CBSE Class 11 Thermodynamics — Getting the Sign Conventions Right
In Class 11 Thermodynamics, most students do not lose marks because they forgot a formula. They lose marks because a minus sign went the wrong way. The physics of the chapter is short; the bookkeeping is where the paper is won or lost. This guide fixes the sign convention once, and then shows it working in numericals with every step of the arithmetic printed.
System, surroundings, and the one rule to remember
The system is the part of the universe you are studying — usually the reaction mixture. Everything else is the surroundings. Every sign in this chapter is written from the system's point of view. That single sentence is the rule. Ask yourself: is the system gaining or losing? Gaining is positive.
q < 0 → heat is released by the system (exothermic)
w > 0 → work is done on the system (compression)
w < 0 → work is done by the system (expansion)
The first law, written the NCERT way
Internal energy change equals heat added to the system plus work done on the system. Both inputs are "things the system receives", which is why they are added.
An honest note about textbooks. Not every book writes it this way. The IUPAC convention followed by NCERT is ΔU = q + w with w = work done on the system. Many older books, and many engineering texts, write ΔU = q − w with w = work done by the system. The physics is identical; only the labelling differs. If you are using a reference book alongside NCERT, check its definition of w on the first page of the chapter before you copy any solved example. Mixing the two conventions in one answer is the most common source of a wrong sign.
Pressure–volume work
For expansion or compression against a constant external pressure:
The minus sign is doing the thinking for you. Expansion means ΔV is positive, so w comes out negative — the system spent energy pushing the surroundings back. Compression means ΔV is negative, so w comes out positive — the surroundings pushed energy in. For a rigid closed container ΔV = 0, so w = 0 and ΔU = q.
For a reversible isothermal expansion of an ideal gas:
Useful unit conversion: 1 L bar = 100 J and 1 L atm ≈ 101.3 J.
Worked example 1. A gas expands from 2.0 L to 10.0 L against a constant external pressure of 1.0 bar. During the expansion it absorbs 300 J of heat. Find w and ΔU.
ΔV = 10.0 − 2.0 = 8.0 L
w = − pext ΔV = − (1.0 bar)(8.0 L) = − 8.0 L bar
w = − 8.0 × 100 = − 800 J (negative: the gas did work on the surroundings)
q = + 300 J (absorbed)
ΔU = q + w = 300 + (− 800) = − 500 J
Reading the answer: the gas gained 300 J as heat but spent 800 J pushing outward, so its internal energy fell by 500 J. The signs tell a story — always check that the story makes sense.
Enthalpy, and where Δng comes in
Enthalpy is defined as H = U + pV. For a reaction at constant temperature involving gases:
Δng = (moles of gaseous products) − (moles of gaseous reactants)
Only gaseous species count in Δng; solids and liquids are ignored because their volume change is negligible. When Δng = 0, ΔH = ΔU exactly.
Worked example 2. For N2(g) + 3H2(g) → 2NH3(g), ΔH = − 92.4 kJ mol−1 at 298 K. Find ΔU.
Δng = 2 − (1 + 3) = − 2
ΔngRT = (− 2)(8.314 J K−1 mol−1)(298 K) = − 4955 J mol−1 = − 4.955 kJ mol−1
ΔU = ΔH − ΔngRT = (− 92.4) − (− 4.955) = − 92.4 + 4.955 = − 87.4 kJ mol−1
Both are negative, as they must be for an exothermic reaction. The gas count fell, so the surroundings did work on the system, and ΔU is the smaller magnitude of the two.
Entropy and Gibbs energy — the spontaneity signs
ΔG < 0 → spontaneous · ΔG = 0 → equilibrium · ΔG > 0 → non-spontaneous
ΔG° = − 2.303 R T log K
ΔS is positive when disorder increases — melting, boiling, dissolving, or any reaction that produces more gas molecules than it consumes. The unit trap here is brutal: ΔH is quoted in kJ mol−1 while ΔS is quoted in J K−1 mol−1. Convert before you subtract.
Worked example 3. A reaction has ΔH = + 30.0 kJ mol−1 and ΔS = + 100 J K−1 mol−1. Is it spontaneous at 298 K? Above what temperature does it become spontaneous?
TΔS = 298 × 100 = 29 800 J mol−1 = 29.8 kJ mol−1
ΔG = ΔH − TΔS = 30.0 − 29.8 = + 0.2 kJ mol−1 → positive, so not spontaneous at 298 K, but only just.
At the crossover, ΔG = 0, so T = ΔH / ΔS = 30 000 J ÷ 100 J K−1 = 300 K. Above 300 K the TΔS term wins and the reaction becomes spontaneous.
Mistakes that flip your sign
- Mixing conventions. Decide you are using ΔU = q + w with w = work done on the system, and never switch mid-answer.
- Forgetting the minus in w = − p ΔV. Expansion must give negative work. If your expanding gas produced positive work, you dropped the sign.
- kJ minus J. In ΔG = ΔH − TΔS, convert ΔS to kJ K−1 mol−1 (divide by 1000) or convert ΔH to J. Never subtract 29 800 from 30.
- Counting solids in Δng. Only gases count.
- Temperature in °C. Every T in this chapter is kelvin.
- Reading "spontaneous" as "fast". ΔG tells you whether a reaction can happen, never how quickly. Rate is kinetics, not thermodynamics — diamond turning to graphite is spontaneous and unimaginably slow.
- Assuming exothermic means spontaneous. It usually helps, but a large negative ΔS at high temperature can still make ΔG positive.
One-page sign summary
| Situation | Sign | Meaning |
|---|---|---|
| Heat absorbed by system | q positive | Endothermic |
| Heat released by system | q negative | Exothermic |
| Gas expands | w negative | System does work |
| Gas compressed | w positive | Work done on system |
| Rigid container | w = 0 | ΔU = q |
| Disorder increases | ΔS positive | Melting, boiling, more gas moles |
| Reaction can proceed | ΔG negative | Spontaneous at that temperature |
| K greater than 1 | ΔG° negative | Products favoured at equilibrium |
Practise by writing the sign before you calculate the number. If the number then disagrees with your prediction, you have found an arithmetic slip while there is still time to fix it.
Verify your ΔG working instantly. The free Gibbs Free Energy calculator takes ΔH, ΔS and T, handles the kJ/J conversion for you, and reports whether the reaction is spontaneous — ideal for checking a full exercise after you have solved it by hand.
Open the Gibbs Free Energy Calculator →Thermodynamics rewards steady, guided practice more than any other Class 11 chapter. ABC Chemistry teaches Class 11–12 chemistry at its Gurugram centre and online across India, with weekly numerical practice and correction — details at abcchemistry.in.