What is a thermodynamic process?
A process is any change that takes a system from one equilibrium state (P₁, V₁, T₁) to another (P₂, V₂, T₂). On a P–V graph each state is a point and the process is a path joining them.
A quasi-static process is one that happens so slowly that the gas is almost in equilibrium at every moment. Only then do P and T have one clear value, so we can draw the path.
Work done by the gas = area under the path on the P–V graph.
Isothermal process
Isothermal means the temperature stays constant. The gas is kept in contact with a large heat reservoir (like a big water bath) and changed slowly.
- Since T is constant: PV = constant (Boyle's law). The graph is a curve called an isotherm.
- U of an ideal gas depends only on T, so ΔU = 0. By the first law, Q = W: all heat taken in becomes work.
- Work done: W = nRT ln(V₂/V₁) = 2.303 nRT log₁₀(V₂/V₁).
Derivation of isothermal work
For a tiny step, dW = P dV and P = nRT/V. So W = ∫ nRT dV/V from V₁ to V₂ = nRT [ln V] = nRT ln(V₂/V₁). Expansion (V₂ > V₁) gives positive W.
Adiabatic process
Adiabatic means no heat enters or leaves: Q = 0. The gas is insulated, or the change is so fast that heat has no time to flow.
- First law: 0 = ΔU + W, so W = −ΔU. Expanding gas does work from its own internal energy, so it cools. Compressed gas heats up.
- Rule: PV^γ = constant, where γ = Cp/Cv (1.67 for monatomic, 1.4 for diatomic gases like air). Also TV^(γ−1) = constant.
- Work done: W = (P₁V₁ − P₂V₂)/(γ − 1) = nR(T₁ − T₂)/(γ − 1).
Because γ > 1, an adiabatic curve is steeper than an isotherm through the same point (its slope is γ times bigger).
Isobaric and isochoric processes
Isobaric (constant pressure): the path is a horizontal line. W = PΔV = nRΔT. Heat Q = nCpΔT; part raises U, part does work. Example: water boiling in an open pan.
Isochoric (constant volume): the path is a vertical line. No volume change, so W = 0 and Q = ΔU = nCvΔT. All heat raises the temperature. Example: heating gas in a closed steel cylinder.
| Process | Fixed | Rule | Work W |
|---|---|---|---|
| Isothermal | T | PV = const | nRT ln(V₂/V₁) |
| Adiabatic | Q = 0 | PV^γ = const | nR(T₁ − T₂)/(γ − 1) |
| Isobaric | P | V/T = const | PΔV |
| Isochoric | V | P/T = const | 0 |
Reversible and irreversible processes
A reversible process can be run backwards so that both the system and the surroundings return exactly to their first states, with no change left anywhere. For this the process must be quasi-static (very slow) and free of friction, leaks or other losses.
An irreversible process cannot be undone completely. Real processes are irreversible because:
- they happen fast, so the gas is not in equilibrium in between (sudden expansion, explosion);
- there are losses such as friction, viscosity and heat flowing across a finite temperature difference.
Examples of irreversible processes: a gas rushing into a vacuum (free expansion), heat flowing from a hot cup to a cool room, a ball bouncing to a stop. Reversible processes are an ideal limit that real ones can only get close to.
Cyclic process
In a cyclic process the system goes through several changes and returns to its starting state. On a P–V graph the path is a closed loop.
- U is a state variable, so after a full cycle ΔU = 0.
- First law: net heat in = net work done, Q = W.
- Net work = area enclosed by the loop. Clockwise loop → net work done by the gas (engine). Anticlockwise → net work done on the gas (refrigerator).
Exam tip: expect a 3–5 mark question on isothermal vs adiabatic (comparison table, derivation of work, graph slopes) and numericals using the loop area.
Try it
Block the nozzle of a bicycle pump with your thumb and push the handle in fast ten times. Feel the pump bottom: warm (adiabatic compression). Now press a deodorant can for two seconds and feel the can: cold (fast expansion). In the 3D, choose "adiabatic" and "isothermal" from the same start point and see which curve falls faster.
Key formulas and definitions
- Isothermal: PV = const, ΔU = 0, Q = W = nRT ln(V₂/V₁)
- Adiabatic: Q = 0, PV^γ = const, TV^(γ−1) = const
- Adiabatic work: W = (P₁V₁ − P₂V₂)/(γ − 1) = nR(T₁ − T₂)/(γ − 1)
- Isobaric: W = PΔV = nRΔT, Q = nCpΔT
- Isochoric: W = 0, Q = ΔU = nCvΔT
- Cyclic: ΔU = 0, Q = W = area of loop
- γ = Cp/Cv; slope(adiabatic) = γ × slope(isothermal)
Worked examples
1. A gas at 1 × 10⁵ Pa and 6 L is compressed isothermally to 2 L. Find the final pressure.
Isothermal: P₁V₁ = P₂V₂. P₂ = 1 × 10⁵ × 6/2 = 3 × 10⁵ Pa.
2. 1 mol of ideal gas at 300 K expands isothermally from 10 L to 20 L. Find the work done (R = 8.31, ln 2 = 0.693).
W = nRT ln(V₂/V₁) = 1 × 8.31 × 300 × 0.693 = 1728 J ≈ 1.73 kJ. Since ΔU = 0, the gas also absorbs 1728 J of heat.
3. At constant pressure 2 × 10⁵ Pa, 0.5 mol of gas is heated from 300 K to 400 K. Find the work done (R = 8.3).
W = nRΔT = 0.5 × 8.3 × 100 = 415 J.
4. Air (γ = 1.4) at 1 × 10⁵ Pa is compressed adiabatically to 1/8 of its volume. Find the final pressure. (8^1.4 ≈ 18.4)
P₁V₁^γ = P₂V₂^γ ⇒ P₂ = P₁ (V₁/V₂)^γ = 1 × 10⁵ × 8^1.4 ≈ 1.84 × 10⁶ Pa, about 18 times. (Isothermally it would only be 8 times.)
5. 2 mol of a monatomic gas (γ = 5/3) expands adiabatically and cools from 400 K to 300 K. Find the work done (R = 8.3).
W = nR(T₁ − T₂)/(γ − 1) = 2 × 8.3 × 100/(2/3) = 1660 × 1.5 = 2490 J. The gas does 2490 J of work and its U falls by 2490 J.
6. Air at 27 °C (300 K) is compressed adiabatically to 1/4 of its volume (γ = 1.4, 4^0.4 ≈ 1.74). Find the final temperature.
TV^(γ−1) = const ⇒ T₂ = T₁ (V₁/V₂)^(γ−1) = 300 × 4^0.4 ≈ 300 × 1.74 = 522 K (about 249 °C). This is why diesel engines need no spark.
7. A gas goes round a rectangle on the P–V graph: P between 1 × 10⁵ and 3 × 10⁵ Pa, V between 1 L and 4 L, clockwise. Find the net work and net heat in one cycle.
Area = ΔP × ΔV = 2 × 10⁵ × 3 × 10⁻³ = 600 J. Clockwise, so the gas does 600 J net work. ΔU = 0, so net heat absorbed = 600 J.
Common mistakes
- Thinking "isothermal" means no heat flows. Heat does flow; it is adiabatic that has no heat flow.
- Thinking "adiabatic" means the temperature stays the same. In an adiabatic process T changes; only Q is zero.
- Using ln with base 10 values: W = 2.303 nRT log₁₀(V₂/V₁) when you use log tables.
- Forgetting the sign of the loop: clockwise on P–V means net work by the gas; anticlockwise means net work on the gas.