System, surroundings and boundary
Thermodynamics is the study of energy changes, mainly heat and work. We pick the part we want to study and call it the system. Everything else is the surroundings. The real or imaginary wall that separates them is the boundary. System + surroundings = the universe.
Three types of system
- Open system: both matter and energy can cross the boundary. Example: hot tea in an open cup (steam leaves, heat leaves).
- Closed system: only energy can cross, not matter. Example: soup in a sealed steel box. It can cool down, but nothing escapes.
- Isolated system: neither matter nor energy can cross. Example: tea in an ideal vacuum flask.
Special kinds of process
- Isothermal: temperature stays constant (ΔT = 0).
- Adiabatic: no heat crosses the boundary (q = 0).
- Isobaric: pressure constant. Isochoric: volume constant.
- Reversible: done in tiny, very slow steps, so the system stays almost in balance all the time. Irreversible: done in one quick step (every real process).
State of a system and state functions
We describe a system by a few measurable values: pressure p, volume V, temperature T and amount n. These are state variables. Once they are fixed, the system is in a definite state.
A state function depends only on the start state and the end state, not on how you got there. Examples: U, H, S, G, p, V, T. A path function depends on the route. Heat q and work w are path functions.
Hill example: your height gain from the foot to the top of a hill is fixed (state function). The distance you walk depends on the road you choose (path function).
That is why we write ΔU (a change between two states) but just q and w (amounts along one path), never Δq or Δw.
Internal energy, heat and work
Internal energy (U) is the total energy stored inside the system: motion of particles, energy in chemical bonds, attractions between particles and more. We cannot measure U itself, only the change ΔU = U₂ − U₁.
Two ways to change U
- Heat (q): energy that flows because of a temperature difference. Heat in: q is positive. Heat out: q is negative.
- Work (w): energy moved by a force through a distance, for example a piston. Work done ON the system: w positive. Work done BY the system: w negative.
Pressure–volume work
When a gas expands against a constant outside pressure p_ext: w = −p_ext (V₂ − V₁) = −p_ext ΔV. Expansion (ΔV > 0) gives negative w. Compression gives positive w.
- Free expansion (into vacuum, p_ext = 0): w = 0.
- Reversible isothermal expansion of an ideal gas: w_rev = −2.303 nRT log(V₂/V₁) = −2.303 nRT log(p₁/p₂). This is the largest work a gas can do between the same two states.
- Units: 1 L bar = 100 J; 1 L atm = 101.3 J; R = 8.314 J K⁻¹ mol⁻¹.
First law of thermodynamics: ΔU = q + w
The first law is the law of conservation of energy: energy of an isolated system stays constant; it can change form but cannot be made or destroyed. For any closed system: ΔU = q + w.
Special cases
- Isolated system: q = 0, w = 0 → ΔU = 0.
- Adiabatic change: q = 0 → ΔU = w_ad. Adiabatic compression heats a gas.
- Constant volume (no p–V work): ΔU = q_V. This is how a bomb calorimeter measures ΔU.
- Isothermal change of an ideal gas: ΔU = 0 → q = −w. All heat taken in is given out as work.
- Free isothermal expansion of an ideal gas: q = 0, w = 0, ΔU = 0.
In the exam
Expect 1-mark questions on definitions and sign conventions, 2–3 mark numericals on ΔU = q + w and w = −p_ext ΔV, and a derivation of reversible work. The unit Thermodynamics carries about 9 marks in CBSE Class 11.
Try it at home: the bicycle pump test
Close the nozzle of a bicycle pump with your thumb and push the handle down quickly 10 times. Touch the bottom of the pump. It feels warm. You did work on the air (w > 0) so fast that heat had no time to escape (almost adiabatic, q ≈ 0). So ΔU = w > 0 and the air got hotter. Now in the 3D, set q = 0 and slide w to +60 J: the particles speed up in the same way.
Key formulas and definitions
- ΔU = q + w (first law, IUPAC signs)
- q > 0 heat absorbed; q < 0 heat released
- w > 0 work done on system; w < 0 work done by system
- w = −p_ext ΔV (expansion against constant pressure)
- w_rev = −2.303 nRT log(V₂/V₁) = −2.303 nRT log(p₁/p₂) (isothermal, ideal gas)
- Isothermal ideal gas: ΔU = 0, q = −w; Adiabatic: q = 0, ΔU = w
- 1 L bar = 100 J; 1 L atm = 101.3 J; R = 8.314 J K⁻¹ mol⁻¹
Worked examples
1. A gas absorbs 200 J of heat and 150 J of work is done on it. Find ΔU.
q = +200 J (heat in). w = +150 J (work on the gas). ΔU = q + w = 200 + 150 = +350 J. The internal energy rises by 350 J.
2. A system gives out 60 J of heat and does 40 J of work on the surroundings. Find ΔU.
Heat out: q = −60 J. Work done by the system: w = −40 J. ΔU = −60 + (−40) = −100 J. The system loses 100 J.
3. A gas expands from 2 L to 7 L against a constant outside pressure of 1.5 bar. Find the work in joules.
ΔV = 7 − 2 = 5 L. w = −p_ext ΔV = −1.5 × 5 = −7.5 L bar. 1 L bar = 100 J, so w = −750 J. Negative because the gas does work.
4. In the same expansion (Example 3), the gas absorbs 1000 J of heat. Find ΔU.
q = +1000 J, w = −750 J. ΔU = q + w = 1000 − 750 = +250 J.
5. 2 mol of an ideal gas expand isothermally and reversibly at 300 K from 10 L to 100 L. Find w, q and ΔU.
w = −2.303 nRT log(V₂/V₁) = −2.303 × 2 × 8.314 × 300 × log(10). log 10 = 1, so w = −11 488 J ≈ −11.49 kJ. For an ideal gas at constant T, ΔU = 0. So q = −w = +11.49 kJ.
6. 1 mol of an ideal gas expands isothermally at 300 K from 1 L to 5 L (a) into vacuum, (b) against a constant 2 bar, (c) reversibly. Compare the work.
(a) Free expansion: p_ext = 0 → w = 0. (b) w = −2 × (5 − 1) = −8 L bar = −800 J. (c) w = −2.303 × 1 × 8.314 × 300 × log 5 = −5744 × 0.699 ≈ −4015 J. The reversible path gives the most work (in size): 4015 J > 800 J > 0. In all three ΔU = 0 because T is constant, so q = −w each time.
7. In an adiabatic process 500 J of work is done on a gas. Find q and ΔU. Will the gas get hotter or cooler?
Adiabatic means q = 0. w = +500 J. ΔU = 0 + 500 = +500 J. U rises, so the gas gets hotter (like the bicycle pump).
Common mistakes
- Using the old physics sign (ΔU = q − w). In CBSE chemistry, use IUPAC: ΔU = q + w, with work done BY the gas negative.
- Forgetting to convert L bar to joules: multiply by 100 (1 L bar = 100 J).
- Writing Δq or Δw. Heat and work are not state functions, so there is no 'change in q'.
- Thinking ΔU = 0 means no heat flows. In an isothermal ideal-gas expansion ΔU = 0, but q = −w is not zero.