Internal energy
Every gas is made of tiny molecules that are always moving. Internal energy U is the total energy of all these molecules (energy of motion and, in real gases, energy of their pulls on each other). It does not include the motion of the whole container.
For an ideal gas, U depends only on temperature. Hotter gas → faster molecules → bigger U.
U is a state variable: it depends only on the present state, not on the past.
Heat and work: two ways to change U
Heat (Q) is energy that flows between the system and surroundings because of a temperature difference. A flame warms the gas.
Work (W) is energy moved by a force through a distance, without needing a temperature difference. A piston pushed in by your hand warms the gas.
Heat and work are not "stored" in a body. They are energy in transit. A body has internal energy; it does not "have" heat or work. That is why Q and W are not state variables: they depend on the path.
Joule showed that a fixed amount of work always makes the same warming as a fixed amount of heat. So heat is just a form of energy, measured in joules (1 cal = 4.186 J).
The first law of thermodynamics
The first law is the law of conservation of energy for heat and work:
ΔQ = ΔU + ΔW
- ΔQ = heat given to the system
- ΔU = rise in internal energy of the system
- ΔW = work done by the system on the surroundings
Sign rules (NCERT style)
- Heat given to the system: Q positive. Heat given out: Q negative.
- Work done by the gas (expansion): W positive. Work done on the gas (compression): W negative.
- Temperature rises: ΔU positive.
(Some chemistry books write ΔU = q + w with w = work done on the gas. It is the same law with a different sign for W.)
Special cases: in a cycle ΔU = 0 so Q = W. If no heat enters (insulated), ΔU = −W.
Work done by a gas: W = PΔV
Let gas at pressure P push a piston of area A through a small distance Δx. Force = PA, so work = PA × Δx. But A × Δx is the increase in volume ΔV. So
W = P ΔV (when P stays constant)
If P changes, W = the area under the P–V curve. Expansion (ΔV > 0) → gas does positive work. Compression → negative work.
Specific heats of a gas and Cp − Cv = R
Molar specific heat = heat needed to raise the temperature of 1 mol by 1 K. For a gas it depends on how we heat it.
- Cv (constant volume): no work is done, so all heat raises U: ΔQ = ΔU = nCvΔT.
- Cp (constant pressure): the gas also expands and does work, so more heat is needed: ΔQ = nCpΔT.
Derivation of Cp − Cv = R (Mayer's relation)
Heat 1 mol by ΔT at constant pressure. First law: CpΔT = ΔU + PΔV.
ΔU is the same as at constant volume (U depends only on T): ΔU = CvΔT.
From PV = RT at constant P: PΔV = RΔT.
So CpΔT = CvΔT + RΔT, which gives Cp − Cv = R.
Exam tip: this derivation (2–3 marks) and numericals on ΔQ = ΔU + ΔW are asked almost every year.
Try it at home
Rub your palms fast for 10 seconds. They feel warm. You did work, and it became internal energy. Now pump a bicycle tyre ten times and touch the pump's bottom: warm again. Work done on air raises its U, just like heat does.
Key formulas and definitions
- ΔQ = ΔU + ΔW (first law)
- W = PΔV (constant pressure)
- W = area under the P–V curve
- ΔU = nCvΔT (ideal gas, any process)
- ΔQ = nCpΔT (constant pressure)
- Cp − Cv = R
- 1 cal = 4.186 J
Worked examples
1. A gas absorbs 500 J of heat and does 200 J of work. Find the change in internal energy.
ΔU = ΔQ − ΔW = 500 − 200 = 300 J. Internal energy rises by 300 J.
2. 800 J of work is done on a gas and it gives out 300 J of heat. Find ΔU.
Work on the gas: ΔW = −800 J. Heat given out: ΔQ = −300 J. ΔU = ΔQ − ΔW = −300 − (−800) = +500 J.
3. A gas at a constant pressure of 2 × 10⁵ Pa expands from 1 L to 4 L. Find the work done by it.
ΔV = 3 L = 3 × 10⁻³ m³. W = PΔV = 2 × 10⁵ × 3 × 10⁻³ = 600 J.
4. In the example above the gas also received 1500 J of heat. By how much did its internal energy change?
ΔU = ΔQ − ΔW = 1500 − 600 = 900 J.
5. 2 mol of an ideal gas is heated from 300 K to 350 K at constant volume. Cv = 20.8 J mol⁻¹ K⁻¹. Find the heat supplied.
At constant volume W = 0, so ΔQ = ΔU = nCvΔT = 2 × 20.8 × 50 = 2080 J.
6. The same 2 mol is heated from 300 K to 350 K at constant pressure. Find the heat needed and the work done (R = 8.3).
Cp = Cv + R = 29.1 J mol⁻¹ K⁻¹. ΔQ = nCpΔT = 2 × 29.1 × 50 = 2910 J. ΔU = 2080 J (same as before). W = ΔQ − ΔU = 830 J. Check: nRΔT = 2 × 8.3 × 50 = 830 J. ✓
7. A gas goes through a full cycle and returns to its starting state. It absorbs 1200 J in total. How much net work does it do?
In a cycle ΔU = 0 (U is a state variable). So ΔW = ΔQ = 1200 J.
Common mistakes
- Mixing sign rules: in physics (NCERT) W is work done BY the gas. Work done on the gas is negative.
- Thinking a gas "contains heat". It contains internal energy; heat is energy on the move.
- Forgetting to change litres to m³ before using W = PΔV (1 L = 10⁻³ m³).
- Using ΔU = nCpΔT. For an ideal gas ΔU = nCvΔT in every process; Cp is only for heat at constant pressure.