265 out of 320 — your best physics paper, forty clear of target. Sixty-nine right, eleven wrong, nothing left blank. Two things stand out. Easy accuracy is 31 of 32, which breaks a pattern that had held for three papers running. And the losses have moved: they are now concentrated in the hard questions, and above all in figures where two numbers have to be read off the axes before any physics starts. Three such questions on this paper, and all three went.
Where the marks went
Difficulty
Right
Out of
Accuracy
Easy
31
32
97%
97%
Moderate
27
32
84%
84%
Hard
11
16
69%
69%
For the first time the profile is the right way up: strongest on the easy questions, weakest on the hard ones. That is what a normal paper looks like, and it is good news — the careless losses on easy questions have largely gone.
Easy accuracy, four papers running
Paper
Easy
Moderate
Hard
Morphology of Flowering Plants
26 / 32
—
16 / 16
Anatomy of Flowering Plants
26 / 32
26 / 32
14 / 16
Kinetic Theory
26 / 32
25 / 32
15 / 16
Thermodynamics (this paper)
31 / 32
27 / 32
11 / 16
Three papers stuck at exactly 26 of 32, then 31. Whatever you did differently on this paper, that is the thing to keep doing.
Section by section
Section
Right
Out of
Lost
Thermal Equilibrium, Heat and Work, State and Path
8
8
—
The First Law of Thermodynamics
10
14
4 wrong
Specific Heat Capacities and Mayer's Relation
8
11
3 wrong
State Variables and Quasi-static Processes
5
5
—
The Isothermal Process
10
11
1 wrong
The Adiabatic Process
14
14
—
Isobaric, Isochoric and Cyclic Processes, and Free Expansion
8
11
3 wrong
The Second Law, Reversible and Irreversible Processes
6
6
—
The first law is the weak block at ten of fourteen. The adiabatic process, the second law and the opening section were perfect.
The habit that cost the most
Three errors · 12 marks · all three hard
Reading one number off the graph instead of two
Three questions gave you a P–V diagram and asked for something that depends on both coordinates of a point. In each one you took a single number off the axes and went straight into the arithmetic. This is the same habit the Anatomy paper showed on two-clue figures: the first clue is read, and the second is never collected. The fix is mechanical: write down the coordinates of every marked point before you start. Four numbers on the page, then the physics.
Q9 — you chose (A), the answer is (C)figure Ā· two coordinates to be read first Ā· look-alike: heat supplied vs change in internal energy Ā· hard
In the step B → C the pressure stays constant. Taking the rise in internal energy printed on the figure, the heat taken in by the gas during that step is
(A) 2250 JYOUR ANSWER
(B) 3750 J
(C) 3150 JCORRECT
(D) 1350 J
Work at constant pressure is PΔV, not PV. The step runs from 2 litre to 5 litre, so ΔV = 3 litre and W = 900 J, giving Q = 2250 + 900 = 3150 J. Your 2250 J is ΔU alone — you used the internal-energy rise printed on the figure and never added the work.
On any constant-pressure step, write ΔV = V₂ − V₁ explicitly before using it. The symbol Δ is the whole instruction.
Q39 — you chose (C), the answer is (D)figure Ā· needs a second clue Ā· two coordinates to be read first Ā· look-alike: a ratio written the right way round vs upside down Ā· hard
Reading the pressure and the volume of each marked state off the axes, the ratio of the temperature at X to the temperature at Y is
(A) 1 : 4
(B) 4 : 1
(C) 3 : 4YOUR ANSWER
(D) 4 : 3CORRECT
Temperature follows the product PV, so both coordinates matter. At X, PV = 1 × 4 = 4; at Y, PV = 3 × 1 = 3; so TX : TY = 4 : 3. You answered 3 : 4 — the right two numbers, the ratio inverted. Y looks more energetic because it sits higher up the pressure axis, and that is the trap.
Q69 — you chose (A), the answer is (C)figure Ā· two coordinates to be read first Ā· hard
Step AB lies on a straight line through the origin, so the pressure is constant along it. Reading the temperature and the volume at A, and taking one mole with R = 8.3 J mol−1 K−1, that constant pressure is
(A) 6.22 × 105 PaYOUR ANSWER
(B) 24.90 × 105 Pa
(C) 12.45 × 105 PaCORRECT
(D) 3.11 × 105 Pa
Both numbers had to come from the same point. Reading A gives T = 300 K and V = 2 litre, so P = nRT/V = 2490 ÷ (2 × 10−3) = 12.45 × 105 Pa. Your 6.22 × 105 is exactly half — the temperature from A paired with the volume from B.
Label the point you are using, then take both numbers from it without looking away.
Cp and Cv
The same pair that cost you four questions on the Kinetic Theory paper cost two more here. It is the single most reliable source of lost marks across both scripts.
Q33 — you chose (B), the answer is (D)look-alike: Cp vs Cv Ā· moderate
Two moles of a diatomic ideal gas are warmed through 50 K with the pressure held constant. The heat that must be supplied is
(A) 830 J
(B) 2075 JYOUR ANSWER
(C) 1453 J
(D) 2905 JCORRECT
The pressure is held constant, so the heat is nCpΔT. Cp = 3.5 × 8.3 = 29.05, giving Q = 2 × 29.05 × 50 = 2905 J. Your 2075 J is the same sum with Cv = 20.75 in place of Cp.
The phrase at constant pressure is not scene-setting — it picks the capacity. Constant pressure → Cp. Constant volume → Cv.
Q74 — you chose (C), the answer is (B)negative stem Ā· look-alike: heat supplied vs change in internal energy Ā· hard
One mole of a diatomic ideal gas is warmed through 100 K at constant pressure. Each of the following is then correct EXCEPT one. Identify the exception.
(A) the internal energy of the gas rises by 2075 J
(B) the heat supplied is equal to the rise in the internal energyCORRECT
(C) the work done by the gas is 830 JYOUR ANSWER
(D) the heat supplied to the gas is 2905 J
Three of the four options carry correct numbers (W = 830 J, Q = 2905 J, ΔU = 2075 J), so the eye slides past them looking for a wrong figure. The exception is the one that is a sentence, not a number: at constant pressure the heat supplied is not equal to the rise in internal energy, because 2905 is not 2075 — the difference is the work.
On a negative stem, check the worded options as carefully as the numerical ones. The exception likes to hide among correct arithmetic.
Q23 — you chose (B), the answer is (C)reference shifted Ā· look-alike: molar heat capacity vs heat capacity per gram Ā· hard
NCERT states Mayer's relation as Cp − Cv = R for one mole of an ideal gas. For oxygen, whose molar mass is 32 g, the difference between the two specific heats taken per gram is
(A) 0.519 J gā»Ā¹ Kā»Ā¹
(B) 8.300 J gā»Ā¹ Kā»Ā¹YOUR ANSWER
(C) 0.259 J gā»Ā¹ Kā»Ā¹CORRECT
(D) 3.855 J gā»Ā¹ Kā»Ā¹
Mayer's relation gives 8.3 J per mole per kelvin. The question asks per gram, and one mole of oxygen is 32 g, so the answer is 8.3 ÷ 32 = 0.259. You answered 8.3 — the right number in the wrong unit.
One sign, one carried answer, one swap
Q21 — you chose (B), the answer is (C)decided by a sign Ā· moderate
The internal energy of a system falls by 300 J while the system does 500 J of work on its surroundings. The heat exchanged, counted positive when it is taken in, is
(A) 800 J
(B) −800 JYOUR ANSWER
(C) 200 JCORRECT
(D) −200 J
Falls by 300 J means ΔU = −300 J. Then ΔQ = ΔU + W = −300 + 500 = +200 J, taken in. You took the fall as +300 and arrived at −800.
Write the sign on the page the moment you read the word falls, released, done on. Do not carry it in your head.
Q17 — you chose (B), the answer is (C)pair 1, second question Ā· moderate
[Pair 1, second question] During that same expansion 2600 J of heat is supplied to the gas. Using your answer to the previous question, the rise in the internal energy of the gas is
(A) 2000 J
(B) 3600 JYOUR ANSWER
(C) 1600 JCORRECT
(D) 1000 J
The first half of this pair you got right: W = 1000 J. Then ΔU = Q − W = 2600 − 1000 = 1600 J. Your 3600 J is 2600 + 1000 — the work added instead of subtracted. The gas does the work, so it comes out of the heat.
ΔU = Q − W when W is work done BY the gas. Say which way the work goes before you put a sign on it.
Q28 — you chose (C), the answer is (A)match Ā· planted swap in rows C & D Ā· easy
Match Column I with Column II.Column IColumn IIAA monatomic gasI1.40BA diatomic gasII8.3 J molā»Ā¹ Kā»Ā¹CA polyatomic gasIII1.67DCp − Cv for one moleIV1.33
(A) A-III, B-I, C-IV, D-IICORRECT
(B) A-III, B-IV, C-I, D-II
(C) A-III, B-I, C-II, D-IVYOUR ANSWER
(D) A-I, B-III, C-IV, D-II
γ is 1.67 for monatomic, 1.40 for diatomic and 1.33 for polyatomic. You placed C and D correctly in spirit but swapped the last two items, matching the polyatomic gas to 8.3 and Cp − Cv to 1.33. The planted swap in this question sits in exactly those two rows.
In a match question, do the row you are surest of first, cross out every option that contradicts it, and only then work on the rest.
Q72 — you chose (C), the answer is (B)statement count Ā· moderate
How many of the following statements are correct? / A. In an isochoric process the gas does no work. / B. Heating one mole of a monatomic ideal gas through 200 K at constant volume needs 2490 J of heat. / C. In a free expansion the gas does work against the vacuum. / D. Over one complete cycle the net heat taken in by the gas…
(A) Only one
(B) Only twoCORRECT
(C) Only threeYOUR ANSWER
(D) All four
Three statements, not four. Statement C says a gas does work in a free expansion — it does not, because a vacuum pushes back with no pressure. Statement D reads like the standard cycle result, but what is zero round a cycle is ΔU, not Q.
On a how many are correct question, write T or F beside each statement before you look at the counts.
Assertion–Reason
Seven of eight — strong. The one loss is the same shape as both A–R losses on the Kinetic Theory paper: a true reason accepted as the right reason. Across the two papers you answered (A) on thirteen of sixteen; when R really explains A you are ten from ten, and when it does not you are three from six.
Q11 — you chose (A), the answer is (B)assertionāreason Ā· look-alike: internal energy as a state variable vs work as a path variable Ā· moderate
Assertion (A): The relation ΔQ = ΔU + ΔW holds for every process an ideal gas can be taken through. / Reason (R): ΔU depends only on the initial and the final states of the gas.A: T / F — ____ R true on its own? Y / N — ____ A because R? ____
(A) Both A and R are true and R is the correct explanation of AYOUR ANSWER
(B) Both A and R are true but R is NOT the correct explanation of ACORRECT
(C)A is true but R is false
(D)A is false but R is true
Both are true. But R says ΔU depends only on the end states — that is a fact about internal energy being a state function. A says ΔQ = ΔU + ΔW holds for every process, which is the first law, and it is true for a quite different reason: energy conservation. R is true and relevant and still does not explain A, so the answer is (B).
Nothing left blank — and that was right
You attempted all eighty. Eleven were wrong, which cost eleven marks, and leaving them blank would have cost the same eleven in forgone marks only if you would otherwise have got them right. With four options and negative marking, attempting is correct whenever you can rule out two. The Kinetic Theory paper went the other way — two easy questions left blank for eight marks — so the judgement to make is not attempt or not but have I eliminated two?
The response grid
Page 13 is the printed response grid and it is completely empty, as it is on the Kinetic Theory script. Five papers now with nothing transferred. In the hall the OMR sheet is the only thing marked — start filling it question by question as you answer, so that it is a habit rather than a job at the end.
What to do before the next paper
Four numbers before any physics.
On every P–V figure, write the coordinates of each marked point down the side of the page before you start. That one step covers Q9, Q39 and Q69 — twelve marks, and all three of your hard losses.
Constant pressure picks Cp.
Underline the words at constant pressure or at constant volume in the stem, and write the matching capacity next to them before computing.
Sign on the page.
The moment you read falls, released or done on the gas, write the minus sign down.
Keep whatever you changed.
31 of 32 on the easy questions, after three papers at 26. That is the biggest single improvement in the record so far — hold on to it.
Fill the grid as you go.
Every paper from here.
Marked from the red-pen script. Every tick was read by its vertex and the whole paper was transcribed twice, by two independent methods, with no disagreement on any of the 80 answers.