Thermodynamics: Temperature, Heat, Ideal Gas Laws, First Law, Entropy, Thermodynamic Processes
Thermodynamics questions ask you to follow energy through a process: what is supplied as heat, what leaves as work, and how entropy constrains the direction of the process.
Temperature and heat
T/K = θ/°C + 273.15. Q = c·m·ΔT; phase changes require latent heat. Heat is energy in transit and is not a state quantity - it makes no sense to say a body contains heat.
Ideal gas
p·V = n·R·T = N·k·T with R = 8.314 J/(mol·K) and k = 1.38×10⁻²³ J/K. Boyle (T const), Gay-Lussac (p const, V/T const) and Amontons (V const, p/T const) are special cases. Kinetic theory gives the mean translational energy of a monatomic gas as (3/2)kT.
First law
ΔU = Q + W with W the work done on the gas (the usual German textbook convention). For a quasistatic process W = −p·ΔV. Isochoric: W = 0. Isothermal (ideal gas): ΔU = 0. Isobaric: Q = n·C_p·ΔT. Adiabatic: Q = 0. Some textbooks write ΔU = Q − W, so state your convention.
Second law and entropy
ΔS = Q_rev/T for a reversible step; ΔS_universe ≥ 0, with equality only for reversible processes. Entropy counts accessible microstates (S = k·ln W). The Carnot efficiency η = 1 − T_k/T_h is the upper bound for any heat engine.
Worked example
2.0 L of ideal gas at 300 K and 1.0×10⁵ Pa is heated at constant pressure to 450 K. Isobaric: V₂ = V₁·T₂/T₁ = 3.0 L. n = pV/(RT) = (1.0×10⁵ · 2.0×10⁻³)/(8.314 · 300) = 0.080 mol. Work done by the gas: p·ΔV = 1.0×10⁵ · 1.0×10⁻³ = 100 J. Q = n·(5/2)R·ΔT = 0.080 · 20.8 · 150 = 250 J. Then ΔU = Q − p·ΔV = 150 J, matching n·(3/2)R·ΔT = 150 J.
Traps
- Entering Celsius temperatures in p·V = n·R·T.
- A sign error caused by switching between the two first-law conventions.
- Assuming ΔU = 0 for every process that looks isothermal.
- Using Celsius values in η = 1 − T_k/T_h.
- Claiming the entropy of the system always increases; only the entropy of the universe does.