ΔU = Q - W, where ΔU is change in internal energy, Q is heat ADDED to system, W is work DONE BY system. Statement of energy conservation including thermal energy.
-- NCERT Class 11 Physics, Ch. 11, p. 230First Law Thermodynamics
8 MCQs7 revision cards9-step worked example
Source: NCERT ThermodynamicsPYQ coverage: NEET 2024, 2025Official key: NTA-verifiedLast updated: 24 Sep 2026
First Law Thermodynamics, explained for NEET
The first law of thermodynamics is energy conservation applied to thermal systems. For a closed system: ΔU = Q − W, where Q is heat added to the system and W is work done by the system (NCERT Class 11 Physics Chapter 11, page 229). Internal energy U is a state function — it depends only on the current state, not on the path taken. Heat and work are path functions — their values depend on how the process is carried out.
The trap that costs marks: sign-convention confusion. NEET problems frequently give "work done ON the system" while the formula uses "work done BY the system." If 400 J of work is done ON a gas, then W = −400 J in ΔU = Q − W. Misreading the sign flips your entire answer. Every time you see a first-law problem, your first move is: identify whether W is stated as done BY or done ON the system, and convert.
Cyclic processes are a common NEET pattern (observed in 2024 and 2025 papers). In any complete cycle the system returns to its initial state, so ΔU_cycle = 0. The first law then gives: Q_net = W_net for the full cycle. A frequent distractor computes work for only one segment and presents it as the answer for the whole cycle.
Special cases worth memorising:
- Isochoric (constant volume): W = 0, so ΔU = Q. All heat goes to internal energy.
- Isobaric (constant pressure): W = PΔV. Heat splits between internal energy change and expansion work.
- Adiabatic: Q = 0, so ΔU = −W. The gas does work at the expense of its own internal energy.
Mayer's relation C_p − C_v = R (per mole, ideal gas) connects the two specific heats and appears in problems requiring you to find ΔU = nC_vΔT when only C_p or γ is given.
Watch-out: when a problem says "heat is supplied at constant pressure," the internal energy change is NOT equal to Q — part of Q goes to PΔV work. Only at constant volume does ΔU = Q.
Can you answer these First Law Thermodynamics MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In the first law of thermodynamics ΔU = Q − W, the quantity ΔU is:
Show answer and why every option is right or wrong
Answer: C. Internal energy is a state function — its change depends only on the initial and final thermodynamic states, not the path (NCERT Class 11 Physics Chapter 11, page 229).
Why A is wrong: A is wrong because internal energy change depends only on initial and final states, not on the path. Heat and work are path functions, not ΔU.
Why B is wrong: B is wrong because ΔU = Q only when W = 0 (isochoric process). In general, ΔU = Q − W, so heat supplied is split between internal energy change and work.
Why D is wrong: D is wrong because ΔU = −W only when Q = 0 (adiabatic process). In general, work done is not equal to the internal energy change.
MCQ 2Easy RecallPractice
For a cyclic process performed on a closed system, which of the following is correct?
Show answer and why every option is right or wrong
Answer: A. In a cyclic process the system returns to its initial state, so ΔU = 0 (state function). The first law then gives Q_net = W_net (NCERT Class 11 Physics Chapter 11, page 229).
Why B is wrong: B is wrong because while ΔU = 0 for a cycle, Q_net and W_net are each non-zero in general — they are equal to each other, not individually zero.
Why C is wrong: C is wrong because ΔU = 0 for any complete cycle (internal energy is a state function and returns to its starting value).
Why D is wrong: D is wrong because W_net is generally non-zero for a cycle — only ΔU is zero. The net work equals the area enclosed by the cycle on a P-V diagram.
MCQ 3Easy RecallPractice
Mayer's relation for one mole of an ideal gas is:
Show answer and why every option is right or wrong
Answer: B. Mayer's relation states C_p − C_v = R for one mole of an ideal gas, where R = 8.314 J mol⁻¹ K⁻¹ (NCERT Class 11 Physics Chapter 11, page 233).
Why A is wrong: A is wrong because Mayer's relation involves the difference of specific heats, not their sum. C_p + C_v has no standard thermodynamic identity.
Why C is wrong: C is wrong because the product C_p × C_v has no standard relation equal to R. Mayer's relation is about their difference.
Why D is wrong: D is wrong because C_p / C_v defines the adiabatic index γ, not Mayer's relation. γ is always greater than 1 for ideal gases, while R ≈ 8.314 J mol⁻¹ K⁻¹.
MCQ 4Direct ApplicationPractice
A gas in a closed container (rigid walls) is heated and absorbs 500 J of heat. The change in internal energy of the gas is:
Show answer and why every option is right or wrong
Answer: A. Rigid walls mean constant volume, so W = 0. By the first law: ΔU = Q − W = 500 − 0 = 500 J. The gas type is irrelevant here (NCERT Class 11 Physics Chapter 11, page 229).
Why B is wrong: B is wrong because ΔU = 0 would require Q = W. Here W = 0 (rigid container), so ΔU = Q = 500 J, not zero.
Why C is wrong: C is wrong because a negative ΔU would mean the gas lost internal energy. The gas absorbed heat at constant volume, so all 500 J went into increasing internal energy.
Why D is wrong: D is wrong because at constant volume, W = 0 regardless of gas type, giving ΔU = Q directly. No additional information is needed.
MCQ 5Direct ApplicationPractice
In an isobaric expansion, a gas absorbs 800 J of heat and does 300 J of work on its surroundings. The change in internal energy is:
Show answer and why every option is right or wrong
Answer: B. Apply ΔU = Q − W = 800 − 300 = 500 J. The heat is split: 300 J goes to expansion work, 500 J goes to raising internal energy (NCERT Class 11 Physics Chapter 11, page 229).
Why A is wrong: A is wrong because 1100 J results from adding Q + W instead of subtracting. The first law is ΔU = Q − W, not Q + W. (Sign-convention trap: W here is work done BY the system, so it is subtracted.)
Why C is wrong: C is wrong because 300 J is the work done, not the internal energy change. Only part of the absorbed heat goes to work; the rest (500 J) increases internal energy.
Why D is wrong: D is wrong because 800 J is the total heat absorbed. At constant pressure, not all absorbed heat goes to internal energy — some is used to do PΔV expansion work.
MCQ 6Direct ApplicationPractice
200 J of work is done ON an ideal gas during a process in which the gas also loses 150 J of heat to the surroundings. The change in internal energy is:
Show answer and why every option is right or wrong
Answer: D. Convert to the BY convention: work done BY the system W = −200 J; heat added Q = −150 J (heat lost). ΔU = Q − W = (−150) − (−200) = +50 J (NCERT Class 11 Physics Chapter 11, page 229).
Why A is wrong: A is wrong because −350 J results from incorrectly treating both quantities as negative and adding them (−150 + (−200)). The sign convention requires W done ON to be converted: W_by = −200 J, then ΔU = Q − W_by. (Sign-convention trap.)
Why B is wrong: B is wrong because +350 J results from adding absolute values (150 + 200). This ignores that heat was lost (Q < 0). The correct calculation accounts for the opposing effects of work input and heat loss.
Why C is wrong: C is wrong because −50 J results from reversing the signs: ΔU = (−200) − (−150) = −50. This swaps which quantity is Q and which is W. Remember: ΔU = Q − W, not W − Q. (Sign-convention trap.)
MCQ 7CalculationPractice
An ideal gas completes a cyclic process consisting of two steps: in step 1, 600 J of heat is absorbed by the gas and 200 J of work is done by the gas; in step 2, the gas is returned to its initial state at constant volume. The heat exchanged in step 2 is:
Show answer and why every option is right or wrong
Answer: B. B is correct. Step 1: ΔU₁ = Q₁ − W₁ = 600 − 200 = +400 J. Over a complete cycle the gas returns to its starting state, so ΔU_cycle = 0 and therefore ΔU₂ = −400 J. Step 2 is at constant volume, so no work is done: W₂ = 0. The first law for step 2 then gives Q₂ = ΔU₂ + W₂ = −400 + 0 = −400 J, and the negative sign means 400 J is RELEASED by the gas. The constant-volume statement is what makes the step solvable — without it, only the combination Q₂ − W₂ = −400 J is fixed, and the heat alone could take any value.
Why A is wrong: A is wrong because the gas must lose internal energy in step 2 to return to its initial state (ΔU₂ = −400 J). Absorbing 400 J more heat would further increase internal energy, moving away from the initial state. (Cycle-closure trap: forgetting ΔU_cycle = 0.)
Why C is wrong: C is wrong because 200 J released accounts only for the work done in step 1, not the full internal energy change. ΔU₁ = 600 − 200 = 400 J, so the gas must shed 400 J of internal energy in step 2, not 200 J. (Single-segment trap: computing only one part of the cycle.)
Why D is wrong: D is wrong because 600 J released mirrors the 600 J absorbed in step 1, as though heat alone had to balance over the cycle. It is the WORK and heat together that balance: the gas keeps 400 J of the absorbed 600 J as internal energy in step 1, and it is that 400 J it must shed in step 2.
MCQ 8CalculationPractice
An ideal monatomic gas (C_v = 3R/2) at temperature 300 K absorbs 1000 J of heat at constant pressure. If the gas contains 2 mol, the rise in temperature is closest to:
Show answer and why every option is right or wrong
Answer: C. C is correct. For a monatomic ideal gas C_p = C_v + R = 3R/2 + R = 5R/2 (Mayer's relation, NCERT Class 11 Physics Chapter 11, page 233). At constant pressure Q = nC_pΔT, so ΔT = Q/(nC_p) = 1000/(2 × 5 × 8.314/2) = 1000/41.57 ≈ 24 K. The starting temperature, 300 K, does not enter: for an ideal gas the heat needed depends only on the change.
Why A is wrong: A is wrong because 48 K comes from using n = 1 instead of n = 2: ΔT = 1000/(1 × 5R/2) = 48 K. The C_p is right; the number of moles is not.
Why B is wrong: B is wrong because 40 K comes from using C_v instead of C_p: ΔT = 1000/(2 × 3R/2) = 40 K. At constant pressure the gas also does expansion work, so each kelvin costs an extra R per mole. (Trap: confusing C_v and C_p at constant pressure.)
Why D is wrong: D is wrong because 60 K comes from Q = nRΔT, i.e. counting only the expansion work PΔV = nRΔT and leaving out the rise in internal energy, nC_vΔT. The heat has to supply both.
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First Law Thermodynamics: quick recall before you leave
How do you solve a First Law Thermodynamics question? A worked example
Pattern: Cyclic process — net Q = net W (since ΔU_cycle = 0). Compute work/heat over each segment.
- 1
Given
A gas is taken round a three-leg cycle and back to its starting state.• Leg 1: the gas absorbs 600 J of heat and does 200 J of work.• Leg 2: the gas absorbs 100 J of heat and does 400 J of work.• Leg 3: the gas returns to its initial state, and 300 J of work is done ON the gas (so W₃ = −300 J).
- 2
Required
The net work done by the gas in one cycle, and the heat exchanged in leg 3.
- 3
Concept
Internal energy is a state function, so over a complete cycle ΔU_cycle = 0 and therefore Q_net = W_net. Apply the first law, ΔU = Q − W, to each leg with W counted as work done BY the gas.
- 4
Formula
• ΔU = Q − W for each leg• ΔU₁ + ΔU₂ + ΔU₃ = 0• W_net = W₁ + W₂ + W₃ = Q_net
- 5
Substitution
ΔU₁ = 600 − 200 = 400 J
ΔU₂ = 100 − 400 = −300 J
ΔU₃ = −(ΔU₁ + ΔU₂) = −(400 − 300) = −100 J - 6
Calculation
W_net = 200 + 400 + (−300) = 300 J
Q₃ = ΔU₃ + W₃ = −100 + (−300) = −400 J, so the gas gives out 400 J in leg 3.
Check: Q_net = 600 + 100 + (−400) = 300 J = W_net ✓
The values 600, 200, 100, 400 and 300 J are problem-defined exact quantities and do not limit the significant figures. - 7
Final answer
W_net = 300 J, and Q₃ = −400 J (400 J released). Note that the cycle law alone could not have given these: without the work in leg 3, Q_net = W_net is an identity with one unknown on each side. A cyclic-process problem is solvable only when every leg but one has both Q and W given, or the cycle's P–V area is known.
- 8
Common trap
The single-segment trap: computing only W₁ = 200 J (or W₁ + W₂ = 600 J) and reporting it as the cycle's net work. You must sum ALL segments. Also, sign-convention confusion: if W₃ = −300 J (work done on gas) is misread as +300 J, you'd get W_net = 900 J — wrong.
- 9
Similar NEET-style question
A gas completes a three-step cycle. Step A: absorbs 800 J of heat, does 500 J of work. Step B: releases 200 J of heat, has 100 J of work done on it. Step C: returns to initial state with 200 J of work done by the gas. Find Q in step C.
Solution sketch: ΔU_A = 300 J, ΔU_B = −200 − (−100) = −100 J, ΔU_C = −200 J. Q_C = ΔU_C + W_C = −200 + 200 = 0 J. This is an adiabatic return — Q_C = 0.
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What to remember before solving First Law Thermodynamics questions
Which First Law Thermodynamics formulas do you need for NEET?
1 formula — click to collapse
First law of thermodynamics
Change in internal energy = heat ADDED minus work DONE BY the system. Energy conservation including thermal energy.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Delta_U | change in internal energy | J |
| Q | heat added to system | J |
| W | work done BY system | J |
Valid when
- Closed system (no mass exchange)
- Sign convention: Q>0 heat in, W>0 system does work
More in Thermodynamics: 2 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
First Law Thermodynamics questions from past NEET papers
2 questions from NEET 2024, 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
11.8 atm
21.3 atm
31.6 atm
41.4 atm
NTA Answer: Option 3(final)
NEET 2024Revised key
1Zero
230 J
3–90 J
4–60 J
NTA Answer: Option 1(revised_final)
How does NEET ask about First Law Thermodynamics?
1 recurring pattern from past papers — click to collapse
Cyclic process: net Q = net W (since Delta_U cycle = 0). Compute work/heat over each segment.
Multi StepMedium
Common distractors
forgets net zero cycle
Computes only one segment's W
Sources
NCERT refs: Class 11 Physics Chapter 11, p.229
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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