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, p. 4Heat Work Internal Energy
8 MCQs2 revision cards9-step worked example
Source: NCERT ThermodynamicsPYQ coverage: NEET 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
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 4), 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 8).
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.
Practice 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?
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?
MCQ 3Easy RecallPractice
For an ideal gas, Cₚ − Cᵥ equals:
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?
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?
MCQ 6Easy RecallPractice
In the first law equation ΔU = Q − W, which statement about the sign convention is correct?
MCQ 7Direct ApplicationPractice
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?
MCQ 8Direct ApplicationPractice
For a monatomic ideal gas, Cᵥ = (3/2)R. Using Mayer's relation, what is Cₚ?
Quick recall before you leave
Worked Example
- 1
Given
An ideal gas undergoes a cyclic process A → B → C → A. - 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.) ---
Before solving, remember these
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, p. 8Formulas
4 formulas — click to collapse
Adiabatic relations for ideal gas
Relations holding during reversible adiabatic process. gamma = Cp/Cv.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | pressure | Pa |
| V | volume | m^3 |
| T | temperature | K |
| gamma | adiabatic index | - |
Valid when
- Q = 0 (no heat exchange)
- Quasi-static (reversible)
- Ideal gas
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
Work done in isothermal process (ideal gas)
Work done by ideal gas during isothermal expansion. Q = W (since Delta_U = 0). Reverse for compression.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | moles | mol |
| R | gas constant 8.314 | J/mol/K |
| T | temperature | K |
| V_i, V_f | initial/final volume | m^3 |
Valid when
- Ideal gas
- Quasi-static (reversible) isothermal
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
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Graph Interpretation
Student misidentifies which P-V curve is adiabatic (steeper) vs isothermal.
When it triggers
P-V graph showing one or more processes; question asks for process type.
How to avoid
Adiabatic curve is STEEPER than isothermal at the same point (slope ratio = γ). Adiabat: PV^γ; isotherm: PV = const.
Root cause: graph misread
Correction
Adiabatic curve is STEEPER than isothermal at the same P-V point. Slope ratio at same point: adiabatic / isothermal = γ. Memorise: 'adiabatic angles down sharper'.
Past Year Questions
4 questions from NEET 2022, 2023, 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)
NEET 2023
115°C
2100°C
3200°C
427°C
NTA Answer: Option 4(final)
NEET 2022
14
21
32
43
NTA Answer: Option 3(final)
How NEET usually asks this
2 recurring patterns 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
Identify P-V diagram or temperature behaviour of isothermal vs adiabatic process. Adiabatic: PV^gamma; isothermal: PV.
InterpretationMedium
Common distractors
swaps adiabatic isothermal curves
Confuses which curve is steeper
Sources
NCERT refs: Class 11 Physics Chapter 11, p.8
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