The First Law of Thermodynamics in Chemistry
The first law is simply conservation of energy written for a chemical system. It looks harmless — three symbols — and yet it produces more lost marks than almost any other topic, because two different sign conventions are in circulation and students mix them. This article states both conventions plainly, explains what internal energy really is, and computes three standard problems in full.
System, surroundings and internal energy
The system is the part of the universe under study — the reacting mixture in a flask. Everything else is the surroundings. A system may be open (exchanges matter and energy), closed (energy only) or isolated (neither).
Internal energy U is the total energy stored inside the system: the kinetic energy of molecular motion plus all the potential energy locked in bonds and intermolecular forces. Its absolute value cannot be measured, and that does not matter — chemistry only ever needs the change, ΔU = Ufinal − Uinitial.
U is a state function: it depends only on the present state, not on the route taken. Heat q and work w are path functions — they depend entirely on how you got there. That contrast is the single most examinable idea in the chapter.
The law, in both conventions
Older convention (many engineering texts): ΔU = q − W, where W is the work done by the system and W = +pext ΔV
These are not two different laws. They are the same law with w = −W. In the IUPAC form, compression (ΔV negative) gives w positive because energy is being pushed into the system; expansion gives w negative because the system spends energy pushing the surroundings back. In the older form, expansion gives W positive because the system does work.
Rule for exams: read the question, spot which symbol it uses, and stay in that convention for the whole answer. Write the version you are using at the top of the solution. This article uses the IUPAC form throughout.
Sign of q is the same in both: q is positive when heat flows into the system (endothermic) and negative when heat leaves it (exothermic).
Worked example 1 — expansion against a constant pressure
A gas absorbs 500 J of heat and expands from 2.00 L to 5.00 L against a constant external pressure of 1.00 atm. Find ΔU.
ΔV = 5.00 − 2.00 = 3.00 L = 3.00 × 10⁻³ m³
pext = 1.00 atm = 101325 Pa
w = −pextΔV = −(101325)(3.00 × 10⁻³) = −303.975 J ≈ −304 J
ΔU = q + w = 500 + (−304) = +196 J
Cross-check in litre-atmospheres: w = −(1.00 atm)(3.00 L) = −3.00 L·atm, and 1 L·atm = 101.325 J, so w = −3.00 × 101.325 = −303.975 J. Same answer, so the unit conversion was done correctly.
Read the result physically: 500 J went in as heat, 304 J was spent pushing the atmosphere out of the way, and only 196 J was left to raise the internal energy.
Worked example 2 — isothermal reversible expansion of an ideal gas
Reversible work is the maximum work a system can deliver, and for an ideal gas at constant temperature it is:
1.00 mol of an ideal gas expands reversibly and isothermally at 298 K from 5.00 L to 10.00 L. Find w, ΔU and q.
V₂/V₁ = 10.00/5.00 = 2.00, and ln 2.00 = 0.6931
nRT = (1.00)(8.314)(298) = 2477.6 J
w = −2477.6 × 0.6931 = −1717 J ≈ −1.72 kJ
For an ideal gas U depends only on temperature. The process is isothermal, so ΔU = 0.
Therefore q = ΔU − w = 0 − (−1717) = +1717 J. Every joule the gas spent doing work was drawn in from the surroundings as heat. Note also that this reversible expansion delivers far more work than the irreversible one in example 1 — the path matters, which is exactly why w is a path function.
Two special cases you must know by name
- Constant volume (a bomb calorimeter). ΔV = 0, so w = 0 and ΔU = qV. The heat measured in a sealed bomb is the internal energy change.
- Constant pressure (an open beaker). Here the convenient function is enthalpy, H = U + pV, and ΔH = qp. The heat measured in an open vessel is the enthalpy change. This is why enthalpy exists at all — laboratory chemistry is normally done open to the atmosphere.
- Adiabatic. q = 0, so ΔU = w. No heat crosses the boundary; the temperature changes instead.
- Cyclic process. The system returns to its start, so ΔU = 0 and q = −w. A machine that produced net energy over a cycle would violate the first law, which is why a perpetual-motion machine of the first kind is impossible.
Worked example 3 — converting ΔU into ΔH
Suppose a bomb calorimeter gives ΔU = −1200.0 kJ mol⁻¹ at 298 K for a reaction in which Δng = −1. Find ΔH.
RT = (8.314 J K⁻¹ mol⁻¹)(298 K) = 2477.6 J mol⁻¹ = 2.478 kJ mol⁻¹
ΔngRT = (−1)(2.478) = −2.478 kJ mol⁻¹
ΔH = −1200.0 + (−2.478) = −1202.5 kJ mol⁻¹
(The −1200.0 figure is an illustrative number for the arithmetic, not a measured value for any particular compound.) The correction is small — about 0.2% here — but examiners ask for it, and it is the only place where R appears in this chapter. When Δng = 0, ΔH and ΔU are equal. Solids and liquids are not counted in Δng.
Quick reference
| Condition | What is zero | First law becomes | Measured heat equals |
|---|---|---|---|
| Constant volume (isochoric) | w = 0 | ΔU = qV | ΔU |
| Constant pressure (isobaric) | — | ΔU = qp − p ΔV | ΔH |
| Isothermal, ideal gas | ΔU = 0 | q = −w | −w |
| Adiabatic | q = 0 | ΔU = w | — |
| Cyclic | ΔU = 0 | q = −w | — |
| Free expansion into vacuum | pext = 0 so w = 0 | ΔU = q | q |
Common mistakes that cost marks
- Mixing the two conventions inside one answer — writing ΔU = q + w and then using w = +pΔV. Every sign after that is wrong.
- Using the internal pressure of the gas for expansion work. It is the external pressure that the system pushes against. In a free expansion into vacuum, pext = 0, so no work is done at all however much the gas expands.
- Forgetting to convert litres to cubic metres when pressure is in pascals. Either work in Pa and m³, or in atm and L and multiply by 101.325 at the end.
- Treating q or w as state functions. Only their sum ΔU is path-independent. "q depends on the path" is a marking-scheme phrase.
- Assuming ΔU = 0 for any constant-temperature process. That result holds for an ideal gas, where U depends on T alone.
- Counting solids and liquids in Δng in the ΔH–ΔU conversion. Only gases count.
- Saying "heat contained in the system". A system contains internal energy, not heat. Heat is energy in transit across the boundary.
Where the first law appears in exams
| Exam | Typical question |
|---|---|
| CBSE / ICSE Class 11 | State the first law, define state and path functions, compute ΔU from q and w |
| NEET / JEE | Work in isothermal, adiabatic and irreversible expansions; ΔH–ΔU conversion |
| IIT-JAM / CUET-PG | Reversible vs irreversible work, calorimetry, Kirchhoff's relation for ΔH with temperature |
| GATE / CSIR-NET | Joule–Thomson expansion, partial derivatives of U and H, Maxwell relations built on this base |
Get the volumes right before you compute the work. The suite has no dedicated first-law tool, but almost every pΔV problem starts by finding a volume or a temperature from PV = nRT — and that is exactly what the Ideal Gas Law calculator does. Put in any three of p, V, n and T and it returns the fourth, ready for the work step above.
Open the Ideal Gas Law (PV = nRT) Calculator →Thermodynamics is the chapter where careful sign work decides the grade. ABC Chemistry teaches Class 11–12 chemistry at the Gurugram centre and online across India — details at abcchemistry.in.