A heat engine uses heat transfer to do work (car engines, steam turbines). Some heat always goes out to the environment (Q_out is never 0).

\displaystyle W = P\Delta V\ (\text{constant pressure})

\displaystyle W = \text{area under the } PV \text{ curve}

\displaystyle W_{cycle} = \text{area inside the loop}

On a PV diagram, the work done is the area under the curve. Expanding (ΔV > 0) gives positive W; compressing gives negative W.
Work depends on the path: the higher-pressure path between the same two points does more work.
For a cycle (returns to start), net work = area inside the loop. Clockwise loop: net work out (engine). Counterclockwise: net work in (fridge/heat pump).

Process	What stays constant	Key fact	First law becomes
Isobaric	Pressure	W = PΔV	ΔE_int = Q − PΔV
Isochoric	Volume	W = 0 (vertical line on PV)	ΔE_int = Q
Isothermal	Temperature	ΔE_int = 0 (ideal gas)	Q = W
Adiabatic	No heat flow	Q = 0	ΔE_int = −W

Monatomic ideal gas: E_int = (3/2)NkT = (3/2)nRT, so internal energy only changes if temperature changes.
Adiabatic expansion cools the gas (work comes out of internal energy). That's why gas from a spray can or CO₂ cartridge feels cold. Adiabatic happens with good insulation or a very fast process.
Isothermal processes are slow (heat must keep flowing in). From the same start, an isothermal expansion does more work than an adiabatic one.
Reversible process — both the system and its surroundings can return exactly to where they started. Real processes are never fully reversible (friction, turbulence).

\displaystyle E_{int} = \tfrac{3}{2}NkT = \tfrac{3}{2}nRT

\displaystyle \Delta E_{int} = \tfrac{3}{2}nR\,\Delta T\ (\text{monatomic})

Ex. 15.2: a rectangle cycle ABCDA with P_AB = 1.50 × 10⁶ Pa, P_CD = 2.00 × 10⁵ Pa, ΔV = 5.00 × 10⁻⁴ m³: W_AB = 750 J, W_CD = −100 J, the two vertical legs do 0, so W = (1.50 × 10⁶ − 2.00 × 10⁵)(5.00 × 10⁻⁴) = 650 J.