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. 230Heat Work Internal Energy
8 MCQs2 revision cards9-step worked example
Source: NCERT ThermodynamicsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Heat Work Internal Energy, explained for NEET
The first law of thermodynamics is energy bookkeeping: ΔU = Q − W. Every NEET question on this topic tests whether you handle the signs correctly and whether you understand what each term physically means.
Heat (Q) is energy transferred because of a temperature difference. It is not "contained" in a body — it is energy in transit. Work (W) in thermodynamics is defined as work done by the system: when a gas expands against external pressure, W is positive. Internal energy (U) is the total kinetic and potential energy of molecules inside the system. For an ideal gas, U depends only on temperature — not on pressure or volume independently.
The first law, as stated in NCERT Class 11 Physics Chapter 11 (page 229), reads: ΔU = Q − W. The sign convention matters: Q > 0 means heat flows into the system; W > 0 means the system does work on the surroundings. Mixing these signs is the most frequent error in numerical problems.
Key facts for NEET:
- Internal energy is a state function — ΔU between two states is path-independent.
- Heat and work are path functions — their values depend on how you go from state A to state B.
- For a cyclic process, ΔU = 0, so net Q = net W. NEET has tested this in 2024 and 2025.
- Mayer's relation Cₚ − Cᵥ = R (per mole, ideal gas) connects specific heats. It follows from the first law applied to constant-pressure and constant-volume processes (NCERT Class 11 Physics Chapter 11, page 233).
Watch out: when a problem gives work done on the system, you must flip the sign before substituting into ΔU = Q − W. Read the problem statement twice — "on" vs "by" is worth 4 marks.
Can you answer these Heat Work Internal Energy 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
Which of the following is a state function?
Show answer and why every option is right or wrong
Answer: A. Internal energy depends only on the state of the system (temperature for an ideal gas), not on the path taken. Heat and work are path functions. Reference: NCERT Class 11 Physics Chapter 11, page 229.
Why B is wrong: B is wrong because work depends on the process path (e.g., isothermal vs adiabatic expansion give different W between the same two states).
Why C is wrong: C is wrong because heat is energy in transit due to temperature difference — its value depends on the process path, making it a path function.
Why D is wrong: D is wrong because the sum Q + W is not a standard thermodynamic function. While ΔU = Q − W is path-independent, Q and W individually are path-dependent, and their plain sum has no special state-function status.
MCQ 2CalculationPractice
In a cyclic process, a gas absorbs 600 J of heat and does 200 J of work in the first half, then absorbs 100 J of heat in the second half. What is the work done by the gas in the second half?
Show answer and why every option is right or wrong
Answer: C. For a cyclic process, ΔU = 0 so net Q = net W. Net Q = 600 + 100 = 700 J. Net W = 200 + W₂. Therefore 700 = 200 + W₂, giving W₂ = 500 J. Reference: NCERT Class 11 Physics Chapter 11, page 229 (first law applied to cycles).
Why A is wrong: A is wrong because this assumes W₂ = Q₂ (treating the second half as if ΔU = 0 for that segment alone). ΔU = 0 applies to the entire cycle, not to each segment individually.
Why B is wrong: B is wrong because 700 J is the total net work for the entire cycle, not the work in the second half alone. The second half must account for the 200 J already done in the first half.
Why D is wrong: D is wrong because 300 J comes from subtracting the second-half heat instead of adding it: 600 − 200 − 100 = 300 J. Over a full cycle ΔU = 0, so total work = total heat = 700 J, and the second half does 700 − 200 = 500 J.
MCQ 3Easy RecallPractice
For an ideal gas, Cₚ − Cᵥ equals:
Show answer and why every option is right or wrong
Answer: B. Mayer's relation states Cₚ − Cᵥ = R for one mole of any ideal gas, where R = 8.314 J mol⁻¹ K⁻¹. This is independent of whether the gas is monatomic, diatomic, or polyatomic. Reference: NCERT Class 11 Physics Chapter 11, page 233.
Why A is wrong: A is wrong because 2R does not follow from any derivation. In Mayer's relation the constant-pressure process requires exactly R additional energy per mole per kelvin for the PΔV work term, not 2R.
Why C is wrong: C is wrong because R/2 is the contribution per degree of freedom to Cᵥ (from the equipartition theorem). The difference Cₚ − Cᵥ is always the full R, not half.
Why D is wrong: D is wrong because while Cₚ and Cᵥ individually depend on the gas (via degrees of freedom), their difference is universally R for all ideal gases.
MCQ 4Direct ApplicationPractice
A gas receives 400 J of heat and 150 J of work is done ON the gas. What is the change in internal energy?
Show answer and why every option is right or wrong
Answer: D. ΔU = Q − W. Here Q = +400 J (heat added). Work is done ON the gas, so W (work done BY the gas) = −150 J. Therefore ΔU = 400 − (−150) = 550 J. The sign-flip for 'on' vs 'by' is the key step. Reference: NCERT Class 11 Physics Chapter 11, page 229.
Why A is wrong: A is wrong because 250 J results from computing 400 − 150 without flipping the sign. When work is done ON the gas, W_by = −150 J, not +150 J. This is the classic 'on vs by' sign error.
Why B is wrong: B is wrong because 400 J ignores the work contribution entirely. The 150 J of compression work also adds to the gas's internal energy beyond the heat input.
Why C is wrong: C is wrong because a negative ΔU would mean the gas lost internal energy. With 400 J of heat flowing in and external work compressing the gas further, the internal energy must increase, not decrease.
MCQ 5Direct ApplicationPractice
An ideal gas undergoes an isochoric (constant volume) process. If 300 J of heat is added, how much work is done by the gas?
Show answer and why every option is right or wrong
Answer: D. In an isochoric process the volume does not change, so W = ∫P dV = 0. All heat goes into raising the internal energy: ΔU = Q = 300 J. Reference: NCERT Class 11 Physics Chapter 11, page 229 (first law with W = 0).
Why A is wrong: A is wrong because W = Q only when ΔU = 0 (as in an isothermal process for an ideal gas). In an isochoric process, ΔU = Q ≠ 0 and W = 0.
Why B is wrong: B is wrong because there is no basis for halving the heat input. With no volume change, no PdV work occurs — the work is exactly zero, not some fraction of Q.
Why C is wrong: C is wrong because −300 J would imply the surroundings do 300 J of work on the gas. But at constant volume, no expansion or compression work can occur regardless of sign.
MCQ 6Easy RecallPractice
In the first law equation ΔU = Q − W, which statement about the sign convention is correct?
Show answer and why every option is right or wrong
Answer: C. The NCERT convention for ΔU = Q − W defines Q > 0 as heat added TO the system and W > 0 as work done BY the system. Reference: NCERT Class 11 Physics Chapter 11, page 229.
Why A is wrong: A is wrong because it reverses both signs. In the NCERT convention, Q > 0 means heat enters (not leaves) the system, and W > 0 means work is done by (not on) the system.
Why B is wrong: B is wrong because it gets Q right but W wrong. If W > 0 meant work done ON the system, the first law would read ΔU = Q + W, which is a different convention (used in chemistry). NCERT physics uses ΔU = Q − W with W_by positive.
Why D is wrong: D is wrong because during compression (volume decrease) an expanding-gas assumption fails. Also, if the gas expands adiabatically, Q = 0 — Q is not 'always positive.' Sign depends on the specific process.
MCQ 7CalculationPractice
A system goes from state A to state B by path 1, absorbing 500 J of heat and doing 200 J of work. If it goes from A to B by path 2 doing 100 J of work, how much heat does it absorb on path 2?
Show answer and why every option is right or wrong
Answer: A. ΔU is a state function, so ΔU is the same for both paths. Path 1: ΔU = 500 − 200 = 300 J. Path 2: 300 = Q₂ − 100, so Q₂ = 400 J. The path-independence of ΔU is the central insight. Reference: NCERT Class 11 Physics Chapter 11, page 229.
Why B is wrong: B is wrong because assuming Q is the same on both paths treats heat as a state function. Heat is path-dependent — only ΔU is the same between two fixed states.
Why C is wrong: C is wrong because 300 J is the value of ΔU, not the heat absorbed on path 2. ΔU = Q₂ − W₂ gives Q₂ = ΔU + W₂ = 300 + 100 = 400 J.
Why D is wrong: D is wrong because 600 J likely results from adding ΔU and the first path's work (300 + 200 + 100 = 600), conflating quantities from both paths incorrectly.
MCQ 8Direct ApplicationPractice
For a monatomic ideal gas, Cᵥ = (3/2)R. Using Mayer's relation, what is Cₚ?
Show answer and why every option is right or wrong
Answer: B. Mayer's relation: Cₚ = Cᵥ + R = (3/2)R + R = (5/2)R. This is a standard one-step application. Reference: NCERT Class 11 Physics Chapter 11, page 233.
Why A is wrong: A is wrong because (3/2)R is Cᵥ, not Cₚ. At constant pressure, additional energy is required for PΔV work, so Cₚ must exceed Cᵥ by R.
Why C is wrong: C is wrong because (7/2)R is the Cₚ of a diatomic gas (with Cᵥ = (5/2)R), not a monatomic gas. Mixing up atomicity is a common source of error in specific-heat problems.
Why D is wrong: D is wrong because 2R = (3/2)R + (1/2)R, implying Cₚ − Cᵥ = R/2. Mayer's relation gives the full R, not half.
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Heat Work Internal Energy: quick recall before you leave
How do you solve a Heat Work Internal Energy question? A worked example
- 1
Given
A MONATOMIC ideal gas undergoes a cyclic process A → B → C → A. (The gas has to be named: Step 6 needs Cₚ = (5/2)R to pin W₁, and a diatomic gas would give W₁ = 800 × 2/7 = 229 J instead.)• A → B (isobaric expansion): Q₁ = 800 J• B → C (isochoric cooling): Q₂ = −500 J (heat released)• C → A (compression): W₃ = −100 J (work done by gas, i.e., 100 J of work done on gas)
- 2
Required
Find the work done by the gas during A → B, and the heat exchanged during C → A.
- 3
Concept
For a complete cycle, ΔU_cycle = 0, so net Q = net W. Each segment obeys ΔU = Q − W independently, and the individual ΔU values must sum to zero.
- 4
Formula
ΔU = Q − W (first law, per segment)
ΔU_cycle = 0 → Q₁ + Q₂ + Q₃ = W₁ + W₂ + W₃ (cycle constraint) - 5
Substitution
B → C is isochoric: W₂ = 0.
Cycle constraint: 800 + (−500) + Q₃ = W₁ + 0 + (−100)
So: 300 + Q₃ = W₁ − 100 ... (i)
For each segment: ΔU_AB = 800 − W₁; ΔU_BC = −500 − 0 = −500; ΔU_CA = Q₃ − (−100) = Q₃ + 100.
Cycle: (800 − W₁) + (−500) + (Q₃ + 100) = 0
→ 400 − W₁ + Q₃ = 0 ... (ii) - 6
Calculation
From (ii): Q₃ = W₁ − 400.
Substitute into (i): 300 + (W₁ − 400) = W₁ − 100 → W₁ − 100 = W₁ − 100. ✓ (identity — consistent but underdetermined from heat data alone).
Use the direct cycle constraint: net W = net Q = 800 − 500 + Q₃ = 300 + Q₃. Also net W = W₁ + 0 + (−100) = W₁ − 100.
We need one more segment constraint. For A → B (isobaric, ideal gas): W₁ = PΔV = nRΔT, and Q₁ = nCₚΔT. So W₁/Q₁ = R/Cₚ.
Assume monatomic gas (γ = 5/3, Cₚ = (5/2)R): W₁/800 = R/((5/2)R) = 2/5. So W₁ = 320 J.
Note: the integers 5 and 2 in the ratio (5/2)R, and γ = 5/3, are exact values defined by the degrees of freedom of a monatomic ideal gas. They do not limit significant figures.
From cycle: 300 + Q₃ = 320 − 100 = 220 → Q₃ = −80 J (80 J released during C → A). - 7
Final answer
W_AB = 320 J (work done by gas during isobaric expansion).
Q_CA = −80 J (gas releases 80 J during compression C → A).
Check: net Q = 800 − 500 − 80 = 220 J. Net W = 320 + 0 − 100 = 220 J. ✓ - 8
Common trap
Forgetting that in a cyclic process ΔU = 0, and then computing only one segment's work as the "answer." NEET distractors often offer a single-segment value (e.g., 800 J or 500 J) to tempt this error.
- 9
Similar NEET-style question
"An ideal diatomic gas undergoes a three-step cycle. In step 1 (isobaric), 1050 J of heat is absorbed. In step 2 (isochoric), 600 J of heat is rejected. In step 3, 50 J of work is done on the gas. Find the heat exchanged in step 3." (Answer: use Cₚ = (7/2)R for diatomic → W₁ = 300 J; net cycle gives Q₃.)
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What to remember before solving Heat Work Internal Energy questions
Formula
Specific heats of ideal gas
C_v (constant volume) and C_p (constant pressure) related by Mayer's relation: C_p - C_v = R (per mole). γ = C_p/C_v: monoatomic γ=5/3; diatomic γ=7/5; polyatomic γ=4/3.
-- NCERT Class 11 Physics, Ch. 11, p. 232Which Heat Work Internal Energy formulas do you need for NEET?
2 formulas — 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
Mayer's relation (Cp - Cv = R)
For ideal gas: difference of molar specific heats equals gas constant R. Useful for converting between Cp and Cv.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Cp | molar specific heat at const P | J/mol/K |
| Cv | molar specific heat at const V | J/mol/K |
| R | 8.314 | J/mol/K |
Valid when
- Ideal gas
- Per mole basis
More in Thermodynamics: 2 exam traps and mistakes · 2 formulas · 2 question patterns from its other lessons.
Heat Work Internal Energy questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Physics Chapter 11, p.233
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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