ΔU = q + w, where q is heat added to system, w is work done ON system (chemistry sign convention; physics uses w = work done BY). Energy is conserved.
-- NCERT Class 11 Chemistry, Ch. 5, p. 4First Law Thermo
8 MCQs9-step worked example
Source: NCERT ThermodynamicsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The first law of thermodynamics is a conservation law: energy cannot be created or destroyed, only transferred between a system and its surroundings as heat or work.
In the chemistry sign convention (NCERT Class 11 Chemistry Chapter 5, page 4), the law is written as:
ΔU = q + w
where ΔU is the change in internal energy, q is heat exchanged, and w is work done on the system. This sign convention is the single detail that trips up NEET aspirants on this topic — physics writes w as work done by the system (ΔU = q − w), and mixing the two conventions flips the sign of the work term.
How the signs work in chemistry convention:
- Heat absorbed by system → q is positive.
- Heat released by system → q is negative.
- Work done on system (compression) → w is positive.
- Work done by system (expansion) → w is negative.
For an ideal gas expanding against external pressure, w = −P_ext ΔV. If the gas expands (ΔV > 0), the system does work on the surroundings, so w is negative — the system loses energy through work.
Three special cases you must recognize instantly:
- Adiabatic process (q = 0): ΔU = w. All energy change comes from work.
- Isochoric process (ΔV = 0, so w = 0): ΔU = q_v. Heat at constant volume directly measures ΔU.
- Cyclic process (system returns to initial state): ΔU = 0, so q = −w.
The NEET-relevant confusion: when a question states "work done by the gas is 200 J," you must enter w = −200 J into ΔU = q + w (chemistry convention). Writing +200 J is the sign-convention trap.
Internal energy is a state function — its change depends only on initial and final states, not the path. Heat and work are path functions. This distinction appears in conceptual NEET questions that ask which quantity is path-independent.
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 1Direct ApplicationPractice
According to the first law of thermodynamics (chemistry convention), if a system absorbs 500 J of heat and has 200 J of work done on it, what is ΔU?
MCQ 2Direct ApplicationPractice
A gas expands against a constant external pressure and does 150 J of work on the surroundings. In the chemistry sign convention, what is the value of w?
MCQ 3Easy RecallPractice
For a cyclic process, which of the following is true?
MCQ 4Easy RecallPractice
Which of the following quantities is a state function?
MCQ 5Direct ApplicationPractice
In an adiabatic process, a gas is compressed and 300 J of work is done on it. What is ΔU?
MCQ 6Direct ApplicationPractice
For a process at constant volume, a system releases 400 J of heat to its surroundings. What is ΔU for the system?
MCQ 7Direct ApplicationPractice
A system absorbs 1000 J of heat and expands, doing 400 J of work on the surroundings. What is ΔU of the system (chemistry convention)?
MCQ 8Concept TrapPractice
An ideal gas undergoes free expansion into vacuum. Which of the following is correct?
Worked Example
- 1
Given
A gas absorbs 2.5 kJ of heat at constant pressure. During this process, the gas expands and does 0.8 kJ of work on the surroundings.
- 2
Required
Find the change in internal energy (ΔU) of the system.
- 3
Concept
First law of thermodynamics in chemistry convention: ΔU = q + w, where w is work done on the system. Since the gas does work on the surroundings, the work done on the system is the negative of that value.
- 4
Formula
ΔU = q + w
- 5
Substitution
q = +2.5 kJ (heat absorbed → positive) Work done by system on surroundings = 0.8 kJ → w = −0.8 kJ (chemistry convention) ΔU = 2.5 + (−0.8)
- 6
Calculation
ΔU = 2.5 − 0.8 = 1.7 kJ
- 7
Final answer
**ΔU = +1.7 kJ** The system's internal energy increases by 1.7 kJ. The heat absorbed (2.5 kJ) is split: 0.8 kJ goes into expansion work against the surroundings, and the remaining 1.7 kJ raises the internal energy. Note on exact values: The quantities 2.5 kJ and 0.8 kJ are given values in the problem statement; the arithmetic 2.5 − 0.8 = 1.7 is exact within the precision of the given data.
- 8
Common trap
Writing w = +0.8 kJ (forgetting to negate when the gas does work on surroundings). This gives ΔU = 2.5 + 0.8 = 3.3 kJ — the classic sign-convention error between physics and chemistry conventions. Always ask: "Is this work done ON the system or BY the system?"
- 9
Similar NEET-style question
A system at constant pressure absorbs 5.0 kJ of heat while expanding against external pressure, doing 1.2 kJ of work. Calculate ΔU. (Answer: ΔU = 5.0 + (−1.2) = 3.8 kJ) ---
Before solving, remember these
Formulas
Enthalpy and ΔH at constant P
Enthalpy = internal energy + PV. At constant pressure, heat absorbed equals enthalpy change.
| Symbol | Quantity | SI Unit |
|---|---|---|
| H | enthalpy | J |
| U | internal energy | J |
| P | pressure | Pa |
| V | volume | m^3 |
Valid when
- At constant pressure
- Closed system
First law of thermodynamics (chemistry)
Internal energy change = heat added to system + work done ON system. Note: chemistry uses w as work done ON system; physics uses w as work done BY.
| Symbol | Quantity | SI Unit |
|---|---|---|
| ΔU | internal energy change | J |
| q | heat | J |
| w | work on system | J |
Valid when
- Closed system
- Sign convention chosen consistently
Gibbs free energy
Spontaneity criterion. ΔG<0: spontaneous. ΔG=0: equilibrium. ΔG>0: non-spontaneous.
| Symbol | Quantity | SI Unit |
|---|---|---|
| ΔG | Gibbs energy change | kJ |
| ΔH | enthalpy | kJ |
| ΔS | entropy | kJ/K |
| T | temperature | K |
Valid when
- Constant T and P
ΔG° and K
Standard free energy change relates to equilibrium constant. K>1: ΔG°<0; K<1: ΔG°>0.
| Symbol | Quantity | SI Unit |
|---|---|---|
| ΔG° | standard free energy | J/mol |
| R | gas constant 8.314 | J/mol/K |
| T | temp | K |
| K | equilibrium constant | - |
Valid when
- Standard state
- Equilibrium
Hess's law
Total enthalpy change is independent of path. Useful for computing ΔH of reactions not directly measurable.
| Symbol | Quantity | SI Unit |
|---|---|---|
| ΔH | enthalpy change | J or kJ |
Valid when
- State function (path-independent)
- Same initial/final states
Standard enthalpy of reaction
From standard formation enthalpies. ΔH°_f of element in standard state = 0.
| Symbol | Quantity | SI Unit |
|---|---|---|
| ΔH°_f | standard formation enthalpy | kJ/mol |
Valid when
- Standard state (1 bar, 298 K)
- Stoichiometric coefficients applied
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.
Category: Sign Convention
When reversing a reaction, ΔH changes sign. Multiply: same as multiply ΔH.
When it triggers
Hess's law problem with combination of multiple reactions.
How to avoid
If reaction is reversed: ΔH → -ΔH. If multiplied by factor n: ΔH → n×ΔH. Apply systematically when combining.
Root cause: sign error
Correction
K > 1: ln K > 0 → ΔG° < 0 (forward favoured). K < 1: ln K < 0 → ΔG° > 0 (reverse favoured). At equilibrium K=1, ΔG°=0.
Root cause: sign error
Correction
Reversing a reaction: ΔH → -ΔH. Multiplying by n: ΔH → n·ΔH. Apply consistently before summing.
Root cause: concept gap
Correction
Spontaneity from ΔG = ΔH - TΔS. Endothermic reactions can be spontaneous if TΔS > ΔH (ice melting at room T).
Past Year Questions
8 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
1A, B and C only
2A and B only
3A and C only
4B, C and D only
NTA Answer: Option 1(final)
NEET 2024Revised key
1A and C
2A, B and D
3A, C and D
4C and D
NTA Answer: Option 3(revised_final)
NEET 2024Revised key
1A-IV, B-III, C-II, D-I
2A-IV, B-II, C-III, D-I
3A-I, B-II, C-III, D-IV
4A-II, B-III, C-IV, D-I
NTA Answer: Option 4(revised_final)
NEET 2024Revised key
10 calorie
2–413.14 calories
3413.14 calories
4100 calories
NTA Answer: Option 2(revised_final)
NEET 2023
Which amongst the following molecules on polymerization produces neoprene? Cl |
1H C C–CHCH
2H CCH–CCH 2 2 2 CH 3 |
3H C C –CHCH
4H CCH–CHCH 2 2 2 2
NTA Answer: Option 1(final)
NEET 2023
1–137.26 cal
2–1381.80 cal
3–13.73 cal
41372.60 cal
NTA Answer: Option 2(final)
NEET 2022
1zero order (y = rate and x = concentration), first order (y = rate and x = t ) ½
2zero order (y = concentration and x = time), first order (y = t and x = concentration) ½
3zero order (y = concentration and x = time), first order (y = rate constant and x = concentration)
4zero order (y = rate and x = concentration), first order (y = t and x = concentration) ½
NTA Answer: Option 4(final)
NEET 2021
10, = 0
2= 0, = 0 total total ΔU ≠ ΔS ≠ ΔU ΔS ≠
30, 0
4= 0, 0 total total
NTA Answer: Option 4(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Combine multiple thermochemical equations using Hess's law to find ΔH of target reaction.
Multi StepMedium
Common distractors
wrong sign when reversing
Forgets to negate ΔH when reversing equation
Predict spontaneity from ΔH and ΔS at given T using ΔG = ΔH - TΔS.
InterpretationMedium
Common distractors
ignores temperature
Uses ΔH alone for spontaneity
Sources
NCERT refs: Class 11 Chemistry Chapter 5, p.4
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