Thermodynamics — Formula Sheet

$W$: Work done by the gas (J) · $P$: Pressure of the gas (Pa) · $V_i,V_f$: Initial and final volume (m³)

Area under the P–V curve. Positive for expansion, negative for compression. Work depends on the path, not only on the initial and final states. Unit: J.

$Q$: Heat supplied to the system (J); negative if heat is given out · $\Delta U$: Change in internal energy (J) · $W$: Work done by the system (J); negative if work is done on it

Energy conservation for a thermodynamic system. Heat supplied (Q) partly raises the internal energy (ΔU) and partly does external work (W). Equivalent form: ΔU = Q − W.

$n$: Number of moles · $C_V$: Molar specific heat at constant volume (J mol⁻¹ K⁻¹) · $\Delta T$: Change in temperature (K)

Internal energy of an ideal gas depends only on temperature. This formula holds for ANY process, not only at constant volume.

$T$: Constant absolute temperature (K) · $V_i,V_f$: Initial and final volume (m³ or L, same unit for both)

Temperature is constant, so ΔU = 0 and Q = W. Because P_iV_i = P_fV_f, also W = 2.303 nRT log₁₀(P_i/P_f). Isothermal changes are slow and take place in a container with conducting walls.

$\Delta V$: Change in volume (m³) · $C_P$: Molar specific heat at constant pressure (J mol⁻¹ K⁻¹)

Pressure stays constant. All three terms of the first law are non-zero: Q = nC_PΔT, ΔU = nC_VΔT and W = nRΔT.

Volume is constant, so the gas does no work. All the heat supplied goes into internal energy, which raises the temperature and the pressure.

$\gamma$: Ratio of specific heats C_P/C_V

No heat exchange (Q = 0), so ΔU = −W. Expansion cools the gas; compression heats it. Happens in insulated containers or in very rapid processes. γ = C_P/C_V.

$P_i,V_i,T_i$: Initial pressure, volume and temperature · $P_f,V_f,T_f$: Final pressure, volume and temperature

Positive when the gas expands and cools (T_f < T_i). Negative for adiabatic compression, where work is done on the gas.

The system returns to its initial state, so the internal energy is unchanged. The net heat absorbed equals the net work done in one cycle.

A gas expands suddenly into a vacuum in an insulated container. No work is done because there is no opposing pressure. For an ideal gas the temperature does not change. It is not a quasi-static process.

$\eta$: Efficiency (fraction or %) · $Q_H$: Heat absorbed from the hot reservoir (J) · $|Q_C|$: Magnitude of the heat rejected to the cold reservoir (J) · $W$: Net work done per cycle (J)

The engine absorbs Q_H from the hot reservoir, does work W, and rejects heat |Q_C| to the cold reservoir, with W = Q_H − |Q_C|. In the textbook sign convention Q_H is positive and Q_C (heat rejected) is negative, so use its magnitude. The second law (Kelvin–Planck) says Q_C can never be zero, so η < 1.

$K$: Coefficient of performance (no unit) · $Q_C$: Heat extracted from the cold reservoir (J) · $|W|$: Magnitude of the work done on the refrigerant (J)

Heat extracted from the cold region per unit work done on the refrigerant. For a refrigerator Q_C > 0 while Q_H and W are negative, so their magnitudes are used. K can be greater than 1. For a heat pump the useful output is the heat delivered to the hot side, so its coefficient of performance is |Q_H|/|W|.

$T_H$: Temperature of the hot reservoir (K) · $T_C$: Temperature of the cold reservoir (K)

Holds only for a reversible (Carnot) cycle: isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression.

$T_H$: Source temperature (K) · $T_C$: Sink temperature (K)

The maximum possible efficiency of any engine working between T_H and T_C. It depends only on the reservoir temperatures (in kelvin), not on the working substance. It is 100% only if T_C = 0 K, which cannot be reached.

The smaller the temperature difference between inside and outside, the higher the coefficient of performance. Temperatures in kelvin.