperature T=tC+273T = t_C + 273, and the extra 0.150.15 never changes an option here
11 atm 1.013×1051.013 \times 10^{5} Pa
molar volume at STP 22.422.4 litres per mole
useful roots 2=1.414\sqrt{2} = 1.414, 3=1.732\sqrt{3} = 1.732, 8/π=1.596\sqrt{8/\pi} = 1.596
monatomic gas Cv=32R=12.47C_v = \frac{3}{2}R = 12.47, Cp=52R=20.79C_p = \frac{5}{2}R = 20.79 J/(mol K), γ=53\gamma = \frac{5}{3}
rigid diatomic gas Cv=52R=20.79C_v = \frac{5}{2}R = 20.79, Cp=72R=29.10C_p = \frac{7}{2}R = 29.10 J/(mol K), γ=75\gamma = \frac{7}{5}
rigid non-linear polyatomic Cv=3R=24.94C_v = 3R = 24.94, Cp=4R=33.26C_p = 4R = 33.26 J/(mol K), γ=43\gamma = \frac{4}{3}

Three housekeeping notes on those.

The symbols hold to this chapter's convention throughout. nn is the number density, molecules per cubic metre, and μ\mu is the number of moles — the reverse of the previous chapter, and the single commonest source of lost marks in the first week of this one. So every gas law below reads PV=μRTPV = \mu RT and every heat reads ΔQ=μCvΔT\Delta Q = \mu C_v \Delta T, never nRTnRT. NN is the number of molecules, NAN_A the Avogadro number, and μ=NNA=MM0\mu = \frac{N}{N_A} = \frac{M}{M_0}.

M0M_0 is the molar mass in kilograms per mole. Oxygen is 0.0320.032, not 3232. Putting grams into vrms=3RTM0v_{rms} = \sqrt{\frac{3RT}{M_0}} makes the answer too small by a factor of 100031.6\sqrt{1000} \approx 31.6, and at least one option in every speed question below is exactly that number, waiting.

Every temperature entering a ratio or a speed formula is in kelvin. Write the K next to it the moment it goes onto your page. Where only a difference ΔT\Delta T appears, Celsius and kelvin give the same number and no conversion is wanted; knowing which of those two situations you are in is worth several marks on this set alone.

[Important] The 30-minute limit is the entire exercise. Most students can get 27 of these right given an hour — and an hour is exactly what the real paper will not give you. Finishing in 30 minutes with 24 correct puts you in far better shape than taking 55 minutes to get 27. Keep the timer where you can see it, and the moment a question passes 60 seconds, mark your best surviving option and move on.

Here is the arithmetic that makes that instruction safe. A blind guess among four options is worth 434=+0.25\frac{4 - 3}{4} = +0.25, essentially nothing. But once you have eliminated two options, a guess between the survivors is worth 412=+1.5\frac{4 - 1}{2} = +1.5 marks on average. Eliminate first, then commit. Leave blank only what you could not narrow down at all.

What this set covers

Topic spread, marking scheme and constants for the thirty question kinetic theory drill

Topic Questions How many
Molecular nature, scales and Brownian motion Q1 to Q3 3
The ideal gas equation in its four forms Q4, Q5, Q7 3
The gas laws, Dalton's law and real gases Q6, Q8, Q9 3
Assumptions and the pressure formula Q10 to Q12 3
Temperature, kinetic energy and vrmsv_{rms} Q13 to Q16 4
The Maxwell distribution and the three speeds Q17 to Q19 3
Degrees of freedom and equipartition Q20, Q21 2
Specific heats, γ\gamma and Dulong-Petit Q22 to Q24 3
Mean free path, collisions and diffusion Q25 to Q27 3
Assertion-reason and column matching Q28 to Q30 3

Two more items, Q8 and Q9, also use the matching and assertion-reason formats, so five of the thirty are in a special format — about the share the real paper carries. And notice where the weight sits: molecular speeds and the quantities built from them supply a third of the set, because they supply about a third of what this chapter is asked about.

Mark It Honestly, Then Read Your Own Answer Sheet

Score with the real scheme: +4+4 for every correct answer, 1-1 for every wrong one, 00 for every blank. No half marks for "I nearly had that one". The number you end up with is the number that means something.

Pacing line and four self scoring bands for the kinetic theory drill

The bands

Your score (out of 120) Verdict What to do next
100 to 120 Exam ready. Over 80% on a full-length set, inside the time. This chapter is now free marks for you. Revisit only the items you missed, then move to the next chapter.
78 to 99 Fast but leaky. You know the material; something leaks on the way to the answer sheet. Almost always a molar mass left in grams, or a Celsius value used where kelvin was required — not a gap in knowledge. Redo every wrong question without the explanation first, and count how many you fix alone.
48 to 77 Recall gaps. The speed is not the problem; the lookup is. Go back to Section 14's formula cards and the degrees-of-freedom table and learn them as flashcards. Then re-attempt this set cold.
Below 48 Rebuild first. Work Sections 1 to 10 properly, then Section 11's worked problems, then Section 14. Re-attempting this set today would teach you nothing except the answer key.

Sort your mistakes into three piles

Do this before you read a single explanation. It is the most useful ten minutes in this section.

  1. Did not know it. A formula you could not recall, whether vˉ\bar{v} or vrmsv_{rms} is the larger, whether ff for carbon dioxide is 5 or 6, whether γ\gamma for a monatomic gas is 1.401.40 or 1.671.67. Cheapest to fix — it is a memory job, and it takes an evening.
  2. Knew it, computed it wrong. You left M0M_0 in grams per mole, used 2727 where 300300 K belonged, dropped the 2\sqrt{2} out of the mean free path, or squared a radius that was meant to be a diameter. Slow down for four seconds on the final line.
  3. Knew it, answered a different question. You gave vˉ\bar{v} when it asked for vrmsv_{rms}, the collision time when it asked for the collision frequency, the molar specific heat when it asked for the per-kilogram one, the translational energy EE when it asked for the whole internal energy UU. The distractors here are built specifically to reward this mistake.

Key Point: Two students both score 88. The first has four pile-1 mistakes and a syllabus gap that revision closes in a day. The second has nine pile-3 mistakes and a reading habit that will follow them into the exam hall. Pile 3 is the expensive one — count it before you explain it away.

The fourteen facts this set keeps testing

  • nn is number density and μ\mu is moles, and PV=μRT=NkBTPV = \mu RT = Nk_BT, P=nkBTP = nk_BT, P=ρRTM0P = \frac{\rho RT}{M_0} are one equation in four costumes.
  • One mole of any ideal gas fills 22.422.4 litres at STP and contains 6.022×10236.022 \times 10^{23} molecules.
  • Boyle: P1VP \propto \frac{1}{V} at fixed TT. Charles: VTV \propto T at fixed PP. The pressure law: PTP \propto T at fixed VV — all three in kelvin, always.
  • Dalton: P=P1+P2+P = P_1 + P_2 + \cdots, because P=(n1+n2+)kBTP = (n_1 + n_2 + \cdots)k_BT.
  • A real gas is closest to ideal at low pressure and high temperature, where its molecules are far apart.
  • P=13nmv2=13ρv2P = \frac{1}{3}nm\overline{v^2} = \frac{1}{3}\rho\,\overline{v^2}, and v2\overline{v^2} is the mean of the squares, never the square of the mean.
  • 12mv2=32kBT\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_BT: the average translational kinetic energy per molecule depends on TT and on nothing else — not the gas, not the pressure, not the mass.
  • vrms=3RTM0v_{rms} = \sqrt{\frac{3RT}{M_0}} with M0M_0 in kg/mol, so vTv \propto \sqrt{T} and v1M0v \propto \frac{1}{\sqrt{M_0}}.
  • vmp<vˉ<vrmsv_{mp} < \bar{v} < v_{rms} in the ratio 2:8/π:3\sqrt{2} : \sqrt{8/\pi} : \sqrt{3}, that is 1:1.128:1.2251 : 1.128 : 1.225 — the peak, the average, the root mean square, in that order.
  • Heating flattens the distribution and slides it right; the area underneath never changes, because the molecules are still all there.
  • ff is 3 monatomic, 5 rigid diatomic, 5 rigid linear polyatomic, 6 rigid non-linear polyatomic, 7 vibrating diatomic — a vibration counts twice because it carries both a kinetic and a potential quadratic term.
  • U=f2μRTU = \frac{f}{2}\mu RT, Cv=f2RC_v = \frac{f}{2}R, Cp=Cv+RC_p = C_v + R, γ=1+2f\gamma = 1 + \frac{2}{f}, and CpCv=RC_p - C_v = R holds for every ideal gas whatever its shape.
  • Dulong-Petit: a simple solid has C3R24.9C \approx 3R \approx 24.9 J/(mol K) per mole of atoms.
  • l=12nπd2=kBT2πd2Pl = \frac{1}{\sqrt{2}\,n\pi d^2} = \frac{k_BT}{\sqrt{2}\,\pi d^2 P} with dd the DIAMETER, collision frequency fcoll=vˉlf_{\text{coll}} = \frac{\bar{v}}{l}, and Graham: r1r2=M0,2M0,1\frac{r_1}{r_2} = \sqrt{\frac{M_{0,2}}{M_{0,1}}}.

[Important] If you got fewer than 24 right, count how many of your errors were a molar mass left in grams or a Celsius value that should have been kelvin. In this chapter those two between them usually account for more lost marks than everything else put together, and both are the cheapest mistakes here to fix: write the kg/mol conversion on its own line, and write the letter K beside every temperature the moment it enters a formula.